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# What does $sin^{-1} \frac{8}{17} +sin^{-1} \frac{3}{5}$ reduce to?

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A)
Toolbox:
• $$sin^{-1}x=tan^{-1}\frac{x}{\sqrt{1-x^2}}$$
• $$tan^{-1}x+tan^{-1}y = tan^{-1} \bigg( \frac{x+y}{1-xy} \bigg)$$ xy<1
Given $sin^{-1} \frac{8}{17} +sin^{-1} \frac{3}{5}$
We know that $$sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$
By taking $x=$$$\frac{8}{17} \rightarrow \:\large \frac{x}{\sqrt{1-x^2}}=\frac{\frac{8}{17}}{\sqrt{1-\frac{64}{289}}}=\frac{8}{15}$$
$$\Rightarrow\:sin^{-1}\frac{8}{17}=tan^{-1}\frac{8}{15}$$
Similarly by taking $x=$$$\frac{3}{5},\large \frac{x}{\sqrt{1-x^2}}=\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}=\frac{3}{4}$$
$$\Rightarrow sin^{-1}\frac{3}{5}=tan^{-1}\frac{3}{4}$$
$\Rightarrow sin^{-1} \frac{8}{17} +sin^{-1} \frac{3}{5}$ $$= tan^{-1}\frac{8}{15}+tan^{-1}\frac{3}{4}$$
We know that $$tan^{-1}x+tan^{-1}y = tan^{-1} \bigg( \frac{x+y}{1-xy} \bigg)$$
Given $$tan^{-1}\frac{8}{15}+tan^{-1}\frac{3}{4},$$ let us take $x=\frac{8}{17}\:and\:y=\frac{3}{4}$
$x+y =\frac{8}{17}+\frac{3}{4} = \frac{8 \times 4 + 3 \times 15}{15 \times 4} = \frac{77}{60}$
$1 - xy = 1 - \frac{8}{17}\times \frac{3}{4} = 1 - \frac{24}{60} = \frac{36}{60}$
$\Rightarrow \large \frac{x+y}{1-xy} = \Large \frac{ \frac{77}{60}}{\frac{36}{60}}$$= \frac{77}{36}$
Therefore, $tan^{-1} \bigg( \frac{x+y}{1-xy} \bigg)$ $$= tan^{-1} \frac{77}{36}$$