Prove that:
$\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$
To Prove: $\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$
Proof:
$\mathrm{LHS}=\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}} \ldots$ (1)
Let $\sin \theta=\frac{1}{\sqrt{17}}$
Therefore $\theta=\sin ^{-1} \frac{1}{\sqrt{1} 7} \ldots$ (2)
From the figure, $\tan \theta=\frac{1}{4}$
$\Rightarrow \theta=\tan ^{-1} \frac{1}{4} \ldots$ (3)
From $(2)$ and $(3)$,
$\sin ^{-1} \frac{1}{\sqrt{17}}=\tan ^{-1} \frac{1}{4} \ldots$ (3)
Now, let $\cos \theta=\frac{9}{\sqrt{85}}$
Therefore $\theta=\cos ^{-1} \frac{9}{\sqrt{8} 5} \ldots$ (4)
From the figure, $\tan \theta=\frac{2}{9}$
$\Rightarrow \theta=\tan ^{-1} \frac{2}{9} \ldots$ (5)
From (4) and (5),
$\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{2}{9} \ldots$ (6)
Substituting (3) and (6) in (1), we get
$\mathrm{LHS}=\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$
$=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4} \times \frac{2}{9}\right)}\right)$
$=\tan ^{-1}\left(\frac{9+8}{36-2}\right)$
$=\tan ^{-1} \frac{17}{34}$
$=\tan ^{-1} \frac{1}{2}$
$=\operatorname{RHS}$
Therefore, LHS = RHS
Hence proved.
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