4. Trigonometrical Ratios#

From Geometry, we know that an acute angle is an angle whose measure is between 0∘0^\circ and 90∘90^\circ. Consider the following figure:

trignometrical ratios

This picture contains two similar triangles △OMP\triangle OMP and △OM′P′.\triangle OM'P'. We are interested in ∠MOP\angle MOP or ∠M′OP′\angle M'OP'. In the △MOP\triangle MOP and △M′OP′,OP,OP′\triangle M'OP', OP, OP' are called hypotenuses i.e sides opposite to right angle, PM,P′M′PM, P'M' are called perpendiculars i.e. sides opposite to angle of interest and OM,OM′OM, OM' are called bases i.e the third angle.

Hypotenuses are usually denoted by h,h, perpendiculars by pp and bases by b.b. Let OM=b,OM′=b′,PM=b,P′M′=b′,OP=h,OP′=h′.OM = b, OM' = b', PM = b, P'M'=b', OP= h, OP'=h'. Since the two triangles are similar pp′=bb′=hh′\frac{p}{p'} = \frac{b}{b'} = \frac{h}{h'} Thus the ratio of any two sides is dependent purely on ∠O\angle O or ∠MOP\angle MOP or ∠M′OP′\angle M'OP'

Since there are 33 sides we can choose 22 in 3C2{}^3C_2 i.e 33 ways and for each combination a there will be two permutations where a side can be in either numerator or denominator. From this we can conclude that there will be six ratios(these are called trigonometrical ratios). These six trigonometrical ratios or functions are given below:

MPOP\frac{MP}{OP} or perp.hyp.\frac{\text{perp.}}{\text{hyp.}} or ph\frac{p}{h} is called the Sine of the ∠MOP\angle MOP

OMOP\frac{OM}{OP} or basehyp.\frac{\text{base}}{\text{hyp.}} or bh\frac{b}{h} is called the Cosine of the ∠MOP\angle MOP

MPOM\frac{MP}{OM} or perp.base\frac{\text{perp.}}{\text{base}} or pb\frac{p}{b} is called the Tangent of the ∠MOP\angle MOP

OMMP\frac{OM}{MP} or baseperp.\frac{\text{base}}{\text{perp.}} or bp\frac{b}{p} is called the Cotangent of the ∠MOP\angle MOP

OPOM\frac{OP}{OM} or hyp.base\frac{\text{hyp.}}{\text{base}} or hb\frac{h}{b} is called the Secant of the ∠MOP\angle MOP

OPMP\frac{OP}{MP} or xhyp.perp.\frac{x\text{hyp.}}{\text{perp.}} or hp\frac{h}{p} is called the Cosecant of the ∠MOP\angle MOP

1−cos⁡MOP1 - \cos MOP is called Versed Sine of ∠MOP\angle MOP and 1−sin⁡MOP1 - \sin MOP is called Coversed sine of ∠MOP\angle MOP. These two are rarely used in trigonometry.

It should be noted that the trigonometrical ratios are all numbers.

The name of all trigonometric ration are wriiten for, for brevity, sin⁡MOP,cos⁡MOP,tan⁡MOP,cot⁡MOP,sec⁡MOP,cosec⁡MOP,vers MOP,coverse MOP.\sin MOP, \cos MOP, \tan MOP, \cot MOP, \sec MOP, \cosec MOP, \text{vers~} MOP, \text{coverse~} MOP.

4.1. Relationship betweeen Trigonometrical Functions or Ratios#

Let us represent the ∠MOP\angle MOP with θ,\theta, we observe from previous section that

sin⁡θ=1cosec⁡θ,cos⁡θ=1sec⁡θ,tan⁡θ=1cot⁡θ\sin \theta = \frac{1}{\cosec\theta}, \cos\theta = \frac{1}{\sec\theta}, \tan\theta = \frac{1}{\cot\theta}

Also that cosec⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=1tan⁡θ\cosec\theta = \frac{1}{\sin\theta}, \sec\theta = \frac{1}{\cos\theta}, \cot\theta = \frac{1}{\tan\theta}

We also observe that tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}

Form Pythagoras theorem in Geometry, we know that hypotenuse2=perpendicular2+base2\text{hypotenuse}^2 = \text{perpendicular}^2 + \text{base}^2 or h2=p2+b2h^2 = p^2 + b^2

