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Prove `Sin^(-1) 8/17 + Sin^(-1) 3/5 = Tan^(-1) 77/36` - Mathematics

Prove `sin^(-1)  8/17 + sin^(-1)  3/5 = tan^(-1)  77/36`

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Solution

Let sin^(-1) 8/17 =  x. Then `sinx = 8/17 => cos x =  sqrt(1 - (8/17)^2) = sqrt(225/289) = 15/17`

`:. tanx = 8/15 => x = tan^(-1)  8/15`

`:. sin^(-1)  8/17 =  tan^(-1)  8/15`  ....(1)

Now let `sin^(-1)  3/5 = y. " Then " sin y = 3/5 => cos y = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5`

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

`:. sin^(-1)  3/5 = tan^(-1)  3/4`    .... 2

Now, we have:

L.H.S = `sin^(-1)  8/17 + sin^(-1)  3/5`

`= tan^(-1)  8/15 + tan^(-1)  3/4 `   [Using 1 and 2]

`= tan^(-1)  (8/15 +  3/4)/(1 - 8/15 xx 3/4)`

`= tan^(-1) ((32+45)/(60 - 24))  ` `"                            "[tan^(-1) x + tan^(-1) y = tan^(-1)  (x+y)/(1-xy)] `

`= tan^(-1)  77/36` = R.H.S

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APPEARS IN

NCERT Class 12 Maths
Chapter 2 Inverse Trigonometric Functions
Q 4 | Page 51
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