7. Trigonometrical Ratios of Angle and Sign Solutions#

  1. Let us solve these one by one:

    1. cos⁡2A=cos⁡60∘=12\cos 2A = \cos 60^\circ = \frac{1}{2}

      cos⁡2A−sin⁡2A=(32)2−(12)2=34−14=12\cos^2A - \sin^2A = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}

      2cos⁡2−1=2.(32)2−1=2.34−1=122\cos^2 - 1 = 2.\left(\frac{\sqrt{3}}{2}\right)^2 - 1 = 2.\frac{3}{4} - 1 = \frac{1}{2}

    2. sin⁡2A=sin⁡60∘=32\sin 2A = \sin 60^\circ = \frac{\sqrt{3}}{2}

      2sin⁡Acos⁡A=2.12.32=322\sin A\cos A = 2.\frac{1}{2}.\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}

    3. cos⁡3A=cos⁡90∘=0\cos 3A = \cos 90^\circ = 0

      4cos⁡3A−3cos⁡A=4.338−332=04\cos^3A - 3\cos A = 4.\frac{3\sqrt{3}}{8} - 3\frac{\sqrt{3}}{2} = 0

    4. sin⁡3A=sin⁡90∘=1\sin 3A = \sin 90^\circ = 1

      3sin⁡A−4sin⁡3A=312−4.123=32−12=13\sin A - 4\sin^3 A = 3\frac{1}{2} - 4. \frac{1}{2^3} = \frac{3}{2} - \frac{1}{2} = 1

    5. tan⁡2A=tan⁡60∘=3\tan 2A = \tan 60^\circ = \sqrt{3}

      2tan⁡A1−tan⁡2A=2.131−13=23.32=3\frac{2\tan A}{1 - \tan^2A} = \frac{2.\frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2}{\sqrt{3}}.\frac{3}{2} = \sqrt{3}

  2. Let us solve these one by one:

    1. sin⁡2A=sin⁡90∘=1\sin 2A = \sin 90^\circ = 1

      2sin⁡Acos⁡A=2sin⁡45∘cos⁡45∘=2.12.12=12\sin A\cos A = 2\sin 45^\circ\cos 45^\circ = 2.\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}} = 1

    2. cos⁡2A=cos⁡90∘=0\cos 2A = \cos 90^\circ = 0

      1−2sin⁡2A=1−2sin⁡245∘=1−2.1(2)2=1−2.12=01 - 2\sin^2A = 1 - 2\sin^245^\circ = 1 - 2.\frac{1}{(\sqrt{2})^2} = 1 - 2.\frac{1}{2} = 0

    3. tan⁡2A=tan⁡90∘=∞\tan 2A = \tan 90^\circ = \infty

      2tan⁡A1−tan⁡2A=2.11−12=20=∞\frac{2\tan A}{1 - \tan^2A} = \frac{2.1}{1 - 1^2} = \frac{2}{0} = \infty

  3. L.H.S. =sin⁡230∘+sin⁡245∘+sin⁡260∘= \sin^230^\circ + \sin^245^\circ + \sin^260^\circ

    =122+1(2)2+(3)222=14+12+34=32== \frac{1}{2^2} + \frac{1}{(\sqrt{2})^2} + \frac{(\sqrt{3})^2}{2^2} = \frac{1}{4} + \frac{1}{2} + \frac{3}{4} = \frac{3}{2} = R.H.S.

  4. L.H.S. =tan⁡230∘+tan⁡245∘+tan⁡260∘= \tan^230^\circ + \tan^245^\circ + \tan^260^\circ

    =1(3)2+1+(3)2=13+1+3=413== \frac{1}{(\sqrt{3})^2} + 1 + (\sqrt{3})^2 = \frac{1}{3} + 1 + 3 = 4\frac{1}{3} = R.H.S.

  5. L.H.S. =sin⁡30∘cos⁡60∘+sin⁡60∘cos⁡30∘= \sin 30^\circ\cos 60^\circ + \sin 60^\circ\cos 30^\circ

    =1212+3232=14+34=1== \frac{1}{2}\frac{1}{2} + \frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 = R.H.S.

  6. L.H.S. =cos⁡45∘cos⁡60∘−sin⁡45∘sin⁡60∘= \cos 45^\circ\cos 60^\circ - \sin 45^\circ\sin 60^\circ

    =1212−1232=122−322= \frac{1}{\sqrt{2}}\frac{1}{2} - \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}}

    =1−322=−3−122== \frac{1 - \sqrt{3}}{2\sqrt{2}} = - \frac{\sqrt{3} - 1}{2\sqrt{2}} = R.H.S.

  7. L.H.S. =cosec⁡245∘.sec⁡230∘.sin⁡290∘.cos⁡60∘= \cosec^245^\circ.\sec^230^\circ.\sin^290^\circ.\cos 60^\circ

    =(2)2.22(3)2.12.12=2.43.1.12=43== (\sqrt{2})^2.\frac{2^2}{(\sqrt{3})^2}.1^2.\frac{1}{2} = 2.\frac{4}{3}.1.\frac{1}{2} = \frac{4}{3} = R.H.S.

  8. L.H.S. =4cot⁡245∘−sec⁡260∘+sin⁡230∘= 4\cot^245^\circ-\sec^260^\circ + \sin^230^\circ

    =4.12−22+122=4−4+14=14== 4.1^2 - 2^2 + \frac{1}{2^2} = 4 - 4 + \frac{1}{4} = \frac{1}{4} = R.H.S.

  9. L.H.S. =sin⁡420∘cos⁡390∘+cos⁡(−300∘)sin⁡(−330∘)= \sin 420^\circ\cos 390^\circ + \cos(-300^\circ)\sin(-330^\circ)

    =sin⁡(360∘+60∘)cos⁡(360∘+30∘)+cos⁡60∘sin⁡30∘= \sin(360^\circ + 60^\circ)\cos(360^\circ + 30^\circ) + \cos 60^\circ\sin30^\circ

    =sin⁡60cos⁡30∘+cos⁡60∘sin⁡30∘=32.33+12.12=34+14=1== \sin 60\cos30^\circ + \cos 60^\circ\sin30^\circ = \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{3} + \frac{1}{2}.\frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 = R.H.S.

  10. L.H.S. =cos⁡570∘sin⁡510∘−sin⁡330∘cos⁡390∘= \cos 570^\circ\sin 510^\circ -\sin 330^\circ\cos 390^\circ

    =cos⁡(360∘+210∘)sin⁡(360∘+150∘)−sin⁡(360∘−30∘)cos⁡(360∘+30∘)= \cos(360^\circ + 210^\circ)\sin(360^\circ + 150^\circ) - \sin(360^\circ - 30^\circ)\cos(360^\circ + 30^\circ)

    =cos⁡(180∘+30∘)sin⁡(180∘−30∘)+sin⁡30∘cos⁡30∘= \cos(180^\circ + 30^\circ)\sin(180^\circ - 30^\circ) + \sin 30^\circ\cos 30^\circ

    =−cos⁡30∘sin⁡30∘+sin⁡30∘cos⁡30∘= -\cos 30^\circ\sin 30^\circ + \sin 30^\circ\cos 30^\circ

    =0== 0 = R.H.S.

  11. cos⁡π3−sin⁡π3=12−32=1−32\cos \frac{\pi}{3} - \sin \frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2}

    tan⁡π3+cot⁡π3=3.+13=43\tan\frac{\pi}{3} + \cot\frac{\pi}{3} = \sqrt{3}. + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}

  12. cos⁡2π3−sin⁡2π3=cos⁡(π−π3))−sin⁡(π−π3)\cos\frac{2\pi}{3} -\sin\frac{2\pi}{3} = \cos\left(\pi - \frac{\pi}{3})\right) - \sin\left(\pi - \frac{\pi}{3}\right)

    =−cos⁡π3−sin⁡π3=−12−32=−1+32= -\cos \frac{\pi}{3} - \sin\frac{\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} = - \frac{1 + \sqrt{3}}{2}

    tan⁡2π3+cot⁡2π3=tan⁡(π−π3)+cot⁡(π−π3)\tan\frac{2\pi}{3} + \cot\frac{2\pi}{3} = \tan\left(\pi - \frac{\pi}{3}\right) + \cot\left(\pi - \frac{\pi}{3}\right)

    =−tan⁡π3.−cot⁡π3=−3−13=−43= -\tan \frac{\pi}{3}. -\cot \frac{\pi}{3} = -\sqrt{3} - \frac{1}{\sqrt{3}} = -\frac{4}{\sqrt{3}}

