5. Trigonometrical Ratios Solutions#

Go through the list of relationship between trigonometrical ratios as the problems are centered around those.

  1. L.H.S. =1−cos⁡A1+cos⁡A= \sqrt{\frac{1- \cos A}{1 + \cos A}}

    Multiplying and dividing with 1−cos⁡A,1 - \cos A, we get

    =(1−cos⁡A)21−cos⁡2A= \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}

    =(1−cos⁡Asin⁡A)2= \sqrt{\left(\frac{1 - \cos A}{\sin A}\right)^2}

    =1−cos⁡Asin⁡A=cosec⁡A−cot⁡A== \frac{1 - \cos A}{\sin A} = \cosec A - \cot A = R.H.S.

  2. L.H.S. 1+tan⁡2A+1+cot⁡2A\sqrt{1 + \tan^2A + 1 + \cot^2A} [∵sec⁡2A=1+tan⁡2A\because \sec^2A = 1 + \tan^2A and cosec⁡2A=1+cot⁡2A\cosec^2A = 1 + \cot^2 A ]

    =tan⁡2A+2tan⁡Acot⁡A+cot⁡2A= \sqrt{\tan^2A + 2\tan A\cot A + \cot^2A} [tan⁡Acot⁡A=1\tan A\cot A = 1 ]

    =(tan⁡A+cot⁡A)2== \sqrt{(\tan A + \cot A)^2} = R.H.S.

  3. L.H.S =(1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)(sin⁡2A+cos⁡2Asin⁡Acos⁡A)= \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right)\left(\frac{\sin^2A + \cos^2A}{\sin A \cos A}\right)

    =cos⁡2Asin⁡Asin⁡2Acos⁡A1sin⁡Acos⁡A=1= \frac{\cos^2A}{\sin A}\frac{\sin^2 A}{\cos A}\frac{1}{\sin A \cos A} = 1

  4. L.H.S. =cos⁡4A−sin⁡4A+1=(cos⁡2A+sin⁡2A)(cos⁡2A−sin⁡2A)+1= \cos^4A - \sin^4A + 1= (\cos^2A + \sin^2A)(\cos^2A - \sin^2A) + 1

    =cos⁡2A−sin⁡2A+1=2cos⁡2A= \cos^2A - \sin^2A + 1 = 2\cos^2A

  5. L.H.S. =(sin⁡A+cos⁡A)(1−sin⁡Acos⁡A)=(sin⁡A+cos⁡A)(sin⁡2A+cos⁡2A−sin⁡Acos⁡A)= (\sin A + \cos A)(1 - \sin A\cos A) = (\sin A + \cos A)(\sin^2A + \cos^2A - \sin A\cos A)

    =sin⁡3A+cos⁡3A== \sin^3A + \cos^3A = R.H.S.

  6. L.H.S =sin⁡A1+cos⁡A+1+cos⁡Asin⁡A=sin⁡2A+(1+cos⁡A)2sin⁡A(1+cos⁡A)= \frac{\sin A}{1 + \cos A}+\frac{1 + \cos A}{\sin A} = \frac{\sin^2A + (1 + \cos A)^2}{\sin A(1 + \cos A)}

    =sin⁡2A+1+2cos⁡A+cos⁡2Asin⁡A(1+cos⁡A)=1+1+2cos⁡A(sin⁡A)(1+cos⁡A))= \frac{\sin^2A + 1 + 2\cos A + \cos^2A}{\sin A(1 + \cos A)} = \frac{1 + 1 + 2\cos A}{(\sin A)(1 + \cos A))}

    =2sin⁡A=2cosec⁡A== \frac{2}{\sin A} = 2\cosec A = R.H.S.

  7. L.H.S. =sin⁡6A−cos6A=(sin⁡2A+cos⁡2A)3−3cos⁡4Asin⁡2A−3cos⁡2Asin⁡4A= \sin^6A - cos^6A = (\sin^2A + \cos^2A)^3 - 3\cos^4A\sin^2A - 3\cos^2A\sin^4A

    =1−3sin⁡2Acos⁡2A(cos⁡2A+sin⁡2A)=1−3sin⁡2Acos⁡2A== 1 - 3\sin^2A\cos^2A(\cos^2A + \sin^2A) = 1 - 3\sin^2A\cos^2A = R.H.S.

  8. L.H.S. =1−sin⁡A1+sin⁡A= \sqrt{\frac{1 - \sin A}{1 + \sin A}}

    Multiplying and dividing with 1−sin⁡A,1 - \sin A, we get

    =(1−sin⁡A)21−sin⁡2A=(1−sin⁡A)2cos⁡2A=(1−sin⁡Acos⁡A)2= \sqrt{\frac{(1 - \sin A)^2}{1 - \sin^2A}} = \sqrt{\frac{(1 - \sin A)^2}{\cos^2A}} = \sqrt{\left(\frac{1 - \sin A}{\cos A}\right)^2}

    =1−sin⁡Acos⁡A=sec⁡A−tan⁡A= \frac{1 - \sin A}{\cos A} = \sec A - \tan A

  9. L.H.S. =cosec⁡Acosec⁡A−1+cosec⁡Acosec⁡A+1=1sin⁡A1sin⁡A−1+1sin⁡A1sin⁡A+1= \frac{\cosec A}{\cosec A - 1} + \frac{\cosec A}{\cosec A + 1} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} - 1} + \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} + 1}

    =11−sin⁡A+11+sin⁡A=1(1−sin⁡A)(1+sin⁡A)=1cos⁡2A=sec⁡2A== \frac{1}{1 - \sin A} + \frac{1}{1 + \sin A} = \frac{1}{(1 - \sin A)(1 + \sin A)} = \frac{1}{\cos^2A} = \sec^2A = R.H.S.

  10. L.H.S. =cosec⁡Atan⁡A+cot⁡A=1sin⁡Asin⁡Acos⁡A+cos⁡Asin⁡A= \frac{\cosec A}{\tan A + \cot A} = \frac{\frac{1}{\sin A}}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}

    =1sin⁡Asin⁡2A+cos⁡2Asin⁡Acos⁡A=1sin⁡A1sin⁡Acos⁡A=cos⁡A== \frac{\frac{1}{\sin A}}{\frac{\sin^2A + \cos^2A}{\sin A\cos A}} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A\cos A}} = \cos A = R.H.S.

  11. L.H.S. =(sec⁡A+cos⁡A)(sec⁡A−cos⁡A)=sec⁡2A−cos⁡2A=1+tan⁡2A−1+sin⁡2A=tan⁡2A+sin⁡2A== (\sec A + \cos A)(\sec A - \cos A) = \sec^2A - \cos^2A = 1 + \tan^2A - 1 + \sin^2A = \tan^2A + \sin^2A = R.H.S.

  12. L.H.S. =1tan⁡A+cot⁡A=1sin⁡Acos⁡A+cos⁡Asin⁡A=1sin⁡2A+cos⁡2Asin⁡Acos⁡A=sin⁡Acos⁡A== \frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2A + \cos^2A}{\sin A\cos A}} = \sin A\cos A = R.H.S.

  13. L.H.S =1−tan⁡A1+tan⁡A=1−1cot⁡A1+1cot⁡A=cot⁡A−1cot⁡A+1== \frac{1 - \tan A}{1 + \tan A} = \frac{1 - \frac{1}{\cot A}}{1 + \frac{1}{\cot A}} = \frac{\cot A - 1}{\cot A + 1} = R.H.S.

  14. L.H.S. =1+tan⁡2A1+cot⁡2A=1+sin⁡2Acos⁡2A1+cos⁡2Asin⁡A= \frac{1 + \tan^2A}{1 + \cot^2A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^A}}

    =cos⁡2A+sin⁡2Acos⁡2Asin⁡2A+cos⁡2Asin⁡2A=sin⁡2Acos⁡2A== \frac{\frac{\cos^2A + \sin^2A}{\cos^2A}}{\frac{\sin^2A + \cos^2A}{\sin^2A}} = \frac{\sin^2A}{\cos^2A} = R.H.S.

  15. L.H.S. =sec⁡A−tan⁡Asec⁡A+tan⁡A= \frac{\sec A - \tan A}{\sec A + \tan A}

    Multiplying and dividing with sec⁡A−tan⁡A,\sec A - \tan A, we get

    (sec⁡A−tan⁡A)2sec⁡2A−tan⁡2A=sec⁡2A−2tan⁡Asec⁡A+tan⁡A=1+tan⁡2A−2tan⁡Asec⁡A+tan⁡2A\frac{(\sec A - \tan A)^2}{\sec^2A - \tan^2A} = \sec^2A - 2\tan A\sec A + \tan^A = 1 + \tan^2A - 2\tan A\sec A + \tan^2A

    =1−2tan⁡Asec⁡A+2tan⁡2A== 1 - 2\tan A\sec A + 2\tan^2A = R.H.S.

