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

ConceptProperties of Inverse Trigonometric Functions

#### 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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