  13. cos⁡5π4−sin⁡5π4=cos⁡(π+π4)+sin⁡(π+π4)\cos\frac{5\pi}{4} - \sin\frac{5\pi}{4} = \cos\left(\pi + \frac{\pi}{4}\right) + \sin \left(\pi + \frac{\pi}{4}\right)

    =−cos⁡π4−sin⁡π4=−12−12=−2= -\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}

    tan⁡5π4+cot⁡5π4=tan⁡(π+π4)+cot⁡(π+π4)\tan \frac{5\pi}{4} + \cot \frac{5\pi}{4} = \tan\left(\pi + \frac{\pi}{4}\right) + \cot\left(\pi + \frac{\pi}{4}\right)

    =tan⁡π4+cot⁡π4=1+1=2= \tan \frac{\pi}{4} + \cot \frac{\pi}{4} = 1 + 1 = 2

  14. cos⁡7π4+cos⁡7π4=cos⁡(π+3π4)+sin⁡(π+3π4)\cos\frac{7\pi}{4} + \cos\frac{7\pi}{4} = \cos \left(\pi + \frac{3\pi}{4}\right) + \sin\left(\pi + \frac{3\pi}{4}\right)

    =−cos⁡3π4−sin⁡3π4=−cos⁡(π−π4)−sin⁡(π−π4)= -\cos\frac{3\pi}{4} - \sin\frac{3\pi}{4} = -\cos\left(\pi - \frac{\pi}{4}\right) -\sin \left(\pi - \frac{\pi}{4}\right)

    =cos⁡π4−sin⁡π4=12−12=0= \cos \frac{\pi}{4} -\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0

    tan⁡7π4+cos⁡7π4=tan⁡(π+3π4)+cos⁡(π+3π4)\tan \frac{7\pi}{4} + \cos \frac{7\pi}{4} = \tan\left(\pi + \frac{3\pi}{4}\right) + \cos \left(\pi + \frac{3\pi}{4}\right)

    =tan⁡3π4+cot⁡3π4=tan⁡(π−π4)+cot⁡(π−π4)= \tan \frac{3\pi}{4} + \cot \frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) + \cot \left(\pi - \frac{\pi}{4}\right)

    =−tan⁡π4−tan⁡π4=−2= -\tan \frac{\pi}{4} - \tan \frac{\pi}{4} = -2

  15. cos⁡11π3+sin⁡11π3=cos⁡(2π+π+2π3)+sin⁡(2π+π+2π3)\cos\frac{11\pi}{3} + \sin \frac{11\pi}{3} = \cos\left(2\pi + \pi + \frac{2\pi}{3}\right) + \sin\left(2\pi + \pi + \frac{2\pi}{3}\right)

    =cos⁡(π+2π3)+sin⁡(π+2π3)=−cos⁡2π3−sin⁡2π3= \cos\left(\pi + \frac{2\pi}{3}\right) + \sin \left(\pi + \frac{2\pi}{3}\right) = -\cos\frac{2\pi}{3} - \sin\frac{2\pi}{3}

    =−cos⁡(π−π3)−sin⁡(π−π3)= -\cos\left(\pi - \frac{\pi}{3}\right) - \sin\left(\pi - \frac{\pi}{3}\right)

    =cos⁡π3−sin⁡π3=1−32=\cos \frac{\pi}{3} -\sin \frac{\pi}{3} = \frac{1 - \sqrt{3}}{2}

    tan⁡11π3+cot⁡11π3=tan⁡(2π+π+2π3)+cot⁡(2π+π+2π3)\tan\frac{11\pi}{3} + \cot\frac{11\pi}{3} = \tan\left(2\pi + \pi + \frac{2\pi}{3}\right) + \cot\left(2\pi + \pi + \frac{2\pi}{3}\right)

    =tan⁡(π+2π3)+cot⁡(π+2π3)= \tan\left(\pi + \frac{2\pi}{3}\right) + \cot\left(\pi + \frac{2\pi}{3}\right)

    =tan⁡2π3+cot⁡2π3=tan⁡(π−π3)+cot⁡(π−π3)= \tan\frac{2\pi}{3} + \cot\frac{2\pi}{3} = \tan\left(\pi - \frac{\pi}{3}\right) + \cot(\pi - \frac{\pi}{3})

    =−tan⁡π3−cot⁡π3=−3−13=−43= -\tan\frac{\pi}{3} -\cot\frac{\pi}{3} = -\sqrt{3} - \frac{1}{\sqrt{3}} = - \frac{4}{\sqrt{3}}

  16. sin⁡\sin of an angle is positive in first and second quadrant. Also, sin⁡45∘=12,\sin 45^\circ = \frac{1}{\sqrt{2}}, therefore the angles will be 45∘45^\circ and 135∘.135^\circ.

