Prove that cos^{- 1} 12/13 + sin^{- 1} 3/5 = sin^{- 1} 56/65

Solution:

Let cos^{- 1} 12/13 = y

⇒ cos y = 12/13

Then,

sin y = √1 - (12/13)²

= 5/13

Therefore,

tan y = 5/12

y = tan^{- 1} 5/12

cos^{- 1} 12/13 = tan^{- 1} 5/12 ....(1)

Now, let sin^{- 1} 3/5 = x

⇒ sin x = 3/5

Then,

cos x = √1 - (3/5)²

= 4/5

Therefore,

tan x = 3/4

x = tan^{- 1} 3/4

sin^{- 1} 3/5 = tan^{- 1} 3/4 ....(2)

Now, let sin^{- 1} 56/65 = z

⇒ sin z = 56/65

Then,

cos z = √1 - (56/65)²

= 33/65

Therefore,

tan z = 56/33

z = tan^{- 1} 56/33

sin^{- 1} 56/65 = tan^{- 1} 56/33 ....(3)

Thus, by using (1) and (2)

LHS = cos^{- 1} 12/13 + sin^{- 1} 3/5

= tan^{- 1}5/12 + tan^{- 1} 3/4

= tan^{- 1} [(5/12.3/4)/(1 - 5/12.3/4)]

= tan^{- 1} [((20 + 36)/48)/((48 - 15)/48)]

= tan^{- 1} (56/33)

sin^{- 1} 56/65 [using (3)]

= RHS

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 6

Summary:

Hence we have proved by using inverse trigonometric functions that cos^{- 1} 12/13 + sin^{- 1} 3/5 = sin^{- 1} 56/65

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Prove that cos- 1 12/13 + sin- 1 3/5 = sin- 1 56/65