  16. L.H.S. =1sec⁡A−tan⁡A= \frac{1}{\sec A - \tan A}

    Multiplying and dividing with sec⁡A+tan⁡A,\sec A + \tan A, we get

    =sec⁡A+tan⁡Asec⁡2A=tan⁡2A=sec⁡A+tan⁡A== \frac{\sec A + \tan A}{\sec^2A = \tan^2A} = \sec A + \tan A = R.H.S.

  17. L.H.S. =tan⁡A1−cot⁡A+cot⁡A1−tan⁡A= \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}

    =sin⁡Acos⁡A1−cos⁡Asin⁡A+cos⁡Asin⁡A1−sin⁡Acos⁡A= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}

    =sin⁡2Acos⁡A(sin⁡A−cos⁡A)+cos⁡2Asin⁡A(cos⁡A−sin⁡A)= \frac{\sin^2A}{\cos A(\sin A - \cos A)} + \frac{\cos^2A}{\sin A(\cos A - \sin A)}

    =sin⁡3A−cos⁡3Asin⁡Acos⁡A(sin⁡A−cos⁡A)=sin⁡2A+cos⁡2A+sin⁡Acos⁡Asin⁡Acos⁡A= \frac{\sin^3A - \cos^3A}{\sin A\cos A(\sin A - \cos A)} = \frac{\sin^2A + \cos^2A + \sin A\cos A}{\sin A\cos A}

    =1+sin⁡Acos⁡Asin⁡Acos⁡A=1+cosec⁡Asec⁡A== \frac{1 + \sin A \cos A}{\sin A\cos A} = 1 + \cosec A\sec A = R.H.S.

  18. L.H.S. =cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡A1−sin⁡Acos⁡A+sin⁡A1−cos⁡Asin⁡A= \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}

    =cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡A=cos⁡2A−sin⁡2Acos⁡A−sin⁡A= \frac{\cos^2A}{\cos A - \sin A} + \frac{\sin^2A}{\sin A - \cos A} = \frac{\cos^2A - \sin^2A}{\cos A - \sin A}

    =cos⁡A+sin⁡A== \cos A + \sin A = R.H.S.

  19. L.H.S. =(sin⁡A+cos⁡A)(tan⁡A+cot⁡A)=(sin⁡A+cos⁡A)(sin⁡Acos⁡A+cos⁡Asin⁡A)= (\sin A + \cos A)(\tan A + \cot A) = (\sin A + \cos A)(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})

    =(sin⁡A+cos⁡A)sin⁡2A+cos⁡2Asin⁡Acos⁡A=sin⁡A+cos⁡Asin⁡Acos⁡A=sec⁡A+cosec⁡A== (\sin A + \cos A)\frac{\sin^2A + \cos^2A}{\sin A\cos A} = \frac{\sin A + \cos A}{\sin A\cos A} = \sec A + \cosec A = R.H.S.

  20. L.H.S. =sec⁡4A−sec⁡2A=(1+tan⁡2A)−1−tan⁡2A=1+2tan⁡2A+tan⁡4A−1−tan⁡2A=tan⁡4A+tan⁡2A== \sec^4A - \sec^2A = (1 + \tan^2A) - 1 - \tan^2A = 1 + 2\tan^2A + \tan^4A - 1 - \tan^2A = \tan^4A + \tan^2A = R.H.S.

  21. L.H.S. =cot⁡4A+cot⁡2A=(cosec⁡2A−1)1+cosec⁡2A−1=cosec⁡4A−2cosec⁡2A+1+cosec⁡2A−1=cosec⁡4A−cosec⁡2A== \cot^4A + \cot^2A = (\cosec^2A - 1)^1 + \cosec^2A - 1 = \cosec^4A - 2\cosec^2A + 1 + \cosec^2A -1 = \cosec^4A - \cosec^2A = R.H.S.

  22. L.H.S. =cosec⁡2A−1=cot⁡2A=cot⁡A=cos⁡Asin⁡A=cos⁡Acosec⁡A== \sqrt{\cosec^2A - 1} = \sqrt{\cot^2A} = \cot A = \frac{\cos A}{\sin A} = \cos A \cosec A = R.H.S.

  23. L.H.S. =sec⁡2Acosec⁡2A=(1+tan⁡2A)(1+cot⁡2A)=1+tan⁡2A+cot⁡2A+tan⁡2Acos⁡2A=2+tan⁡2A+cot⁡2A== \sec^2A\cosec^2A = (1 + \tan^2A)(1 + \cot^2A) = 1 + \tan^2A + \cot^2A + \tan^2A\cos^2A = 2 + \tan^2A + \cot^2A = R.H.S.

  24. L.H.S. =tan⁡2A−sin⁡2A=sin⁡2Acos⁡2A−sin⁡2A=sin⁡2A−sin⁡2Acos⁡2Acos⁡2A=sin⁡2A(1−cos⁡2A)sec⁡2A=sin⁡4Asec⁡2A== \tan^2A - \sin^2A = \frac{\sin^2A}{\cos^2A} - \sin^2A = \frac{\sin^2A - \sin^2A\cos^2A}{\cos^2A} = \sin^2A(1 - \cos^2A)\sec^2A = \sin^4A\sec^2A = R.H.S.

  25. L.H.S. =(1+cot⁡A−cosec⁡A)(1+tan⁡A+sec⁡A)=(1+cos⁡Asin⁡A−1sin⁡A)(1+sin⁡Acos⁡A+1cos⁡A)= (1 + \cot A - \cosec A)(1 + \tan A + \sec A) = \left(1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}\right)\left(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}\right)

    =(sin⁡A+cos⁡A−1)(sin⁡A+cos⁡A+1)sin⁡Acos⁡A=(sin⁡A+cos⁡A)2−1sin⁡Acos⁡A= \frac{(\sin A + \cos A - 1)(\sin A + \cos A + 1)}{\sin A\cos A} = \frac{(\sin A + \cos A)^2 - 1}{\sin A\cos A}

    =sin⁡2A+cos⁡2A+sin⁡Acos⁡A−1sin⁡Acos⁡A=1+2sin⁡Acos⁡A−1sin⁡Acos⁡A=2== \frac{\sin^2A + \cos^2A + \sin A\cos A - 1 }{\sin A\cos A} = \frac{1 + 2\sin A\cos A - 1}{\sin A\cos A} = 2= R.H.S.

  26. L.H.S. =cot⁡Acos⁡Acot⁡A+cos⁡A=cos⁡2Asin⁡Acos⁡Asin⁡A+cos⁡A= \frac{\cot A\cos A}{\cot A + \cos A} = \frac{\frac{\cos^2 A}{\sin A}}{\frac{\cos A}{\sin A} + \cos A}

    =cos⁡2Asin⁡Asin⁡Acos⁡A(1+sin⁡A)=cos⁡A1+sin⁡A= \frac{\cos^2A}{\sin A}{\frac{\sin A}{\cos A(1 + \sin A)}} = \frac{\cos A}{1 + \sin A}

    Proceeding in same way for R.H.S. we obtain 1−sin⁡Acos⁡A\frac{1 - \sin A}{\cos A}

    cos⁡A1+sin⁡A=cos⁡A(1−sin⁡A)1−sin⁡2A=1−sin⁡Acos⁡A=\frac{\cos A}{1 + \sin A} = \frac{\cos A(1 - \sin A)}{1 - \sin^2A} = \frac{1 - \sin A}{\cos A} = R.H.S.

  27. L.H.S. =cot⁡A+tan⁡Bcot⁡B+tan⁡A=1tan⁡A+tan⁡B1tan⁡B+tan⁡A= \frac{\cot A + \tan B}{\cot B + \tan A} = \frac{\frac{1}{\tan A} + \tan B}{\frac{1}{\tan B} + \tan A}

    1+tan⁡Atan⁡Btan⁡A1+tan⁡Atan⁡Btan⁡B=tan⁡Btan⁡A=cot⁡Atan⁡B=\frac{\frac{1 + \tan A\tan B}{\tan A}}{\frac{1 + \tan A\tan B}{\tan B}} = \frac{\tan B}{\tan A} = \cot A\tan B = R.H.S.