  17. cos⁡\cos of an angle is positive in second and third quadrant. Also, cos⁡60∘=12,\cos 60^\circ = \frac{1}{2}, therefore the angles will be 120∘120^\circ and 240∘.240^\circ.

  18. tan⁡\tan of an angle is negative in second and fourth quadrant. Also, tan⁡45∘=1,\tan 45^\circ = 1, therefore the angles will be 135∘135^\circ and 315∘.315^\circ.

  19. cot⁡\cot of an angle is negative in second and fourth quadrant. Also, cot⁡30∘=3,\cot 30^\circ = \sqrt{3}, therefore the angles will be 150∘150^\circ and 330∘.330^\circ.

  20. sec⁡\sec of an angle is negative in second and third quadrant. Also, sec⁡30∘=23,\sec 30^\circ = \frac{2}{\sqrt{3}}, therefore the angles will be 150∘150^\circ and 210∘.210^\circ.

  21. cosec⁡\cosec of an angle is negative in third and fourth quadrant. Also, cosec⁡30∘=2,\cosec 30^\circ = 2, therefore the angles will be 210∘210^\circ and 330∘.330^\circ.

  22. sin⁡(−65∘)=−sin⁡65∘=−cos⁡(90∘−65∘)=−cos⁡25∘\sin(-65^\circ) = -\sin 65^\circ = -\cos(90^\circ - 65^\circ) = -\cos 25^\circ

  23. cos⁡(−84∘)=cos⁡84∘=sin⁡(90∘−84∘)=sin⁡6∘\cos(-84^\circ) = \cos 84^\circ = \sin(90^\circ - 84^\circ) = \sin 6^\circ

  24. tan⁡(137∘)=tan⁡(180∘−43∘)=−tan⁡43∘\tan(137^\circ) = \tan(180^\circ - 43^\circ) = -\tan 43^\circ

  25. sin⁡(168∘)=sin⁡(180∘−12∘)=sin⁡12∘\sin(168^\circ) = \sin(180^\circ - 12^\circ) = \sin 12^\circ

  26. cos⁡(287∘)=cos⁡(180+107∘)=−cos⁡107∘=sin⁡17∘\cos(287^\circ) = \cos(180 + 107^\circ) = -\cos 107^\circ = \sin 17^\circ

  27. tan⁡(−246∘)=−tan⁡(246∘)=−tan⁡(180+66∘)=−tan⁡66∘=−tan⁡(90∘−24∘)=−cot⁡24∘\tan(-246^\circ) = -\tan(246^\circ) = -\tan(180 + 66^\circ) = -\tan 66^\circ = -\tan(90^\circ - 24^\circ) = -\cot 24^\circ

  28. sin⁡843∘=sin⁡(2∗360∘+123∘)=sin⁡(123∘)=sin⁡(90∘+33∘)=cos⁡33∘\sin 843^\circ = \sin(2*360^\circ + 123^\circ) = \sin(123^\circ) = \sin(90^\circ + 33^\circ) = \cos 33^\circ

  29. cos⁡(−928∘)=cos⁡(2∗360∘+208∘)=cos⁡(180∘+28∘)=−cos⁡28∘\cos(-928^\circ) = \cos(2*360^\circ + 208^\circ) = \cos(180^\circ + 28^\circ) = -\cos 28^\circ

  30. tan⁡1145∘=tan⁡(3∗360∘+65∘)=tan⁡(65∘)=tan⁡(90∘−25∘)=cot⁡25∘\tan 1145^\circ = \tan(3*360^\circ + 65^\circ) = \tan(65^\circ) = \tan(90^\circ - 25^\circ) = \cot 25^\circ

  31. cos⁡1410∘=cos⁡(360∗3+330∘)=cos⁡(180+180∘−30∘)=cos⁡40∘\cos 1410^\circ = \cos(360*3 + 330^\circ) = \cos(180^ + 180^\circ - 30^\circ) = \cos 40^\circ

  32. cot⁡(−1054∘)=−cot⁡(3∗360−26∘)=cot⁡26∘\cot(-1054^\circ) = -\cot(3*360 - 26^\circ) = \cot 26^\circ

