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Mathematics Question on Inverse Trigonometric Functions

Prove tan⁡−1211+tan⁡−1724=tan⁡−112\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}=\tan^{-1}\frac{1}{2}

Answer

To prove: tan⁡−1211+tan⁡−1724=tan⁡−112\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}=\tan^{-1}\frac{1}{2}

LHS= tan⁡−1211+tan⁡−1724\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}

= tan⁡−1211+7241−211+724\tan^{-1}\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}+\frac{7}{24}} [tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy]\bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]

= tan⁡−148−77264−14=tan⁡−1125250=tan⁡−112=R.H.S.\tan^{-1}\frac{48-77}{264-14}=\tan^{-1}\frac{125}{250}=\tan^{-1}\frac{1}{2}=R.H.S.