Question #e1f68

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Question #e1f68

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Answer 1 · 2016-12-26T09:43:20

$sin ({tan}^{- 1} u + {sin}^{- 1} v)$ $sin(tan^-1u+sin^-1v)$

$= sin ({cot}^{- 1} (\frac{1}{u}) + {sin}^{- 1} v)$ $=sin(cot^-1(1/u)+sin^-1v)$

$= sin ({csc}^{- 1} \sqrt{1 + \frac{1}{u^{2}}} + {sin}^{- 1} v)$ $=sin(csc^-1sqrt(1+1/u^2)+sin^-1v)$

$= sin ({csc}^{- 1} (\frac{\sqrt{u^{2} + 1}}{u}) + {sin}^{- 1} v)$ $=sin(csc^-1((sqrt(u^2+1))/u)+sin^-1v)$

$= sin ({sin}^{- 1} (\frac{u}{\sqrt{u^{2} + 1}}) + {sin}^{- 1} v)$ $=sin(sin^-1(u/(sqrt(u^2+1)))+sin^-1v)$

$= {sin sin}^{- 1} ((\frac{u}{\sqrt{u^{2} + 1}}) \times \sqrt{1 - v^{2}} + \sqrt{1 - \frac{u^{2}}{u^{2} + 1}} \times v)$ $=sinsin^-1((u/(sqrt(u^2+1)))xxsqrt(1-v^2)+sqrt(1-u^2/(u^2+1))xxv)$

$= (\frac{u \sqrt{1 - v^{2}}}{\sqrt{u^{2} + 1}} + \frac{v}{\sqrt{u^{2} + 1}})$ $=((usqrt(1-v^2))/(sqrt(u^2+1))+v/sqrt(u^2+1))$

$= \frac{u \sqrt{1 - v^{2}} + v}{\sqrt{u^{2} + 1}}$ $=(usqrt(1-v^2)+v)/sqrt(u^2+1)$

Answer 2 · 2016-12-26T10:54:02

$= \frac{u \sqrt{1 - v^{2}} + v}{\sqrt{1 + u^{2}}}$ $=(u sqrt(1-v^2)+v)/sqrt(1+u^2)$

Explanation:

Let $a = tan^(-1)u in Q_1, if u >= 0, and in Q_2 if u <= 0$ .

$sin a = u/sqrt(1+u^2) and cos a= 1/sqrt(1+u^2)$

Let $b = sin^(-1)v in Q_1, if v >= 0, and in Q_2 if v <= 0$ .

$sin b = v and cos v = sqrt(1-v^2)$

In any case, cos a and cos b > 0. sin a < 0, if u < 0.

The given expression is

$sin ( a + b ) = sin a cos b + cos a sin b$

$=(u/sqrt(1+u^2))(sqrt(1-v^2))+(1/sqrt(1+u^2))(v)$

$=(u sqrt(1-v^2)+v)/sqrt(1+u^2)$

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