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Find the value of the given expression. tan(sin^(–1) 3/5 + cot^(–1) 3/2)

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Question

Find the value of the given expression.

`tan(sin^(-1)  3/5 + cot^(-1)  3/2)`

Sum
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Solution

`tan(sin^(-1)  3/5 + cot^(-1)  3/2)`

⇒ `sin^-1  3/5 = x`

⇒ `sin x = 3/5`

⇒ `cos x = sqrt(1 - sin^2 x) = 4/5`

⇒ `sec x = 5/4`

⇒ `tan x = sqrt(sec^2 - 1)`

⇒ `sqrt (25/16 - 1)`

⇒ `3/4`

⇒ `tan^-1  3/4`

⇒ `sin^-1  3/5 = tan^-1  3/4`  ...(1)

⇒ `cot^-1  3/2 = tan^-1  2/3`  ...(2)

⇒ `tan(sin^-1  3/5 + cot^-1  3/2)`

⇒ `tan (tan^-1 (((3/4 + 2/3))/(1 - 3/4 xx 2/3)))`

⇒ `tan (tan^-1 (((9 + 8)/12)/(1 - 1/2)))`

⇒ `tan(tan^-1 (17/6))`

= `17/6`

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Chapter 2: Inverse Trigonometric Functions - EXERCISE 2.2 [Page 30]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 2 Inverse Trigonometric Functions
EXERCISE 2.2 | Q 12. | Page 30

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