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Find the values of  `tan(sin^(-1)  3/5 + cot^(-1)  3/2)` - CBSE (Science) Class 12 - Mathematics

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Question

Find the values of  `tan(sin^(-1)  3/5 + cot^(-1)  3/2)`

Solution

Let `sin^(-1)  3/5 = x. " Then " sin x = 3/5 => cos x = sqrt(1 - sin^2x) = 4/5 => sec x = 5/4`

`:. tan x =  sqrt(sec^2 x - 1) = sqrt(25/16 - 1) = 3/4`

`:. x = tan^(-1) 3/4`

`:. sin^(-1)  3/5 =  tan^(-1) 3/4`  ...(i)

Now `cot^(-1)  3/2 = tan^(-1)  2/3`  ...(ii)        `[tan^(-1)  1/x = cot^(-1)  x]`

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

`= tan (tan^(-1)  3/4 + tan^(-1)  2/3)`    [Using i and ii]

`= tan(tan^(-1)  (3/4 + 2/3)/(1 - 3/4 . 2/3))`      `[tan^(-1) x + tan^(-1) y  = tan^(-1) (x+ y)/1 - xy)]`

`= tan(tan^(-1)  (9 + 8)/(12 - 6))`

`= tan (tan^(-1)  17/6) = 17/6`

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APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 12 (2018 to Current)
Chapter 2: Inverse Trigonometric Functions
Q: 18 | Page no. 48
Solution Find the values of  `tan(sin^(-1)  3/5 + cot^(-1)  3/2)` Concept: Properties of Inverse Trigonometric Functions.
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