  28. L.H.S. =(1sec⁡2A−cos⁡2A+1cosec⁡2A−sin⁡2A)cos⁡2Asin⁡2A= \left(\frac{1}{\sec^2 A - \cos^2A} + \frac{1}{\cosec^2A - \sin^2A}\right)\cos^2A\sin^2A

    =(11cos⁡2A−cos⁡2A+11sin⁡2A−sin⁡2A)cos⁡2Asin⁡2A= \left(\frac{1}{\frac{1}{\cos^2A} - \cos^2A} + \frac{1}{\frac{1}{\sin^2A} - \sin^2A}\right)\cos^2A\sin^2A

    =(cos⁡2A1−cos⁡4A+sin⁡2A1−sin⁡4)cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{1 - \cos^4A} + \frac{\sin^2A}{1 - \sin^4}\right)\cos^2A\sin^2A

    =(cos⁡2A(1−cos⁡2A)(1+cos⁡2A)+sin⁡2A(1−sin⁡2A)(1+sin⁡2A))cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{(1 - \cos^2A)(1 + \cos^2A)} + \frac{\sin^2A}{(1 - \sin^2A)(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos⁡2Asin⁡2A(1+cos⁡2A)+sin⁡2Acos⁡2A(1+sin⁡2A))cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{\sin^2A(1 + \cos^2A)} + \frac{\sin^2A}{\cos^2A(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos⁡4A(1+sin⁡2A)+sin⁡4A(1+cos⁡2A)sin⁡2Acos⁡2A(1+cos⁡2A))(1+sin⁡2A))cos⁡2Asin⁡2A= \left(\frac{\cos^4A(1 + \sin^2A) + \sin^4A(1 + \cos^2A)}{\sin^2A\cos^2A(1 + \cos^2A))(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos⁡4A+sin⁡4A+cos⁡2Asin⁡2A(cos⁡2A+sin⁡2A)(1+cos⁡2A)(1+sin⁡2A))= \left(\frac{\cos^4A + \sin^4A + \cos^2A\sin^2A(\cos^2A + \sin^2A)}{(1 +\cos^2A)(1 + \sin^2A)}\right)

    =((cos⁡2A+sin⁡2A)2−cos⁡2Asin⁡2A1+sin⁡2A+cos⁡2A+sin⁡2Acos⁡2A)= \left(\frac{(\cos^2A + \sin^2A)^2 - \cos^2A\sin^2A}{1 + \sin^2A + \cos^2A + \sin^2A\cos^2A}\right)

    =1−sin⁡2Acos⁡2A2+sin⁡2Acos⁡2A== \frac{1 - \sin^2A\cos^2A}{2 + \sin^2A\cos^2A} = R.H.S.

  29. L.H.S. =sin⁡8A−cos⁡8A=(sin⁡4A−cos⁡4A)(sin⁡4A+cos⁡4A)= \sin^8A - \cos^8A = (\sin^4A - \cos^4A)(\sin^4A + \cos^4A)

    =(sin⁡2A+cos⁡2A)(sin⁡2A−cos⁡2A)((sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A)= (\sin^2A + \cos^2A)(\sin^2A - \cos^2A)((\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A)

    =(sin⁡2A−cos⁡2A)(1−2sin⁡2Acos⁡2A)== (\sin^2A - \cos^2A)(1 - 2\sin^2A\cos^2A) = R.H.S.

  30. L.H.S. =cos⁡Acosec⁡A−sin⁡Asec⁡Acos⁡A+sin⁡A=cos⁡Asin⁡A−sin⁡Acos⁡Acos⁡A+sin⁡A= \frac{\cos A\cosec A - \sin A\sec A}{\cos A + \sin A} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\cos A}}{\cos A + \sin A}

    =cos⁡2A−sin⁡2Asin⁡Acos⁡A(cos⁡A+sin⁡A)=cos⁡Asin⁡Asin⁡Acos⁡A=cosec⁡A−sec⁡A== \frac{\cos^2A - \sin^2A}{\sin A\cos A(\cos A + \sin A)} = \frac{\cos A \sin A}{\sin A\cos A} = \cosec A - \sec A = R.H.S.

  31. Given, 1cosec⁡A−cot⁡A−1sin⁡A=1sin⁡A−1cosec⁡A+cot⁡A\frac{1}{\cosec A - \cot A} - \frac{1}{\sin A} = \frac{1}{\sin A} - \frac{1}{\cosec A + \cot A}

    Therefore alternatively we can prove that 1cosec⁡A−cot⁡A+1cosec⁡A+cot⁡A=2sin⁡A\frac{1}{\cosec A - \cot A} + \frac{1}{\cosec A + \cot A} = \frac{2}{\sin A}

    L.H.S. =11sin⁡A−cos⁡Asin⁡A+11sin⁡A+cos⁡Asin⁡A= \frac{1}{\frac{1}{\sin A} - \frac{\cos A}{\sin A}} + \frac{1}{\frac{1}{\sin A} + \frac{\cos A}{\sin A}}

    =sin⁡A1−cos⁡A−sin⁡A1+cos⁡A=sin⁡A(1+cos⁡A+1−cos⁡A)1−cos⁡2A=2sin⁡Asin⁡2A=2sin⁡A== \frac{\sin A}{1 - \cos A} - \frac{\sin A}{1 + \cos A} = \frac{\sin A(1 + \cos A + 1 - \cos A)}{1 - \cos^2 A} = \frac{2\sin A}{\sin^2A} = \frac{2}{\sin A} = R.H.S.

  32. L.H.S. =tan⁡A+sec⁡A−1tan⁡A−sec⁡A+1=sin⁡A+1−cos⁡Acos⁡Asin⁡A+cos⁡A−1)cos⁡A= \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{\frac{\sin A + 1 - \cos A}{\cos A}}{\frac{\sin A + \cos A - 1)}{\cos A}}

    =1+sin⁡A−cos⁡Asin⁡A+cos⁡A−1=(1+sin⁡A−cos⁡A)(sin⁡A+cos⁡A−1)(sin⁡A+cos⁡A−1)2= \frac{1 + \sin A - \cos A}{\sin A + \cos A - 1} = \frac{(1 + \sin A - \cos A)(\sin A + \cos A - 1)}{(\sin A + \cos A - 1)^2}

    Solving this yields =1+sin⁡Acos⁡A== \frac{1 + \sin A}{\cos A} = R.H.S.

  33. L.H.S. =(tan⁡A+cosec⁡B)2−(cot⁡B−sec⁡A)2=tan⁡2A+cosec⁡2B+2tan⁡Acosec⁡B−cot⁡2B−sec⁡2A+2sec⁡Acot⁡B= (\tan A + \cosec B)^2 - (\cot B - \sec A)^2 = \tan^2A + \cosec^2B + 2\tan A\cosec B - \cot^2B - \sec^2A+ 2\sec A\cot B

    =(cosec⁡2B−cot⁡2B)−(sec2A−tan2A)+2tan⁡Acot⁡B(cosec⁡Bcot⁡B+sec⁡Atan⁡A)= (\cosec^2B - \cot^2B) - (sec^2A - tan^2A) + 2\tan A\cot B\left(\frac{\cosec B}{\cot B} + \frac{\sec A}{\tan A}\right)

    =1−1+2tan⁡Acot⁡B(sec⁡B+cosec⁡A)== 1 - 1 + 2\tan A\cot B\left(\sec B + \cosec A\right) = R.H.S.

  34. L.H.S. =2sec⁡2A−sec⁡4A−2cosec⁡2A+cosec⁡4A=2(1+tan⁡2A)−(1+tan⁡2A)2−2(1+cot⁡2A)+(1+cot⁡2A)2= 2\sec^2 A - \sec^4A - 2\cosec^2A + \cosec^4A = 2(1 + \tan^2A) - (1 + \tan^2A)^2 - 2(1 + \cot^2A) + (1 + \cot^2A)^2

    =2+2tan⁡2A−1−2tan⁡2A+tan⁡4A−2−2cot⁡2A+1+2cot⁡2A+cot⁡4A=cot⁡4A−tan⁡4A== 2 + 2\tan^2A - 1 - 2\tan^2A + \tan^4A - 2 - 2\cot^2A + 1 + 2\cot^2A + \cot^4A = \cot^4A - \tan^4A = R.H.S.