  33. sec⁡1327∘=sec⁡(3∗360∘+247∘)=sec⁡(180∘+67∘)=−sec⁡67∘=−sec⁡(90∘−23∘)=−cosec⁡23∘\sec 1327^\circ = \sec(3*360^\circ + 247^\circ) = \sec(180^\circ + 67^\circ) = -\sec 67^\circ = -\sec(90^\circ - 23^\circ) = -\cosec 23^\circ

  34. cosec⁡(−756∘)=−cosec⁡(2∗360∘+36∘)=−cosec⁡36∘\cosec (-756^\circ) = -\cosec(2*360^\circ + 36^\circ) = -\cosec 36^\circ

  35. sin⁡140∘+cos⁡140∘=sin⁡(90∘+50∘)+cos⁡(180∘−40∘)=cos⁡50∘−cos⁡40∘\sin 140^\circ + \cos 140^\circ = \sin(90^\circ + 50^\circ) + \cos(180^\circ - 40^\circ) = \cos 50^\circ - \cos 40^\circ

    cos⁡40∘>cos⁡50∘\cos 40^\circ > \cos 50^\circ therefore sign will be negative.

  36. sin⁡278∘+cos⁡278∘=sin⁡(180∘+98∘)+cos⁡(180∘+98∘)\sin 278^\circ + \cos 278^\circ = \sin(180^\circ + 98^\circ) + \cos(180^\circ + 98^\circ)

    =−sin⁡(98∘)−cos⁡(98∘)=−cos⁡8∘+cos⁡82∘= -\sin(98^\circ) - \cos(98^\circ) = -\cos 8^\circ + \cos 82^\circ

    cos⁡8∘>cos⁡82∘\cos 8^\circ > \cos 82^\circ therefore sign will be negative.

  37. sin⁡(−356∘)+cos⁡(−356∘)=−sin⁡(180∘+180∘−4∘)+cos⁡(180∘+180∘−4∘)\sin(-356^\circ) + \cos(-356^\circ) = -\sin(180^\circ + 180^\circ - 4^\circ) + \cos(180^\circ + 180^\circ - 4^\circ)

    sin⁡4∘+cos⁡4∘\sin 4^\circ + \cos 4^\circ which will yield a positive sign.

  38. sin⁡(−1125∘)+cos⁡(−1125∘)=−sin⁡(3∗360∘+45∘)+cos⁡(3∗360∘+445∘)\sin(-1125^\circ) + \cos(-1125^\circ) = -\sin(3*360^\circ + 45^\circ) + \cos(3*360^\circ + 445^\circ)

    =−sin⁡45∘+cos⁡45∘=0= -\sin 45^\circ + \cos 45^\circ = 0 which is neither negative nor positive.

  39. sin⁡215∘−cos⁡215∘=sin⁡(180∘+35∘)−cos⁡(180∘+35∘)\sin 215^\circ - \cos 215^\circ = \sin(180^\circ + 35^\circ) - \cos(180^\circ + 35^\circ)

    =−sin⁡35∘+cos⁡35∘= -\sin 35^\circ + \cos 35^\circ

    ∵cos⁡35∘>sin⁡35∘\because \cos 35^\circ > \sin 35^\circ the sign will be positive.

  40. sin⁡825∘−cos⁡825∘=sin⁡(2∗360∘+105∘)−cos⁡(2∗360+105∘)\sin 825^\circ - \cos 825^\circ = \sin(2*360^\circ + 105^\circ) - \cos(2*360 + 105^\circ)

    =cos⁡15∘+sin⁡15∘= \cos 15^\circ + \sin 15^\circ for which sign will be positive.

  41. sin⁡(−634∘)−cos⁡(−634)∘=−sin⁡(360∘+274∘)−cos⁡(360∘+274∘)\sin(-634^\circ) - \cos(-634)^\circ = -\sin (360^\circ + 274^\circ) - \cos (360^\circ + 274^\circ)

    =−sin⁡(180∘+90∘+4∘)−cos⁡(180∘+90∘+4∘)= -\sin(180^\circ + 90^\circ + 4^\circ) - \cos(180^\circ + 90^\circ + 4^\circ)

    =cos⁡4∘−sin⁡4∘= \cos 4^\circ - \sin 4^\circ whic will have positive sign.

  42. sin⁡(−457∘)−cos⁡(−457∘)=−sin⁡(360∘+90∘+7∘)−cos⁡(360∘+90∘+7∘)\sin(-457^\circ) - \cos(-457^\circ) = -\sin(360^\circ + 90^\circ + 7^\circ) -\cos(360^\circ + 90^\circ + 7^\circ)

    =−cos⁡7∘+sin⁡7∘=-\cos 7^\circ + \sin 7^\circ which will have negative sign.

  43. cos⁡135∘=−12\cos 135^\circ = -\frac{1}{\sqrt{2}} then given tan⁡A=−12\tan A = -\frac{1}{\sqrt{2}}

    sin⁡A=±13,cos⁡A=±23\sin A = \pm \frac{1}{\sqrt{3}}, \cos A = \pm \frac{\sqrt{2}}{\sqrt{3}}

  44. sin⁡(270∘+A)=sin⁡(180∘+90∘+A)=−sin⁡(90∘+A)=−cos⁡A\sin(270^\circ + A) = \sin(180^\circ + 90^\circ + A) = -\sin(90^\circ + A) = -\cos A

    tan⁡(270∘+A)=tan⁡(180∘+90∘+A)=tan⁡(90∘+A)=−cot⁡A\tan(270^\circ + A) = \tan(180^\circ + 90^\circ + A) = \tan(90^\circ + A) = -\cot A

  45. cos⁡(270∘−A)=cos⁡(180∘+90∘−A)=−cos⁡(90∘−A)=−sin⁡A\cos(270^\circ - A) = \cos(180^\circ + 90^\circ - A) = -\cos(90^\circ - A) = -\sin A

    cot⁡(270∘−A)=cot⁡(180∘+90∘−A)=cot⁡(90∘−A)=tan⁡A\cot(270^\circ - A) = \cot(180^\circ + 90^\circ - A) = \cot(90^\circ - A) = \tan A

  46. L.H.S. =cos⁡A+sin⁡(270∘+A)−sin⁡(270∘−A)+cos⁡(180∘+A)= \cos A + \sin(270^\circ + A) - \sin(270^\circ - A) + \cos(180^\circ + A)

    Using results from previous problems we get

    =cos⁡A+−cos⁡A+cos⁡A−cos⁡A=0= \cos A + -\cos A + \cos A - \cos A = 0

  47. L.H.S. =sec⁡(270∘−A)sec⁡(90∘−A)−tan⁡(270∘−A)tan⁡(90∘+A)+1= \sec(270^\circ - A)\sec(90^\circ - A) - \tan(270^\circ - A)\tan(90^\circ + A) + 1

    =sec⁡(180∘+90∘−A)cosec⁡A+tan⁡(180∘+90∘−A)cot⁡A+1= \sec(180^\circ + 90^\circ - A)\cosec A + \tan(180^\circ + 90^\circ - A)\cot A + 1

    =−cosec⁡2A+cot⁡2A+1=−1+1=0== -\cosec^2A + \cot^2A + 1 = -1 + 1 = 0 = R.H.S.

  48. L.H.S. =cot⁡A+tan⁡(180∘+A)+tan⁡(90∘+A)+tan⁡(360∘−A)= \cot A + \tan(180^\circ + A) + \tan(90^\circ + A) + \tan(360^\circ - A)

    =cot⁡A+tan⁡A−cot⁡A−tan⁡A=0== \cot A + \tan A - \cot A - \tan A = 0 = R.H.S.

  49. Given, 3tan⁡245∘−sin⁡260∘−12cot⁡230∘+18sec⁡245∘3\tan^245^\circ - \sin^260^\circ - \frac{1}{2}\cot^230^\circ + \frac{1}{8}\sec^245^\circ

    =3.12−(32)2−12(3)2+18(2)2= 3.1^2 - \left(\frac{\sqrt{3}}{2}\right)^2 - \frac{1}{2}(\sqrt{3})^2 + \frac{1}{8}(\sqrt{2})^2

    =3−34−32+28=1= 3 - \frac{3}{4} - \frac{3}{2} + \frac{2}{8} = 1

  50. Given, =sin⁡(2π−60∘).tan⁡(2π−30∘).sec⁡(2π+60∘)tan⁡(π−45∘).sin⁡(π+30∘).sec⁡(2π−45∘)= \frac{\sin(2\pi - 60^\circ).\tan(2\pi - 30^\circ).\sec(2\pi + 60^\circ)}{\tan(\pi - 45^\circ).\sin(\pi + 30^\circ).\sec(2\pi - 45^\circ)}