  1. Dividing both sides by h2,h^2, we get

    p2h2+b2h2=1\frac{p^2}{h^2} + \frac{b^2}{h^2} = 1

    sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2\theta = 1

    We can rewrite this as sin⁡2θ=1−cos⁡2θ,cos⁡2θ=1−sin⁡2θ,sin⁡θ=1−cos⁡2θ,cos⁡θ=1−sin⁡2θ\sin^2\theta = 1 - \cos^2\theta, \cos^2\theta = 1 - \sin^2\theta, \sin\theta = \sqrt{1 - \cos^2\theta}, \cos\theta = \sqrt{1 - \sin^2\theta}

  2. Divising both sides by b2,b^2, we get

    h2b2=p2b2+1\frac{h^2}{b^2} = \frac{p^2}{b^2} + 1

    sec⁡2θ=tan⁡2θ+1\sec^2\theta = \tan^2\theta + 1

    We can rewrite this as sec⁡2θ−tan⁡2θ=1,tan⁡2θ=sec⁡2θ−1,sec⁡θ=1+tan⁡2θ,tan⁡θ=sec⁡2θ−1\sec^2\theta - \tan^2\theta = 1, \tan^2\theta = \sec^2\theta - 1, \sec\theta = \sqrt{1 + \tan^2\theta}, \tan\theta = \sqrt{\sec^2\theta - 1}

  3. Divising both sides by p2,p^2, we get

    h2p2=1+b2p2\frac{h^2}{p^2} = 1 + \frac{b^2}{p^2}

    cosec⁡2θ=1+cot⁡2θ\cosec^2\theta = 1 + \cot^2\theta

    We can rewrite this as cosec⁡2θ−cot⁡2θ=1,cot⁡2θ=cosec⁡2θ−1,cosec⁡θ=1+cot⁡2θ,cot⁡θ=cosec⁡2θ−1\cosec^2\theta - \cot^2\theta = 1, \cot^2\theta = \cosec^2\theta - 1, \cosec\theta = \sqrt{1 + \cot^2\theta}, \cot\theta = \sqrt{\cosec^2\theta - 1}

4.2. Problems#

Prove the following:

  1. 1−cos⁡A1+cos⁡A=cosec⁡A−cot⁡A\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \cosec A - \cot A

  2. sec⁡2A+cosec⁡2A=tan⁡A+cot⁡A\sqrt{\sec^2A + \cosec^2A} = \tan A + \cot A

  3. (cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)(tan⁡A+cot⁡A)=1(\cosec A - \sin A)(\sec A - \cos A)(\tan A + \cot A) = 1

  4. cos⁡4A−sin⁡4A+1=2cos⁡2A\cos^4 A - \sin^4 A + 1 = 2\cos^2 A

  5. (sin⁡A+cos⁡A)(1−sin⁡Acos⁡A)=sin⁡3A+cos⁡3A(\sin A + \cos A)(1 - \sin A\cos A) = \sin^3A + \cos^3A

  6. sin⁡A1+cos⁡A+1+cos⁡Asin⁡A=2cosec⁡A\frac{\sin A}{1 + \cos A}+\frac{1 + \cos A}{\sin A} = 2\cosec A

  7. sin⁡6A−cos6A=1−3cos⁡2Asin⁡2A\sin^6A - cos^6A = 1 - 3\cos^2A\sin^2A

  8. 1−sin⁡A1+sin⁡A=sec⁡A−tan⁡A\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \sec A - \tan A

  9. cosec⁡Acosec⁡A−1+cosec⁡Acosec⁡A+1=2sec⁡2A\frac{\cosec A}{\cosec A - 1} + \frac{\cosec A}{\cosec A + 1} = 2\sec^2 A

  10. cosec⁡Atan⁡A+cot⁡A=cos⁡A\frac{\cosec A}{\tan A + \cot A} = \cos A

  11. (sec⁡A+cos⁡A)(sec⁡A−cos⁡A)=tan⁡2A+sin⁡2A(\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2A