  35. L.H.S. =(sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=sin⁡2A+cosec⁡2A+2sin⁡Acosec⁡A+cos⁡2A+sec⁡2A+2sec⁡Acosec⁡A= (\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = \sin^2A + \cosec^2A + 2\sin A\cosec A + \cos^2A + \sec^2A + 2\sec A\cosec A

    =(sin⁡2A+cos⁡2A)+4+cosec⁡2A+sec⁡2A=1+4+1+cot⁡2A+1+tan⁡2A=cot⁡2A+tan⁡2A+7== (\sin^2A + \cos^2A) + 4 + \cosec^2A + \sec^2A = 1 + 4 + 1 + \cot^2A + 1 + \tan^2A = \cot^2A + \tan^2A + 7 = R.H.S.

  36. L.H.S =(cosec⁡A+cot⁡A)(1−sin⁡A)−(sec⁡A+tan⁡A)(1−cos⁡A)= (\cosec A + \cot A)(1 - \sin A) - (\sec A + \tan A)(1 - \cos A)

    =(1sin⁡A+cos⁡Asin⁡A)(1−sin⁡A)−(1cos⁡A+sin⁡Acos⁡A)(1−cos⁡A)= \left(\frac{1}{\sin A} + \frac{\cos A}{\sin A}\right)(1 - \sin A) - \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \cos A)

    =(1+cos⁡A)(1−sin⁡A)sin⁡A−(1+sin⁡A)(1−cos⁡A)cos⁡A= \frac{(1 + \cos A)(1 - \sin A)}{\sin A} - \frac{(1 + \sin A)(1 - \cos A)}{\cos A}

    =cos⁡A(1+cos⁡A−sin⁡A−cos⁡Asin⁡A)−sin⁡A(1+sin⁡A−cos⁡A−sin⁡Acos⁡A)sin⁡Acos⁡A= \frac{\cos A(1 + \cos A - \sin A - \cos A\sin A) -\sin A(1 + \sin A - \cos A - \sin A\cos A)}{\sin A\cos A}

    =(1−sin⁡Acos⁡A)(cos⁡A−sin⁡A)+(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)sin⁡Acos⁡A= \frac{(1 - \sin A\cos A)(\cos A - \sin A) + (\cos A - \sin A)(\cos A + \sin A)}{\sin A\cos A}

    =(cos⁡A−sin⁡A)(1−cos⁡Asin⁡A+cos⁡A+sin⁡A)sin⁡Acos⁡A= \frac{(\cos A - \sin A)(1 - \cos A\sin A + \cos A + \sin A)}{\sin A\cos A}

    =((cos⁡A−sin⁡A)sin⁡Acos⁡A)(2−1−cos⁡Asin⁡A+cos⁡A+sin⁡A)= \left(\frac{(\cos A - \sin A)}{\sin A\cos A}\right)(2 - 1 - \cos A\sin A + \cos A + \sin A)

    =(cosec⁡A−sec⁡A)(2−(1−cos⁡A)(1−sin⁡A))== (\cosec A - \sec A)(2 - (1 - \cos A)(1 - \sin A)) = R.H.S.

  37. L.H.S =(1+cot⁡A+tan⁡A)(sin⁡A−cos⁡A)=(1+cosec⁡Asec⁡A+sec⁡Acosec⁡A)(1cosec⁡A−1sec⁡A)= (1 + \cot A + \tan A)(\sin A - \cos A) = \left(1 + \frac{\cosec A}{\sec A} + \frac{\sec A}{\cosec A}\right)\left(\frac{1}{\cosec A} - \frac{1}{\sec A}\right)

    =(sec⁡Acosec⁡A+sec⁡A∣cosec⁡Asec⁡Acosec⁡A)(sec⁡A−cosec⁡Asec⁡Acosec⁡a)= \left(\frac{\sec A\cosec A + \sec A | \cosec A}{\sec A\cosec A}\right)\left(\frac{\sec A - \cosec A}{\sec A\cosec a}\right)

    =sec⁡3A−cosec⁡3Asec⁡2Acosec⁡2A=sec⁡Acosec⁡2A−cosec⁡Asec⁡2A== \frac{\sec^3A - \cosec^3A}{\sec^2A\cosec^2A} = \frac{\sec A}{\cosec^2A} - \frac{\cosec A}{\sec^2A} = R.H.S.

  38. Given, 1sec⁡A−tan⁡A−1cos⁡A=1cos⁡A−1sec⁡A+tan⁡A\frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A}

    So alternatively we can prove that 1sec⁡A−tan⁡A+1sec⁡A+tan⁡A=2cos⁡A\frac{1}{\sec A - \tan A} + \frac{1}{\sec A + \tan A} = \frac{2}{\cos A}

    L.H.S. =sec⁡A+tan⁡A+sec⁡A−tan⁡Asec⁡2A−tan2A=2sec⁡A=2cos⁡A== \frac{\sec A + \tan A + \sec A - \tan A}{\sec^2A - tan^2A} = 2\sec A = \frac{2}{\cos A} = R.H.S.

  39. L.H.S. =3(sin⁡A−cos⁡A)4+4(sin⁡6A+cos⁡6A)+6(sin⁡A+cos⁡A)2= 3(\sin A - \cos A)^4 + 4(\sin^6 A + \cos^6 A) + 6(\sin A + \cos A)^2

    =3[(sin⁡A−cos⁡A)2]2+4[(sin⁡2A)3+(cos⁡2A)3]+6(sin⁡2A+cos⁡2A+2cos⁡Asin⁡A)= 3[(\sin A - \cos A)^2]^2 + 4[(\sin^2A)^3 + (\cos^2A)^3] + 6(\sin^2A + \cos^2A + 2\cos A\sin A)

    =3[(sin⁡2A+cos⁡2A−2sin⁡Acos⁡A)]2+4(sin⁡2A+cos⁡2A)(sin⁡4A+cos⁡4A−sin⁡2Acos⁡2A)+6(1+2cos⁡Asin⁡A)= 3[(\sin^2A + \cos^2A - 2\sin A\cos A)]^2 + 4(\sin^2A + \cos^2A)(\sin^4A + \cos^4A - \sin^2A\cos^2A) + 6(1 + 2\cos A\sin A)

    =3(1−2sin⁡Acos⁡A)2+4(sin⁡4A+cos⁡4A−sin⁡2Acos⁡2A)+6(1+2cos⁡Asin⁡A)= 3(1 - 2\sin A\cos A)^2 + 4(\sin^4A + \cos^4A - \sin^2A\cos^2A) + 6(1 + 2\cos A\sin A)

    =3(1+4sin⁡2Acos⁡2A−4cos⁡Asin⁡A)+4[(sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A−sin⁡2Acos⁡2A]+6(1+2sin⁡Acos⁡A)= 3(1 + 4\sin^2A\cos^2A - 4\cos A\sin A) + 4[(\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A - \sin^2A\cos^2A] + 6(1 + 2\sin A\cos A)

    =3+12sin⁡2Acos⁡2A−12cos⁡Asin⁡A+4(1−3sin⁡2Acos⁡2A)+6+12sin⁡Acos⁡A= 3 + 12\sin^2A\cos^2A -12\cos A\sin A + 4(1 - 3\sin^2A\cos^2A) + 6 + 12\sin A\cos A

    =13== 13 = R.H.S.

  40. L.H.S. =1+cos⁡A1−cos⁡A= \sqrt{\frac{1 + \cos A}{1 - \cos A}}

    Multiplying and dividing with 1+cos⁡A,1 + \cos A, we get

    =(1+cos⁡A)21−cos⁡2A=(1+cos⁡Asin⁡A)2=cosec⁡A+cot⁡A== \sqrt{\frac{(1 + \cos A)^2}{1 - \cos^2A}} = \sqrt{\left(\frac{1 + \cos A}{\sin A}\right)^2} = \cosec A + \cot A = R.H.S.

  41. L.H.S. =cos⁡A1+sin⁡A+cos⁡A1−sin⁡A= \frac{\cos A}{1 + \sin A} + \frac{\cos A}{1 - \sin A}

    cos⁡A(1−sin⁡A)+cos⁡A(1+sin⁡A)1−sin⁡2A\frac{\cos A(1 - \sin A) + \cos A(1 + \sin A)}{1 - \sin^2A}

    =2cos⁡Acos⁡2A=2sec⁡A== \frac{2\cos A}{\cos^2A} = 2\sec A = R.H.S.

