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# What does $cos^{-1} \frac {12}{13} +sin^{-1} \frac{3}{5}$ reduce to?

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A)
• $$cos^{-1}x=sin^{-1}\sqrt{1-x^2}$$
• $$sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2} +y\sqrt{1-x^2} \bigg]$$
Given $cos^{-1} \frac {12}{13} +sin^{-1} \frac{3}{5}$
We know that $$cos^{-1}x=sin^{-1}\sqrt{1-x^2}$$
By taking $x=$$$\frac{12}{13},\rightarrow \sqrt{1-x^2}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$$
$$\Rightarrow\:cos^{-1}\frac{12}{13}=sin^{-1}\frac{5}{13}$$
$$\Rightarrow cos^{-1}\frac{12}{13}+sin^{-1}\frac{3}{5}=sin^{-1}\frac{5}{13}+sin^{-1}\frac{3}{5}$$
We know that $$sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2} +y\sqrt{1-x^2} \bigg]$$
Given $sin^{-1}\frac{5}{13}+sin^{-1}\frac{3}{5}$, let's take $x=\frac{5}{13}$ and $y=\frac{3}{5}$
$x \sqrt {1-y^2} = \frac{5}{13} \sqrt {1-(\frac{3}{5})^2} = \frac{5}{13} \sqrt{1-\frac{9}{25}} = \frac{5}{13}.\frac{4}{5} = \frac{20}{65}$
$y \sqrt {1-x^2} = \frac{3}{5} \sqrt{1-(\frac{5}{13})^2} = \frac{3}{5} \sqrt{1-\frac{25}{169}} = \frac{3}{5}.\frac{12}{13} = \frac{36}{65}$
Therefore, $sin^{-1}\frac{5}{13}+sin^{-1}\frac{3}{5} = sin^{-1} ( \frac{20}{65}+ \frac{36}{65}) = \frac{56}{65}$