Prove that:

Question:

Prove that:

$2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$

Solution:

To Prove: $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$

Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Proof:

$\mathrm{LHS}=2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\left(\frac{1}{3} \times \frac{1}{3}\right)}\right)+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left(\frac{6}{9-1}\right)+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\left(\frac{3}{4} \times \frac{1}{7}\right)}\right)$

$=\tan ^{-1}\left(\frac{21+4}{28-3}\right)$

$=\tan ^{-1} \frac{25}{25}$

$=\tan ^{-1} 1$

$=\frac{\pi}{4}$

$=$ RHS

Therefore LHS = RHS

Hence proved.

 

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