    =−sin⁡6−∘.−tan⁡30∘.sec⁡60∘−tan⁡45∘.−sin⁡30∘.sec⁡45∘= \frac{-\sin 6-^\circ. -\tan 30^\circ.\sec 60^\circ}{-\tan 45^\circ. -\sin 30^\circ.\sec 45^\circ}

    =32.13.21.12.2=2= \frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{3}}.2}{1.\frac{1}{\sqrt{2}}.\sqrt{2}} = \sqrt{2}

  51. L.H.S. =tan⁡1∘tan⁡2∘…tan⁡89∘= \tan 1^\circ\tan 2^\circ \ldots \tan 89^\circ

    =(tan⁡1∘.tan⁡(90∘−1∘).(tan⁡2∘.tan⁡(90∘−2∘).….(tan⁡44∘.tan⁡(90∘−44∘).tan⁡45∘= (\tan 1^\circ.\tan(90^\circ - 1^\circ).(\tan 2^\circ.\tan(90^\circ - 2^\circ).\ldots.(\tan 44^\circ.\tan(90^\circ - 44^\circ).\tan 45^\circ

    =(tan⁡1∘.cot⁡1∘).(tan⁡2∘.cot⁡2∘).….(tan⁡44∘.cot⁡44∘).1= (\tan 1^\circ.\cot 1^\circ).(\tan 2^\circ.\cot 2^\circ).\ldots.(\tan 44^\circ.\cot 44^\circ).1

    =1.1.…1.1[∵tan⁡θcot⁡θ=1]= 1.1.\ldots 1.1 [\because \tan\theta\cot\theta = 1]

    =1== 1 = R.H.S.

  52. L.H.S. =(sin⁡25∘+sin⁡285∘)+(sin⁡210∘+sin⁡280∘)+…+(sin⁡240∘+sin⁡250∘)+sin⁡245∘+sin⁡290∘=(\sin^25^\circ + \sin^285^\circ) + (\sin^210^\circ + \sin^280^\circ) + \ldots + (\sin^240^\circ + \sin^250^\circ) + \sin^245^\circ + \sin^290^\circ

    =(sin⁡25∘+cos⁡25∘)+(sin⁡210∘+cos⁡210∘)+…+(sin⁡240∘+cos⁡240∘)+sin⁡245∘+sin⁡290∘= (\sin^25^\circ + \cos^25^\circ) + (\sin^210^\circ + \cos^210^\circ) + \ldots + (\sin^240^\circ + \cos^240^\circ) + \sin^245^\circ + \sin^290^\circ

    =1+1+ 8 times+(12)2+1=912== 1 + 1 +~\text{8 times}+ \left(\frac{1}{\sqrt{2}}\right)^2 + 1 = 9\frac{1}{2} = R.H.S.

  53. Given expression can be rewritten as =cos⁡2π16+cos⁡23π16+cos⁡2(π2−3π16)+cos⁡2(π2−π16)= \cos^2 \frac{\pi}{16} + \cos^2 \frac{3\pi}{16} + \cos^2\left(\frac{\pi}{2} - \frac{3\pi}{16}\right) + \cos^2\left(\frac{\pi}{2} - \frac{\pi}{16}\right)

    =cos⁡2π16+cos⁡23π16+sin⁡23π16+sin⁡2π16= \cos^2 \frac{\pi}{16} + \cos^2 \frac{3\pi}{16} + \sin^2\frac{3\pi}{16} + \sin^2\frac{\pi}{16}

    =1+1=2= 1 + 1 = 2

  54. Substituting the values (23)2(2)2+(3)2.12\left(\frac{2}{\sqrt{3}}\right)^2(\sqrt{2})^2 + (\sqrt{3})^2.1^2

    =43.2+3=173= \frac{4}{3}.2 + 3 = \frac{17}{3}

  55. Substituting the values (3)2−2.122−341(2)2−4.122(\sqrt{3})^2 - 2.\frac{1}{2^2} - \frac{3}{4}\frac{1}{(\sqrt{2})^2} - 4.\frac{1}{2^2}

