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Q)

What does $sin^{-1} \frac{5}{13}+cos^{-1} \frac {3}{5}$ reduce to?

$\begin{array}{1 1} tan^{-1}\frac{63}{16} \\ sin^{-1}\frac{63}{16} \\cos^{-1}\frac{63}{16} \\ cot^{-1}\frac{63}{16} \end{array}$

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A)
Toolbox:
• $$sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$
• $$cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}$$
• $$tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}$$
Given $sin^{-1} \large\frac{5}{13}+cos^{-1} \large\frac {3}{5}$
We know that $$sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$

By taking $x=\large\frac{5}{13} \rightarrow$ $$\large \frac{x}{\sqrt{1-x^2}}$$=$$\Large \frac{\frac{5}{13}}{\sqrt{1-\large\frac{25}{169}}}$$$$=\large\frac{5}{13}.\large\frac{13}{12}=\large\frac{5}{12}$$
$$\Rightarrow\:sin^{-1}\large\frac{5}{13}=tan^{-1}\large\frac{5}{12}$$
We know that $$cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}$$

By taking $x=$$$\frac{3}{5} \rightarrow \large \frac{\sqrt{1-x^2}}{x}=\Large \frac{\sqrt{1-\large\frac{9}{25}}}{\large\frac{3}{5}}$$$$=\large\frac{4}{5}.\large\frac{5}{3}=\large\frac{3}{4}$$

$$\Rightarrow\:cos^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{4}{3}$$
$$\Rightarrow sin^{-1} \large\frac{5}{13}+cos^{-1}\large\frac{3}{5}$$ $$= tan^{-1}\large\frac{5}{12}+tan^{-1}\large\frac{4}{3}$$

$$\large\frac{x+y}{1-xy}=\large\frac{\large\frac{5}{12}+\large\frac{4}{3}}{1-\large\frac{5}{12}.\large\frac{4}{3}}=\large\frac{63}{36}.\large\frac{36}{16}=\large\frac{63}{16}$$
$$= tan^{-1} \bigg[ \large\frac{\large\frac{5}{12}+\large\frac{4}{3}}{1-\large\frac{20}{36}} \bigg]$$
$$= tan^{-1}\large\frac{63}{16}$$
=L.H.S

We know that $$tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}$$

By taking $$x=\large\frac{5}{12}\:and\:y=\large\frac{4}{3}\:in\:tan^{-1}x+tan^{-1}y$$
$x+y = \large\frac{5}{12}+\large\frac{4}{3} = \large\frac{63}{36}$
$1 - xy = 1 - \large\frac{5}{12} \times \frac{4}{3} = \large\frac{16}{36}$
$\large \frac{x+y}{1-xy} =\large \frac{63}{36}.\large\frac{36}{16}=\large\frac{63}{16}$

$\Rightarrow tan^{-1}\large\frac{5}{12}+tan^{-1}\large\frac{4}{3} =$ $$tan^{-1}\large\frac{63}{16}$$