# Prove that:

Question:

Prove that:

$\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$

Solution:

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.