Prove that:
$\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$
To Prove: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$
Proof:
$\mathrm{LHS}=\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2} \ldots$ (1)
Let $\sec \theta=\frac{\sqrt{5}}{2}$
Therefore $\theta=\sec ^{-1} \frac{\sqrt{5}}{2} \cdots$ (2)
From the figure, $\tan \theta=\frac{1}{2}$
$\Rightarrow \theta=\tan ^{-1} \frac{1}{2} \ldots$ (3)
From $(2)$ and $(3)$
$\sec ^{-1} \frac{\sqrt{5}}{2}=\tan ^{-1} \frac{1}{2}$
Substituting in $(1)$, we get
$\mathrm{LHS}=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}$
$=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\left(\frac{1}{3} \times \frac{1}{2}\right)}\right)$
$=\tan ^{-1}\left(\frac{2+3}{6-1}\right)$
$=\tan ^{-1} \frac{5}{5}$
$=\tan ^{-1} 1$
$=\frac{\pi}{4}$
$=\mathrm{RHS}$
Therefore, LHS $=$ RHS
Hence proved.
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