    =98= \frac{9}{8}

  56. Given expression is sec⁡∘(2π+120∘).cosec⁡(2π+210∘).tan⁡(2π−30∘)sin⁡(2π+240∘).cos⁡(2π+300∘).cot⁡(2π+45∘)\frac{\sec\circ(2\pi + 120^\circ).\cosec(2\pi + 210^\circ).\tan(2\pi - 30^\circ)}{\sin(2\pi + 240^\circ).\cos(2\pi + 300^\circ).\cot(2\pi + 45^\circ)}

    =sec⁡(90∘+30∘).cosec⁡(180∘+30∘).−tan⁡30∘sin⁡(180∘+60∘).cos⁡(360∘−60∘).cot⁡45∘= \frac{\sec(90^\circ + 30^\circ).\cosec(180^\circ + 30^\circ).-\tan 30^\circ}{\sin(180^\circ + 60^\circ).\cos(360^\circ - 60^\circ).\cot 45^\circ}

    =−cosec⁡30∘.−cosec⁡30∘.−tan⁡30∘−sin⁡60∘.cos⁡60∘.cot⁡45∘= \frac{-\cosec 30^\circ. -\cosec 30^\circ. -\tan 30^\circ}{-\sin 60^\circ. \cos 60^\circ.\cot 45^\circ}

    =2.2.1332.12.1= \frac{2.2.\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2}.\frac{1}{2}.1}

    =163= \frac{16}{3}

  57. L.H.S. cos⁡630∘+sin⁡630∘=(32)6+126=2764+164=716\cos^630^\circ + \sin^6 30^\circ = \left(\frac{3}{2}\right)^6 + \frac{1}{2}^6 = \frac{27}{64} + \frac{1}{64} = \frac{7}{16}

    R.H.S. =1−3sin⁡230∘cos⁡230∘=1−3.122.322=1−916=716= 1 - 3\sin^230^\circ\cos^230^\circ = 1 - 3.\frac{1}{2^2}.\frac{3}{2^2} = 1 - \frac{9}{16} = \frac{7}{16}

    Thus, L.H.S. = R.H.S.

  58. L.H.S. =(1+1+2)(1+1−2)=4−2=2=cosec⁡2π4= \left(1 + 1 + \sqrt{2}\right)(1 + 1 - \sqrt{2}) = 4 - 2 = 2 = \cosec^2\frac{\pi}{4}

59 and 60 are similar to 52 and 51 respectively and have been left as an exercise.

  1. L.H.S. =sin⁡2π18+sin⁡22π18+…+sin⁡29π18= \sin^2 \frac{\pi}{18} + \sin^2\frac{2\pi}{18} + \ldots + \sin^2\frac{9\pi}{18}

    sin⁡28π18=sin⁡2(π2−π18)=cos⁡2π18\sin^2\frac{8\pi}{18} = \sin^2\left(\frac{\pi}{2} - \frac{\pi}{18}\right) = \cos^2\frac{\pi}{18}

    sin⁡27π18=sin⁡2(π2−2π18)=cos⁡22π18\sin^2\frac{7\pi}{18} = \sin^2\left(\frac{\pi}{2} - \frac{2\pi}{18}\right) = \cos^2\frac{2\pi}{18}

    Following similarly, the original expression can be written as

    (sin⁡2π18+cos⁡2π18)+(sin⁡22π18+cos⁡22π18)+(sin⁡23π18+cos⁡23π18)+(sin⁡24π18+cos⁡24π18)+sin⁡2π2\left(\sin^2 \frac{\pi}{18} + \cos^2\frac{\pi}{18}\right) + \left(\sin^2 \frac{2\pi}{18} + \cos^2\frac{2\pi}{18}\right) + \left(\sin^2 \frac{3\pi}{18} + \cos^2\frac{3\pi}{18}\right) + \left(\sin^2 \frac{4\pi}{18} + \cos^2\frac{4\pi}{18}\right) + \sin^2 \frac{\pi}{2}

    =5== 5 = R.H.S.

  2. 2nα=π22n\alpha = \frac{\pi}{2}

    L.H.S. =tan⁡αtan⁡2αtan⁡3α.….tan⁡(2n−2)αtan⁡(2n−1)α= \tan\alpha\tan2\alpha\tan3\alpha. \ldots .\tan(2n - 2)\alpha\tan(2n - 1)\alpha

    =tan⁡αtan⁡2αtan⁡3α.….cot⁡2α.cot⁡α= \tan\alpha\tan2\alpha\tan3\alpha. \ldots .\cot2\alpha.\cot\alpha

    =1== 1 = R.H.S.