  12. 1tan⁡A+cot⁡A=sin⁡Acos⁡A\frac{1}{\tan A + \cot A} = \sin A\cos A

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

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

  15. sec⁡A−tan⁡Asec⁡A+tan⁡A=1−2sec⁡Atan⁡A+2tan⁡2A\frac{\sec A - \tan A}{\sec A + \tan A} = 1 - 2\sec A\tan A + 2\tan^2 A

  16. 1sec⁡A−tan⁡A=sec⁡A+tan⁡A\frac{1}{\sec A - \tan A} = \sec A + \tan A

  17. tan⁡A1−cot⁡A+cot⁡A1−tan⁡A=sec⁡Acosec⁡A+1\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \sec A\cosec A+ 1

  18. cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=sin⁡A+cos⁡A\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A

  19. (sin⁡A+cos⁡A)(tan⁡A+cot⁡A)=sec⁡A+cosec⁡A(\sin A + \cos A)(\tan A + \cot A) = \sec A + \cosec A

  20. sec⁡4A−sec⁡2A=tan⁡4A+tan⁡2A\sec^4A - \sec^2A = \tan^4A + \tan^2A

  21. cot⁡4A+cot⁡2A=cosec⁡4A−cosec⁡2A\cot^4A + \cot^2A = \cosec^4A - \cosec^2A

  22. cosec⁡2A−1=cos⁡Acosec⁡A\sqrt{\cosec^2A - 1} = \cos A\cosec A

  23. sec⁡2Acosec⁡2A=tan⁡2A+cot⁡2A+2\sec^2A\cosec^2A = \tan^2A + \cot^2A + 2

  24. tan⁡2A−sin⁡2A=sin⁡4Asec⁡2A\tan^2A - \sin^2A = \sin^4A \sec^2A

  25. (1+cot⁡A−cosec⁡A)(1+tan⁡A+sec⁡A)=2(1 + \cot A - \cosec A)(1 + \tan A + \sec A) = 2

  26. cot⁡Acos⁡Acot⁡A+cos⁡A=cot⁡A−cos⁡Acot⁡Acos⁡A\frac{\cot A\cos A}{\cot A + \cos A} = \frac{\cot A - \cos A}{\cot A \cos A}

  27. cot⁡A+tan⁡Bcot⁡B+tan⁡A=cot⁡Atan⁡B\frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B

  28. (1sec⁡2A−cos⁡2A+1cosec⁡2A−sin⁡2A)cos⁡2Asin⁡2A=1−cos⁡2Asin⁡2A2+cos⁡2Asin⁡2A\left(\frac{1}{\sec^2 A - \cos^2A} + \frac{1}{\cosec^2A - \sin^2A}\right)\cos^2A\sin^2A = \frac{1 - \cos^2A\sin^2A}{2 + \cos^2A\sin^2A}

  29. sin⁡8A−cos⁡8A=(sin⁡2A−cos⁡2A)(1−2sin⁡2Acos⁡2A)\sin^8A - \cos^8A = (\sin^2A - \cos^2A)(1 - 2\sin^2A\cos^2A)

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

  31. 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}

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

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

  34. 2sec⁡2A−sec⁡4A−2cosec⁡2A+cosec⁡4A=cot⁡4A−tan⁡4A2\sec^2 A - \sec^4A - 2\cosec^2A + \cosec^4A = \cot^4A - \tan^4A

  35. (sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=tan⁡2A+cot⁡2A+7(\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = \tan^2A + \cot^2A + 7

  36. (cosec⁡A+cot⁡A)(1−sin⁡A)−(sec⁡A+tan⁡A)(1−cos⁡A)=(cosec⁡A−sec⁡A)[2−(1−cos⁡A)(1−sin⁡A)](\cosec A + \cot A)(1 - \sin A) - (\sec A + \tan A)(1 - \cos A) = (\cosec A - \sec A)[2 - (1 - \cos A)(1 - \sin A)]

  37. (1+cot⁡A+tan⁡A)(sin⁡A−cos⁡A)=sec⁡Acosec⁡2A−cosec⁡Asec⁡2A(1 + \cot A + \tan A)(\sin A - \cos A) = \frac{\sec A}{\cosec^2A} - \frac{\cosec A}{\sec^2A}

  38. 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}

  39. 3(sin⁡A−cos⁡A)4+4(sin⁡6A+cos⁡6A)+6(sin⁡A+cos⁡A)2=133(\sin A - \cos A)^4 + 4(\sin^6 A + \cos^6 A) + 6(\sin A + \cos A)^2 = 13