  42. L.H.S. =tan⁡Asec⁡A−1+tan⁡Asec⁡A+1= \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1}

    =tan⁡Asec⁡A+tan⁡A+tan⁡Asec⁡A−tan⁡A(sec⁡2A−1)= \frac{\tan A\sec A + \tan A + \tan A\sec A - \tan A}{(\sec^2A - 1)}

    =2tan⁡Asec⁡Atan⁡2A=2cosec⁡A== \frac{2\tan A\sec A}{\tan^2 A} = 2\cosec A = R.H.S.

  43. L.H.S. =11−sin⁡A−11+sin⁡A= \frac{1}{1 - \sin A} - \frac{1}{1 + \sin A}

    =1+sin⁡A−1+sin⁡A1−sin⁡2A=2sin⁡Acos⁡2A=2sec⁡Atan⁡A== \frac{1 + \sin A - 1 + \sin A}{1 - \sin^2A} = \frac{2\sin A}{\cos^2A} = 2\sec A\tan A = R.H.S.

  44. L.H.S. =1+tan⁡2A1+cot⁡2A=1+sin⁡2Acos⁡2A1+cos⁡2Asin⁡2A= \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^2A}}

    =sin⁡2A+cos⁡2Acos⁡2Asin⁡2A+cos⁡2Asin⁡2A=sin⁡2Acos⁡2A=tan⁡2A= \frac{\frac{\sin^2A + \cos^2A}{\cos^2A}}{\frac{\sin^2A + \cos^2A}{\sin^2A}} = \frac{\sin^2A}{\cos^2A} = \tan^2A

    R.H.S. =(1−tan⁡A1−cot⁡A)2=(1−sin⁡Acos⁡A1−cos⁡Asin⁡A)= \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \left(\frac{1 - \frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}}\right)

    =(cos⁡A−sinAcos⁡Asin⁡A−cos⁡Asin⁡A)2= \left(\frac{\frac{\cos A - sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^2

    =sin⁡2Acos⁡2A=tan⁡2A= \frac{\sin^2A}{\cos^2A} = \tan^2A

    Hence, L.H.S. = R.H.S.

  45. L.H.S. =1+2tan⁡2Acos⁡2A=1+2sin⁡2Acos⁡4A=cos⁡4A+2sin⁡2Acos⁡4A= 1 + \frac{2\tan^2 A}{\cos^2 A} = 1 + \frac{2\sin^2A}{\cos^4A} = \frac{\cos^4A + 2\sin^2A}{\cos^4A}

    =(1−sin⁡2A)2+2sin⁡2Acos⁡4A=1−2sin⁡2A+sin⁡4A+2sin⁡2Acos⁡4A= \frac{(1 - \sin^2A)^2 + 2\sin^2A}{\cos^4A} = \frac{1 - 2\sin^2A + \sin^4A + 2\sin^2A}{\cos^4A}

    =tan⁡4A+sec4A== \tan^4A + sec^4A = R.H.S.

  46. L.H.S. =(1−sin⁡A−cos⁡A)2=1−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A+sin⁡2A+cos⁡2A= (1 - \sin A - \cos A)^2 = 1 - 2\cos A - 2\sin A + 2\sin A\cos A + \sin^2A + \cos^2A

    =2−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A=2(1−cos⁡A)(1−sin⁡A)== 2 - 2\cos A - 2\sin A + 2\sin A\cos A = 2(1 - \cos A)(1 - \sin A) = R.H.S.

        1. =cot⁡A+cosec⁡A−1cot⁡A−cosec⁡A+1= \frac{\cot A + \cosec A - 1}{\cot A - \cosec A + 1}

    =cos⁡A+1−sin⁡Acos⁡A+sin⁡A−1= \frac{\cos A + 1 - \sin A}{\cos A + \sin A - 1}

    Multiplying and dividing with cos⁡A−(1−sin⁡A),\cos A - (1 - \sin A), we get

    =cos⁡2A−1+2sin⁡A−sin⁡2Acos⁡2A+sin⁡2A+1−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A= \frac{\cos^2A - 1 + 2\sin A - \sin^2A}{\cos^2A + \sin^2A + 1 - 2\cos A - 2\sin A + 2\sin A\cos A}

    =2sin⁡A−2sin⁡2A2(1−cos⁡A)(1−sin⁡A)=sin⁡A1−cos⁡A= \frac{2\sin A - 2\sin^2A}{2(1 - \cos A)(1 - \sin A)} = \frac{\sin A}{1 - \cos A}

    Multiplyig numerator and denominator with 1+cos⁡A,1 + \cos A, we get

    =sin⁡A(1+cos⁡A)1−cos⁡2A=1+cos⁡Asin⁡A== \frac{\sin A(1 + \cos A)}{1 - \cos^2A} = \frac{1 + \cos A}{\sin A} = R.H.S.

  48. L.H.S. =(sin⁡A+sec⁡A)2+(cos⁡A+cosec⁡A)2= (\sin A + \sec A)^2 + (\cos A + \cosec A)^2

    =(sin⁡Acos⁡A+1cos⁡A)2+(cos⁡Asin⁡A+1sin⁡A)2= \left(\frac{\sin A\cos A + 1}{\cos A}\right)^2 + \left(\frac{\cos A\sin A + 1}{\sin A}\right)^2

    =(1+sin⁡Acos⁡A)2(1cos⁡2A+1sin⁡2A)= (1 + \sin A\cos A)^2 \left(\frac{1}{\cos^2A} + \frac{1}{\sin^2A}\right)

    =(1+sin⁡Acos⁡A)2(sin⁡2A+cos⁡2Asin⁡2Acos⁡2A)= (1 + \sin A\cos A)^2\left(\frac{\sin^2A + \cos^2A}{\sin^2A\cos^2A}\right)

    =(1+sin⁡Acos⁡Asin⁡Acos⁡A)2=(1+sec⁡Acosec⁡A)2== \left(\frac{1 + \sin A\cos A}{\sin A\cos A}\right)^2 = (1 + \sec A\cosec A)^2 = R.H.S.

  49. L.H.S. =2sin⁡Atan⁡A(1−tan⁡A)+2sin⁡Asec⁡2A(1+tan⁡A)2= \frac{2\sin A\tan A(1 - \tan A) + 2\sin A\sec^2A}{(1 + \tan A)^2}

    =2sin⁡A(tan⁡A−tan⁡2A+sec⁡2A)(1+tan⁡A)2=2sin⁡A(1+tan⁡A)(1+tan⁡A)2= \frac{2\sin A(\tan A - \tan^2A + \sec^2A)}{(1 + \tan A)^2} = \frac{2\sin A(1 + \tan A)}{(1 + \tan A)^2}

    =2sin⁡A1+tan⁡A== \frac{2\sin A}{1 + \tan A} = R.H.S.

  50. Given, 2sin⁡A=2−cos⁡A2\sin A = 2 - \cos A

    cos⁡A=2−2sin⁡A,\cos A = 2 - 2\sin A, squaring both sides we, get cos⁡2A=4−8sin⁡A+4sin⁡2A\cos^2A = 4 - 8\sin A + 4\sin^2A

    1−sin⁡2A=4−8sin⁡A+4sin⁡2A⇒5sin⁡2A−8sin⁡A+3=01 - \sin^2A = 4 - 8\sin A + 4\sin^2A \Rightarrow 5\sin^2A - 8\sin A + 3 = 0

    5sin⁡2A−5sin⁡A−3sin⁡A+3=0⇒sin⁡A=1,355\sin^2A - 5\sin A - 3\sin A + 3 = 0 \Rightarrow \sin A = 1, \frac{3}{5}

  51. Given 8sin⁡A=4+cos⁡A⇒8sin⁡A−4=cos⁡A8\sin A = 4 + \cos A \Rightarrow 8\sin A - 4 = \cos A

    Squaring both sides, we get

    64sin⁡2A−64sin⁡A+16=cos⁡2A=1−sin⁡2A64\sin^2A - 64\sin A + 16 = \cos^2A = 1 - \sin^2A

    65sin⁡2A−64sin⁡A+15=065\sin^2A - 64\sin A + 15 = 0

    65sin⁡2A−39sin⁡A−25sin⁡A+15=065\sin^2A - 39\sin A - 25\sin A + 15 = 0

    sin⁡A=35,513\sin A = \frac{3}{5}, \frac{5}{13}

  52. Given, tan⁡A+sec⁡A=1.5\tan A + \sec A = 1.5

    ⇒1+sin⁡A=1.5cos⁡A⇒2+2sin⁡A=3cos⁡A\Rightarrow 1 + \sin A = 1.5 \cos A \Rightarrow 2 + 2\sin A = 3\cos A

