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Prove that: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$
Prove that: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$
CBSE | Class 12 | Exercise 4C | Inverse Trigonometric Functions | Maths | RS Aggarwal
Prove that: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$

Solution:

To Prove: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Proof:
$\begin{array}{l}
\text { LHS }=\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24} \\
=\tan ^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\left(\frac{2}{11} \times \frac{7}{24}\right)}\right) \\
=\tan ^{-1}\left(\frac{48+77}{264-14}\right) \\
=\tan ^{-1} \frac{125}{250} \\
=\tan ^{-1} \frac{1}{2} \\
=\text { RHS }
\end{array}$
Therefore LHS $=$ RHS
Hence proved.

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