  40. 1+cos⁡A1−cos⁡A=cosec⁡A+cot⁡A\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \cosec A + \cot A

  41. cos⁡A1+sin⁡A+cos⁡A1−sin⁡A=2sec⁡A\frac{\cos A}{1 + \sin A} + \frac{\cos A}{1 - \sin A} = 2\sec A

  42. tan⁡Asec⁡A−1+tan⁡Asec⁡A+1=2cosec⁡A\frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1} = 2\cosec A

  43. 11−sin⁡A−11+sin⁡A=2sec⁡Atan⁡A\frac{1}{1 - \sin A} - \frac{1}{1 + \sin A} = 2\sec A\tan A

  44. 1+tan⁡2A1+cot⁡2A=(1−tan⁡A1−cot⁡A)2\frac{1 + \tan^2 A}{1 + \cot^2 A} = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2

  45. 1+2tan⁡2Acos⁡2A=tan⁡4A+sec4A1 + \frac{2\tan^2 A}{\cos^2 A} = \tan^4 A + sec^4 A

  46. (1−sin⁡A−cos⁡A)2=2(1−sin⁡A)(1−cos⁡A)(1 - \sin A - \cos A)^2 = 2(1 - \sin A)(1 - \cos A)

  47. cot⁡A+cosec⁡A−1cot⁡A−cosec⁡A+1=1+cos⁡Asin⁡A\frac{\cot A + \cosec A - 1}{\cot A - \cosec A + 1} = \frac{1 + \cos A}{\sin A}

  48. (sin⁡A+sec⁡A)2+(cos⁡A+cosec⁡A)2=(1+sec⁡Acosec⁡A)2(\sin A + \sec A)^2 + (\cos A + \cosec A)^2 = (1 + \sec A\cosec A)^2

  49. 2sin⁡Atan⁡A(1−tan⁡A)+2sin⁡Asec⁡2A(1+tan⁡A)2=2sin⁡A1+tan⁡A\frac{2\sin A\tan A(1 - \tan A) + 2\sin A\sec^2A}{(1 + \tan A)^2} = \frac{2\sin A}{1 + \tan A}

  50. If 2sin⁡A=2−cos⁡A,2\sin A = 2 - \cos A, find sin⁡A.\sin A.

  51. If 8sin⁡A=4+cos⁡A,8\sin A = 4 + \cos A, find sin⁡A.\sin A.

  52. If tan⁡A+sec⁡A=1.5,\tan A + \sec A = 1.5, find sin⁡A.\sin A.

  53. If cot⁡A+cosec⁡A=5,\cot A + \cosec A = 5, find cos⁡A.\cos A.

  54. If 3sec⁡4A+8=10sec⁡2A,3\sec^4 A + 8 = 10\sec^2A, find the value of tan⁡A.\tan A.

  55. If tan⁡2A+sec⁡A=5,\tan^2A + \sec A = 5, find cos⁡A.\cos A.

  56. If tan⁡A+cot⁡A=2,\tan A + \cot A = 2, find sin⁡A.\sin A.

  57. If sec⁡2A=2+2tan⁡A,\sec^2A = 2 + 2\tan A, find tan⁡A.\tan A.

  58. If tan⁡A=2x(x+1)2x+1,\tan A = \frac{2x(x + 1)}{2x + 1}, find sin⁡A\sin A and cos⁡A.\cos A.

  59. If 3sin⁡A+5cos⁡A=5,3\sin A + 5\cos A = 5, show that 5sin⁡A−3cos⁡A=±35\sin A - 3\cos A = \pm 3

  60. If sec⁡A+tan⁡A=sec⁡A−tan⁡A\sec A + \tan A = \sec A - \tan A prove that each side is ±1\pm 1

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

    1. sin⁡4A+sin⁡4B=2sin⁡2Asin⁡2B\sin^4A + \sin^4B = 2\sin^2A \sin^2B

    2. cos⁡4Bcos⁡2A+sin⁡4Bsin⁡2A=1,\frac{\cos^4 B}{\cos^2 A} + \frac{\sin^4 B}{\sin^2 A} = 1,