    Squaring both sides, we get

    4+8sin⁡A+4sin⁡2A=9−9sin⁡2A4 + 8\sin A + 4\sin^2A = 9 - 9\sin^2A

    13sin⁡2A+8sin⁡A−5=013\sin^2A + 8\sin A - 5 =0

    sin⁡A=−1,513\sin A = -1, \frac{5}{13}

  53. Given, cot⁡A+cosec⁡A=5⇒1+cos⁡A=5sin⁡A\cot A + \cosec A = 5 \Rightarrow 1 + \cos A = 5\sin A

    Squarig both sides, we get

    1+2cos⁡A+cos⁡2A=25sin⁡2A=25(1−cos⁡2A)1 + 2\cos A + \cos^2A = 25\sin^2A = 25(1 - \cos^2A)

    26cos⁡2A+2cos⁡A−24=026\cos^2A + 2\cos A - 24 = 0

    26cos⁡2A+26cos⁡A−24cos⁡A−24=026\cos^2A + 26\cos A - 24\cos A - 24 = 0

    cos⁡A=−1,1213\cos A= -1, \frac{12}{13}

  54. 3sec⁡4A+8=10sec⁡2A⇒3(sec2A)2+8=10(1+tan⁡2A)3\sec^4 A + 8 = 10\sec^2A \Rightarrow 3(sec^2A)^2 + 8 = 10(1 + \tan^2A)

    ⇒3(1+tan⁡2A)2+8=10+10tan⁡2A\Rightarrow 3(1 + \tan^2A)^2 + 8 = 10 + 10\tan^2A

    3+6tan⁡2A+3tan⁡4A+8=10+10tan⁡2A3 + 6\tan^2A + 3\tan^4A + 8 = 10 + 10\tan^2A

    3tan⁡4A−4tan⁡2A+1=03\tan^4A - 4\tan^2A + 1 = 0

    tan⁡A=±1,±13\tan A = \pm1, \pm\frac{1}{\sqrt{3}}

  55. Given, tan⁡2A+sec⁡A=5⇒sec⁡2A−1+sec⁡A=5\tan^2A + \sec A = 5 \Rightarrow \sec^2A - 1 + \sec A = 5

    sec⁡2A+sec⁡A−6=0\sec^2A + \sec A - 6 = 0

    sec⁡A=−3,2⇒cos⁡A=−13,12\sec A = -3, 2\Rightarrow \cos A = -\frac{1}{3}, \frac{1}{2}

  56. Given tan⁡A+cot⁡A=2⇒tan⁡A+1tan⁡A=2\tan A + \cot A = 2 \Rightarrow \tan A + \frac{1}{\tan A} = 2

    tan⁡2A−2tan⁡A+1=0\tan^2A - 2\tan A + 1 = 0

    tan⁡A=1⇒sin⁡A=12\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}

  57. Given, sec⁡2A=2+2tan⁡A⇒tan⁡2A+1=2+2tan⁡A\sec^2A = 2 + 2\tan A \Rightarrow \tan^2A + 1 = 2 + 2\tan A

    tan⁡2A−2tan⁡A−1=0\tan^2A - 2\tan A - 1 = 0

    tan⁡A=2±4+42=1±2\tan A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}

  58. Given, tan⁡A=2x(x+1)2x+1\tan A = \frac{2x(x + 1)}{2x + 1}

    sin⁡A=2x(x+1)[2x(x+1)]2+(2x+1)2\sin A = \frac{2x(x + 1)}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}

    cos⁡A=2x+1[2x(x+1)]2+(2x+1)2\cos A = \frac{2x + 1}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}

  59. Given, 3sin⁡A+5cos⁡A=5,3\sin A + 5\cos A = 5, let 5sin⁡A−3cos⁡A=x5\sin A - 3\cos A = x

    Squaring and adding, we get

    9sin⁡2A+25cos⁡2A+30sin⁡Acos⁡A+25sin⁡2A+9cos⁡2A−30sin⁡Acos⁡A=25+x29\sin^2A + 25\cos^2A + 30\sin A\cos A + 25\sin^2A + 9\cos^2A - 30\sin A\cos A = 25 + x^2

    9(sin⁡2A+cos⁡2A)+25(cos⁡2A+sin⁡2A)=25+x29(\sin^2A + \cos^2A) + 25(\cos^2A + \sin^2A) = 25 + x^2

    34=25+x2⇒x2=9⇒x=±334 = 25 + x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3

  60. Given, sec⁡A+tan⁡A=sec⁡A−tan⁡A\sec A + \tan A = \sec A - \tan A

    Multiplying both sides with sec⁡A+tan⁡A,\sec A + \tan A, we get

    (sec⁡A+tan⁡A)2=sec⁡2A−tan⁡2A=1(\sec A + \tan A)^2 = \sec^2A - \tan^2A = 1

    sec⁡A+tan⁡A=±1\sec A + \tan A = \pm 1

    We can prove that sec⁡A−tan⁡A\sec A - \tan A to be ±1\pm 1 by multiplying given equation with sec⁡A−tan⁡A\sec A - \tan A

  61. Given, cos⁡4Acos⁡2B+sin⁡4Asin⁡2B=1=sin⁡2A+cos⁡2A\frac{\cos^4 A}{\cos^2 B} + \frac{\sin^4 A}{\sin^2 B} = 1 = \sin^2A + \cos^2A

    ⇒cos⁡4Acos⁡2B−cos⁡2A=sin⁡2A−sin⁡4Asin⁡2B=0\Rightarrow \frac{\cos^4 A}{\cos^2 B} - \cos^2A = \sin^2A - \frac{\sin^4A}{\sin^2B} = 0

    ⇒cos⁡2A(cos⁡2A−cos⁡2B)cos⁡2B=sin⁡2A(sin⁡2B−sin⁡2A)sin⁡2B\Rightarrow \frac{\cos^2A(\cos^2A - \cos^2B)}{\cos^2B} = \frac{\sin^2A(\sin^2B - \sin^2A)}{\sin^2B}

    ⇒cos⁡2A(cos⁡2A−cos⁡2B)cos⁡2B=sin⁡2A(1−cos⁡2B−1+cos⁡2A)sin⁡2B\Rightarrow \frac{\cos^2A(\cos^2A - \cos^2B)}{\cos^2B} = \frac{\sin^2A(1 - \cos^2B - 1 + \cos^2A)}{\sin^2B}

    ⇒(cos⁡2A−cos⁡2B)(cos⁡2Acos⁡2B−sin⁡2Asin⁡2B)=0\Rightarrow (\cos^2A - \cos^2B )\left(\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B}\right) = 0

    When cos⁡2A−cos⁡2B=0,cos⁡2A=cos⁡2B\cos^2A - \cos^2B = 0, \cos^2A = \cos^2B

    When cos⁡2Acos⁡2B−sin⁡2Asin⁡2B=0\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B} = 0

    cos⁡2Asin⁡2B=sin⁡2Acos⁡2B⇒cos⁡2A(1−cos⁡2B)=(1−cos⁡2A)cos⁡2B\cos^2A\sin^2B = \sin^2A\cos^2B \Rightarrow \cos^2A(1 - \cos^2B) = (1 - \cos^2A)\cos^2B

    cos⁡2A=cos⁡2B⇒sin⁡2A=sin⁡2B\cos^2A = \cos^2B \Rightarrow \sin^2A = \sin^2B

    1. L.H.S. =sin⁡4A+sin⁡4B=(sin⁡2A−sin⁡2B)2+2sin⁡2Asin⁡2B=2sin⁡2Asin⁡2B== \sin^4A + \sin^4B = (\sin^2A - \sin^2B)^2 + 2\sin^2A\sin^2B = 2\sin^2A\sin^2B = R.H.S.

    2. L.H.S. =cos⁡4Bcos⁡2A+sin⁡4Bsin⁡2A= \frac{\cos^4 B}{\cos^2 A} + \frac{\sin^4 B}{\sin^2 A}

      =cos⁡4Bcos⁡2B+sin⁡4Bsin⁡2B=cos⁡2B+sin⁡2B=1== \frac{\cos^4 B}{\cos^2 B} + \frac{\sin^4 B}{\sin^2 B} = \cos^2B + \sin^2B = 1 = R.H.S.

