Find the value of tan (sin^{- 1} 3/5 + cot^{- 1} 3/2)

Solution:

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios. 

Here the basic trigonometric function of Sin θ = y, can be changed to θ = sin^-1 y

Let sin^{- 1} 3/5 = x

⇒ sin x = 3/5

Then,

⇒ cos x = √ 1 - sin² x

= 4/5

⇒ sec x = 5/4

Therefore,

tan x = √ sec² x - 1

= √ ( 25/16) - 1

= 3 / 4

x = tan^{- 1} 3/4

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

Now,

cot^{- 1} 3 / 2 = tan^{- 1} 2 / 3 ....(2)

Thus,

by using equation (1) and (2)

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

= tan [tan^{- 1} (3/4 + 2/3) / (1 - (3/4).(2/3)]

= tan (tan^{- 1} 17 / 6)

= 17 / 6

Hence, tan (sin^{- 1} 3/5 + cot^{- 1} 3/2)

= 17 / 6

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 18

Find the value of tan (sin^{- 1} 3/5 + cot^{- 1} 3/2)

Summary:

The value of tan (sin^{- 1} 3/5 + cot^{- 1} 3/2) is 17 / 6. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios