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CBSE Class 12 Math 2008 Solved Paper

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Question : 12 of 29
Marks: +1, -0
Prove the following:
tan⁡−113+tan⁡−115\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} + tan⁡−117+tan⁡−118\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} = π8\frac{\pi}{8}
Solution:  
L.H.S. = tan⁡−113+tan⁡−115\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} + tan⁡−117+tan⁡−118\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8}
= tan⁡−1∣13+151−13⋅15∣\tan^{-1}\left|\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}\cdot\frac{1}{5}}\right| + tan⁡−1∣17+181−17⋅18∣\tan^{-1}\left|\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\cdot\frac{1}{8}}\right|
= tan⁡−1∣5+31515−115∣\tan^{-1}\left|\frac{\frac{5+3}{15}}{\frac{15-1}{15}}\right| + tan⁡−1∣8+75656−156∣\tan^{-1}\left|\frac{\frac{8+7}{56}}{\frac{56-1}{56}}\right|
= tan⁡−1∣8151415∣\tan^{-1}\left|\frac{\frac{8}{15}}{\frac{14}{15}}\right| + tan⁡−1∣15565556∣\tan^{-1}\left|\frac{\frac{15}{56}}{\frac{55}{56}}\right|
= tan⁡−1(814)\tan^{-1}\left(\frac{8}{14}\right) + tan⁡−1(1555)\tan^{-1}\left(\frac{15}{55}\right)
= tan⁡−1(47)+tan⁡−1(311)\tan^{-1}\left(\frac{4}{7}\right)+\tan^{-1}\left(\frac{3}{11}\right)
= tan⁡−1(47+3111−47⋅311)\tan^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7}\cdot\frac{3}{11}}\right)
= tan⁡−1(44+217777−1277)\tan^{-1}\left(\frac{\frac{44+21}{77}}{\frac{77-12}{77}}\right)
= tan⁡−1(6565)\tan^{-1}\left(\frac{65}{65}\right)
= tan⁡−1\tan^{-1} 1
= tan⁡−1(tan⁡π4)\tan^{-1}\left(\tan\frac{\pi}{4}\right)
= π4\frac{\pi}{4}
= R.H.S.
Hence proved.
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