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# Prove that $sin^{-1} \frac{8}{17} +sin^{-1} \frac{3}{5} =tan^{-1} \frac{77}{36}$

Toolbox:
• $sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}$
• $tan^{-1}x+tan^{-1}y = tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg)$ xy<1
Given $sin^{-1} \large\frac{8}{17} +sin^{-1} \large\frac{3}{5} =tan^{-1} \large\frac{77}{36}$
We know that $sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$

By taking $x=$$\large\frac{8}{17} \rightarrow \:\large \frac{x}{\sqrt{1-x^2}}=\large\frac{\large\frac{8}{17}}{\sqrt{1-\large\frac{64}{289}}}=\large\frac{8}{15}$
$\Rightarrow\:sin^{-1}\large\frac{8}{17}=tan^{-1}\large\frac{8}{15}$

Similarly by taking $x=$$\large\frac{3}{5},\large \frac{x}{\sqrt{1-x^2}}=\large\frac{\large\frac{3}{5}}{\sqrt{1-\large\frac{9}{25}}}=\large\frac{3}{4}$
$\Rightarrow sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}$
$\Rightarrow sin^{-1} \large\frac{8}{17} +sin^{-1} \large\frac{3}{5}$ $= tan^{-1}\large\frac{8}{15}+tan^{-1}\large\frac{3}{4}$

We know that $tan^{-1}x+tan^{-1}y = tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg)$
Given $tan^{-1}\large\frac{8}{15}+tan^{-1}\large\frac{3}{4},$ let us take $x=\large\frac{8}{17}\:and\:y=\large\frac{3}{4}$
$x+y =\large\frac{8}{15}+\large\frac{3}{4} = \large\frac{(8 \times 4) +( 3 \times 15)}{15 \times 4} = \large\frac{77}{60}$
$1 - xy = 1 - (\large\frac{8}{15}\times \frac{3}{4} )= 1 - \large\frac{24}{60} = \large\frac{36}{60}$
$\Rightarrow \large \frac{x+y}{1-xy} = \Large \frac{ \large\frac{77}{60}}{\large\frac{36}{60}}$$= \large\frac{77}{36}$

Therefore, $tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg)$ $= tan^{-1} \large\frac{77}{36}$

edited Mar 15, 2013