  62. Given, cos⁡A+sin⁡A=2cos⁡A\cos A + \sin A = \sqrt{2}\cos A

    Squaring both sides

    1+2sin⁡Acos⁡A=2cos⁡2A1 + 2\sin A\cos A = 2\cos^2A

    2−2cos⁡2A=2sin⁡2A=1−2sin⁡Acos⁡A=sin⁡2A+cos⁡2A−2sin⁡Acos⁡A=(cos⁡A−sin⁡A)22 - 2\cos^2A = 2\sin^2A = 1 - 2\sin A\cos A = \sin^2A + \cos^2A - 2\sin A\cos A = (\cos A - \sin A)^2

    cos⁡A−sin⁡A=±2sin⁡A\cos A - \sin A = \pm \sqrt{2}\sin A

  63. Given, acos⁡A−bsin⁡A=c,a\cos A - b\sin A = c, and let asin⁡A+bcos⁡A=xa\sin A + b\cos A = x

    Squaring and adding, we get

    ⇒a2cos⁡2A+b2sin⁡2A−2abcos⁡Asin⁡A+a2sin⁡2A+b2cos⁡2A+2absin⁡Acos⁡A=c2+x2\Rightarrow a^2\cos^2A + b^2\sin^2A - 2ab\cos A\sin A + a^2\sin^2A + b^2\cos^2A + 2ab\sin A\cos A = c^2 + x^2

    ⇒a2(cos⁡2A+sin⁡2A)+b2(sin⁡2A+cos⁡2A)=c2+x2\Rightarrow a^2(\cos^2A + \sin^2A) + b^2(\sin^2A + \cos^2A) = c^2 + x^2

    ⇒a2+b2=c2+x2⇒x=±a2+b2−c2\Rightarrow a^2 + b^2 = c^2 + x^2 \Rightarrow x = \pm \sqrt{a^2 + b^2 - c^2}

  64. Given, 1−sin⁡A=1+sin⁡A1 - \sin A = 1 + \sin A

    Multiplying both sides by 1−sin⁡A,1 - \sin A, we get

    (1−sin⁡A)2=1−sin⁡2A=cos⁡2A(1 - \sin A)^2 = 1 - \sin^2A = \cos^2A

    1−sin⁡A=±cos⁡A1 - \sin A = \pm \cos A

    Similarly if we multiply with 1+sin⁡1 + \sin we can prove that

    1+sin⁡A=±cos⁡A1 + \sin A = \pm \cos A

  65. Let us solve these one by one:

    1. Given, sin⁡4A+sin⁡2A=1⇒sin⁡4A=1−sin⁡2A=cos⁡2\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2

      sin⁡4Acos⁡4A=sec2A\frac{\sin^4A}{\cos^4A} = sec^2A

      1tan⁡4A=cos2A\frac{1}{\tan^4A} = cos^2A

      Also from, sin⁡4A=cos⁡2A⇒tan⁡2A=cosec⁡2A⇒1tan⁡2A=sin⁡2A\sin^4A = \cos^2A \Rightarrow \tan^2A = \cosec^2A \Rightarrow \frac{1}{\tan^2A} = \sin^2A

      ⇒1tan⁡4A+1tan⁡2A=sin⁡2A+cos⁡2A=1\Rightarrow \frac{1}{\tan^4A} + \frac{1}{\tan^2A} = \sin^2A + \cos^2A = 1

    2. Given, sin⁡4A+sin⁡2A=1⇒sin⁡4A=1−sin⁡2A=cos⁡2A\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2A

      ⇒sin⁡2Acos⁡2A=1sin⁡2A⇒tan⁡2A=cosec⁡2A\Rightarrow \frac{\sin^2A}{\cos^2A} = \frac{1}{\sin^2A} \Rightarrow \tan^2A = \cosec^2A

      tan⁡2A=1+cot⁡2A\tan^2A = 1 + \cot^2A

      Multiplying both sides with tan⁡2A,\tan^2A, we get

      tan⁡4A=tan⁡2A+1\tan^4A = \tan^2A + 1

      tan⁡4A−tan⁡2A=1\tan^4A - \tan^2A = 1

  66. Given, cos⁡2−sin⁡2A=tan⁡2B⇒cos⁡2A−sin⁡2Acos⁡2A+sin⁡2A=tan⁡2B\cos^2 - \sin^2 A = \tan^2 B \Rightarrow \frac{\cos^2A - \sin^2A}{\cos^2A + \sin^2A} = \tan^2B

    Dividing both numerator and denominator of L.H.S. with cos⁡2A,\cos^2A, we get

    1−tan⁡2A1+tan⁡2A=tan⁡2B\frac{1 - \tan^2A}{1 + \tan^2A} = \tan^2B

    1−tan⁡2A−tan⁡2B−tan⁡2Atan⁡2B=01 - \tan^2A - \tan^2B - \tan^2A\tan^2B = 0

    1−tan⁡2B=tan⁡2A(1+tan⁡2B)⇒1−tan⁡2B1+tan⁡2B=tan⁡2A1 - \tan^2B = \tan^2A(1 + \tan^2B) \Rightarrow \frac{1 - \tan^2B}{1 + \tan^2B} = \tan^2A

    cos⁡2B−sin⁡2B=tan⁡2A⇒2cos⁡2B−1=tan⁡2A\cos^2B - \sin^2B = \tan^2A \Rightarrow 2\cos^2B - 1 = \tan^2A

  67. We will prove this by induction. Let sin⁡A+cosec⁡A=2,\sin A + \cosec A = 2, thus it is true for n=1n = 1

    Squaring both sides sin⁡2A+cosec⁡2A+2sin⁡Acosec⁡A=22⇒sin⁡2A+cosec⁡2A=2\sin^2A + \cosec^2A + 2\sin A\cosec A = 2^2 \Rightarrow \sin^2A + \cosec^2A = 2

    Thus, it is true for n=2n = 2 as well. Let it be true for n=m−1n = m-1 and n=mn = m

    sin⁡mA+cosec⁡mA=2\sin^mA + \cosec^mA = 2

    Multiplying both sides with sin⁡A+cosec⁡A,\sin A + \cosec A, we get

    sin⁡m+1A+cosec⁡m+1A+sin⁡mAcosec⁡A+cosec⁡mAsin⁡A=22\sin^{m + 1}A + \cosec^{m + 1}A + \sin^mA\cosec A + \cosec^mA\sin A = 2^2

    sin⁡m+1A+cosec⁡m+1A+sin⁡m−1A+cosec⁡m−1A=4\sin^{m + 1}A + \cosec^{m + 1}A + \sin^{m - 1}A + \cosec^{m - 1}A = 4

    sin⁡m+1A+cosec⁡m+1=4−2=2\sin^{m + 1}A + \cosec^{m + 1} = 4 - 2 = 2

    Thus, we have proven it by induction.

  68. L.H.S. =sec⁡A+tan⁡3Acosec⁡A=sec⁡A(1+tan⁡3Acosec⁡Asec⁡A)= \sec A + \tan^3A\cosec A = \sec A\left(1 + \tan^3A\frac{\cosec A}{\sec A}\right)

    =sec⁡A(1+tan⁡3Acot⁡a)=sec⁡A(1+tan⁡2A)=sec⁡3A=(1+1−e2)32=(2−e2)32=\sec A(1 + \tan^3A\cot a) = \sec A(1 + \tan^2A) = \sec^3A = (1 + 1 - e^2)\frac{3}{2} = (2 - e^2)\frac{3}{2}

  69. Cross multiplying given equations, we have

    sec⁡Abr−qc=tan⁡Apc−qr=1qa−pb\frac{\sec A}{br - qc} = \frac{\tan A}{pc - qr} = \frac{1}{qa - pb}

    ∴sec⁡2A−tan⁡2A=1,(br−qcaq−pb)2−(pc−qaaq−pb)2=1\therefore \sec^2A - \tan^2A = 1, \left(\frac{br - qc}{aq - pb}\right)^2 - \left(\frac{pc - qa}{aq - pb}\right)^2 = 1

    (br−qa)2−(pc−ar)2=(qa−pb)2(br - qa)^2 - (pc - ar)^2 = (qa - pb)^2

  70. Given, cosec⁡A−sin⁡A=m⇒1sin⁡A−sin⁡A=m\cosec A - \sin A = m \Rightarrow \frac{1}{\sin A } - \sin A = m

    ⇒1−sin⁡2Asin⁡A=m⇒cos⁡2Asin⁡A=m\Rightarrow \frac{1 - \sin^2A}{\sin A} = m \Rightarrow \frac{\cos^2A}{\sin A} = m