  62. If cos⁡A+sin⁡A=2cos⁡A,\cos A + \sin A = \sqrt{2}\cos A, prove that cos⁡A−sin⁡A=±2sin⁡A\cos A - \sin A = \pm \sqrt{2}\sin A

  63. If acos⁡A−bsin⁡A=c,a\cos A - b\sin A = c, prove that asin⁡A+bcos⁡A=a2+b2−c2a\sin A + b\cos A = \sqrt{a^2 + b ^2 - c^2}

  64. If 1−sin⁡A=1+sin⁡A,1 - \sin A = 1 + \sin A, then prove that value of each side is ±cos⁡A\pm \cos A

  65. If sin⁡4A+sin⁡2A=1,\sin^4 A + \sin^2 A = 1, prove that

    1. 1tan⁡4A+1tan⁡2A=1\frac{1}{\tan^4 A} + \frac{1}{\tan^2A} = 1

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

  66. If cos⁡2A−sin⁡2A=tan⁡2B,\cos^2A - \sin^2 A = \tan^2 B, prove that 2cos⁡2B−1=cos⁡2B−sin⁡2B=tan⁡2A2\cos^2B - 1 = \cos^2B - \sin^2B = \tan^2A

  67. If sin⁡A+cosec⁡A=2,\sin A + \cosec A = 2, then prove that sin⁡nA+cosec⁡nA=2\sin^nA + \cosec^nA = 2

  68. If tan⁡2A=1−e2,\tan^2 A = 1 - e^2, prove that sec⁡A+tan⁡3Acosec⁡A=(2−e2)32\sec A + \tan^3A\cosec A = (2 - e^2)^\frac{3}{2}

  69. Eliminate AA between the equations asec⁡A+btan⁡A+c=0a\sec A + b\tan A + c = 0 and psec⁡A+qtan⁡A+r=0p\sec A + q\tan A + r = 0

  70. If cosec⁡A−sin⁡A=m\cosec A - \sin A = m and sec⁡A−cos⁡A=n,\sec A - \cos A = n, elimiate AA

  71. Is the equation sec⁡2A=4xy(x+y)2\sec^2 A = \frac{4xy}{(x + y)^2} possible for real values of xx and yy?

  72. Show that the equation sin⁡A=x+1x\sin A = x + \frac{1}{x} is imossible for real values of x.x.

  73. If sec⁡A−tan⁡A=p,p≠0,\sec A - \tan A = p, p\neq 0, find tan⁡A,sec⁡A\tan A, \sec A and sin⁡A.\sin A.

  74. If sec⁡A=p+14p,\sec A = p + \frac{1}{4p}, show that sec⁡A+tan⁡A=2p\sec A + \tan A = 2p or 12p.\frac{1}{2p}.

  75. If sin⁡Asin⁡B=p,cos⁡Acos⁡B=q,\frac{\sin A}{\sin B} = p, \frac{\cos A}{\cos B} = q, find tan⁡A\tan A and tan⁡B.\tan B.

  76. If sin⁡Asin⁡B=2,tan⁡Atan⁡B=3,\frac{\sin A}{\sin B} = \sqrt{2}, \frac{\tan A}{\tan B}= \sqrt{3}, find AA and B.B.

  77. If tan⁡A+cot⁡A=2,\tan A + \cot A = 2, find sin⁡A.\sin A.

  78. If m=tan⁡A+sin⁡Am = \tan A + \sin A and n=tan⁡A−sin⁡A,n = \tan A - \sin A, prove that m2−n2=4mn.m^2 - n^2 = 4\sqrt{mn}.

  79. If sin⁡A+cos⁡A=m\sin A + \cos A = m and sec⁡A+cosec⁡A=n,\sec A + \cosec A = n, prove that n(m2−1)=2m.n(m^2 - 1) = 2m.

  80. If xsin⁡3A+ycos⁡3A=sin⁡Acos⁡Ax\sin^3 A + y\cos^3 A = \sin A\cos A and xsin⁡A−ycos⁡A=0,x\sin A - y\cos A = 0, prove that x2+y2=1x^2 + y^2 = 1

  81. Prove that sin⁡2A=(x+y)24xy\sin^2A = \frac{(x + y)^2}{4xy} is possible for real values of xx and yy only when x=yx = y and x,y≠0x,y \neq 0