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प्रश्न
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
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उत्तर
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + se c^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right)\]
= \[2 \tan^{- 1} \left( \frac{1}{5} \right) + \tan^{- 1} \left( \sqrt{\left( \frac{5\sqrt{2}}{7} \right)^2 - 1} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) \left[ \text { Using }se c^{- 1} x = \tan^{- 1} \sqrt{x^2 - 1} \right]\]
\[= 2 \tan^{- 1} \left( \frac{1}{5} \right) + \tan^{- 1} \left( \frac{1}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right)\]
= 2 \[\left( \tan^{- 1} \left( \frac{1}{5} \right) + \tan^{- 1} \left( \frac{1}{8} \right) \right) + \tan^{- 1} \left( \frac{1}{7} \right)\]
\[= 2 \tan^{- 1} \left( \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \times \frac{1}{8}} \right) + \tan^{- 1} \left( \frac{1}{7} \right) \left[\text { Using} \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) \right]\]
\[= 2 \tan^{- 1} \left( \frac{13}{39} \right) + \tan^{- 1} \left( \frac{1}{7} \right)\]
\[= 2 \tan^{- 1} \left( \frac{1}{3} \right) + \tan^{- 1} \left( \frac{1}{7} \right)\]
\[= \tan^{- 1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} \right) + \tan^{- 1} \left( \frac{1}{7} \right) \left[ \text { Using} 2 \tan^{- 1} x = \tan^{- 1} \frac{2x}{1 - x^2}, \text { if } \left| x \right| < 1 \right]\]
\[= \tan^{- 1} \left( \frac{3}{4} \right) + \tan^{- 1} \left( \frac{1}{7} \right)\]
\[= \tan^{- 1} \left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \times \frac{1}{7}} \right)\]
\[= \tan^{- 1} \left( 1 \right)\]
\[ = \frac{\pi}{4}\]
\[ = RHS\]
Hence proved.
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