    Also given, sec⁡A−cos⁡A=n⇒1cos⁡A−cos⁡A=n\sec A - \cos A = n \Rightarrow \frac{1}{\cos A} - \cos A = n

    ⇒1−cos⁡2Acos⁡A=n⇒sin⁡2Acos⁡A=n\Rightarrow \frac{1 - \cos^2A}{\cos A} = n \Rightarrow \frac{\sin^2A}{\cos A} = n

    We have sin⁡A=cos⁡2Am,\sin A = \frac{\cos^2A}{m}, putting this in derived equation

    cos⁡3A=m2n\cos^3A = m^2n

    ∴sin⁡A=(m2n)23m=(mn2)13\therefore \sin A = \frac{(m^2n)^\frac{2}{3}}{m} = (mn^2)\frac{1}{3}

    Thus, sin⁡2A+cos⁡2A=1\sin^2A + \cos^2A = 1 gives us (m2n)23+(mn2)23=1(m^2n)^\frac{2}{3} + (mn^2)^\frac{2}{3} = 1

  71. Given, sec⁡2A=4xy(x+y)2\sec^2A = \frac{4xy}{(x + y)^2}

    ∵sec⁡2A≥1∴4xy(x+y)2≥1\because \sec^2A \geq 1 \therefore \frac{4xy}{(x + y)^2}\geq 1

    ⇒(x+y)2≤4xy⇒(x−y)2≤0\Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0

    But for real xx and y,(x−y)2≮0y, (x - y)^2\nless 0

    ∴(x−y)2=0∴x−y\therefore (x - y)^2 = 0 \therefore x - y

    Also, x+y≠0⇒x≠0,y≠0x + y \neq 0 \Rightarrow x\neq 0, y \neq 0

  72. Given, sin⁡A=x+1x,\sin A = x + \frac{1}{x}, squaring we get

    sin⁡2A=x2+1x2+2≥2\sin^2A = x^2 + \frac{1}{x^2} + 2 \geq 2

    which is not possible since sin⁡A≤1\sin A \leq 1

  73. Given, sec⁡A−tan⁡A=p\sec A - \tan A = p

    ⇒1−sin⁡A=pcos⁡A\Rightarrow 1 - \sin A = p\cos A

    Squaring, we obtain

    1+sin⁡2A−2sin⁡A=p2cos⁡2A=p2(1−sin⁡2A)1 + \sin^2A - 2\sin A = p^2\cos^2A = p^2(1 - \sin^2A)

    (1+p)sin⁡2A−2sin⁡A+1−p2=0(1 + p^)\sin^2A - 2\sin A + 1 - p^2 = 0

    sin⁡A=1±p21+p2\sin A = \frac{1 \pm p^2}{1 + p^2}

    Now tantan and sec⁡A\sec A can be easily found.

  74. sec⁡A+tan⁡A=4p2+14p±sec⁡2A−1\sec A + \tan A = \frac{4p^2 + 1}{4p} \pm \sqrt{\sec^2A - 1}

    =4p±((4p2+1)216p2−1)= 4p \pm \sqrt{\left(\frac{(4p^2 + 1)^2}{16p^2} - 1\right)}

    =2p= 2p or 12p\frac{1}{2p}

  75. Given, sin⁡Asin⁡B=p,cos⁡Acos⁡B=q\frac{\sin A}{\sin B} = p, \frac{\cos A}{\cos B} = q

    Squaring, we get

    sin⁡2Asin⁡2B=p2,cos⁡2Acos⁡2B=q2\frac{\sin^2A}{\sin^2B} = p^2, \frac{\cos^2A}{\cos^2B}= q^2

    sin⁡2A−sin⁡2Bsin⁡2B=p2−1,cos⁡2B−cos⁡2Acos⁡2B=1−q2\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\cos^2B - \cos^2A}{\cos^2B} = 1 - q^2

    sin⁡2A−sin⁡2Bsin⁡2B=p2−1,sin⁡2A−sin⁡2Bcos⁡2B=1−q2\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\sin^2A - \sin^2B}{\cos^2B} = 1 - q^2

    Dividing, we obtain

    tan⁡2B=±1−q2p2−1\tan^2B = \pm \sqrt{\frac{1 - q^2}{p^2 - 1}}

    Dividing original equations

    tan⁡Atan⁡B=±pq1−q2p2−1\frac{\tan A}{\tan B} = \pm \frac{p}{q}\sqrt{\frac{1 - q^2}{p^2 - 1}}

  76. sin⁡A=2sin⁡B,sin⁡A.cos⁡Bcos⁡A.sin⁡B=3\sin A = \sqrt{2}\sin B, \frac{\sin A.\cos B}{\cos A. \sin B} = \sqrt{3}

    Substituting for sin⁡A\sin A

    2sin⁡Bcos⁡B1−2sin⁡2Asin⁡B=3\frac{\sqrt{2}\sin B\cos B}{\sqrt{1 - 2\sin^2A}\sin B} = \sqrt{3}

    Squaring, we get

    2cos⁡2A=3(1−2sin⁡2A)⇒4sin⁡2A−1=02\cos^2 A = 3(1 - 2\sin^2A) \Rightarrow 4\sin^2A - 1 = 0

    ⇒sin⁡A=±12,⇒A=±45∘\Rightarrow \sin A = \pm \frac{1}{2}, \Rightarrow A = \pm 45^\circ

    Thus, B=±30∘B = \pm 30^\circ

  77. Given, tan⁡A+cot⁡A=2,1+tan⁡2A=2tan⁡A⇒(1−tan⁡A)2=0\tan A + \cot A = 2, 1 + \tan^2A = 2\tan A \Rightarrow (1 - \tan A)^2 = 0

    tan⁡A=1⇒sin⁡A=±12\tan A = 1\Rightarrow \sin A = \pm\frac{1}{\sqrt{2}}

  78. m2−n2=(tan⁡A+sin⁡A)2−(tan⁡A−sin⁡A)2=4tan⁡Asin⁡Am^2 - n^2 = (\tan A +\sin A)^2 - (\tan A - \sin A)^2 = 4\tan A\sin A

    4mn=4tan⁡2A−sin⁡2A=4sin⁡2Acos⁡2A−sin⁡2A4\sqrt{mn} = 4\sqrt{\tan^2A - \sin^2A} = 4\sqrt{\frac{\sin^2A}{\cos^2A} - \sin^2A}

    =4sin⁡2A(1−cos⁡2A)cos⁡2A=4sin⁡4Acos⁡2A= 4\sqrt{\frac{\sin^2A(1 - \cos^2A)}{\cos^2A}} = 4\sqrt{\frac{\sin^4A}{\cos^2A}}

    =4tan⁡Asin⁡A= 4\tan A\sin A

  79. m2−1=2sin⁡Acos⁡A⇒n(m2−1)=2sin⁡Acos⁡A(sec⁡A+cosec⁡A)m^2 - 1 = 2\sin A\cos A \Rightarrow n(m^2 - 1) = 2\sin A\cos A(\sec A + \cosec A)

    =2sin⁡A+2cos⁡A=2m= 2\sin A + 2\cos A = 2m

  80. Given, xsin⁡3A+ycos⁡3A=sin⁡Acos⁡Ax\sin^3 A + y\cos^3 A = \sin A\cos A

    (xsin⁡A)sin⁡2A+(ycos⁡A)cos⁡2A=sin⁡Acos⁡A(x\sin A)\sin^2A + (y\cos A)\cos^2A = \sin A\cos A

    Frpm other equation xsin⁡A=ycos⁡A,x\sin A = y\cos A, substituting this in above equation

    (xsin⁡A)sin⁡2A+(xsin⁡A)cos⁡2a=sin⁡Acos⁡A(x\sin A)\sin^2A + (x\sin A)\cos^2a = \sin A\cos A

    xsin⁡A=sin⁡Acos⁡Ax\sin A = \sin A\cos A

    x=cos⁡A∴y=sin⁡Ax = \cos A \therefore y = \sin A

    Thus, x2+y2=1x^2 + y^2 = 1

  81. Given, sin⁡2A=(x+y)24xy\sin^2A = \frac{(x + y)^2}{4xy}

    sin⁡2A≤1⇒(x+y)2≤4xy⇒(x−y)2≤0\sin^2A \leq 1 \Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0

    But for real xx and yy (x−y)2≮0(x - y)^2\nless 0

    ⇒x=y\Rightarrow x = y

    Also, xy≠0,x,y≠0xy \neq 0, x, y\neq 0