cos^{-1}left(frac{15}{17}right) + 2tan^{-1}le | vedclass

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(D) આપેલ પદાવલિ: $\cos^{-1}\left(\frac{15}{17}ight) + 2	an^{-1}\left(\frac{1}{5}ight)$.પ્રથમ,$2	an^{-1}\left(\frac{1}{5}ight)$ ને $\cos^{-1}$

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(D) આપેલ પદાવલિ: $\cos^{-1}\left(\frac{15}{17}ight) + 2	an^{-1}\left(\frac{1}{5}ight)$.પ્રથમ,$2	an^{-1}\left(\frac{1}{5}ight)$ ને $\cos^{-1}$ સ્વરૂપમાં ફેરવવા માટે સૂત્ર $2	an^{-1}(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$ નો ઉપયોગ કરો:$2	an^{-1}\left(\frac{1}{5}ight) = \cos^{-1}\left(\frac{1 - (1/5)^2}{1 + (1/5)^2}ight) = \cos^{-1}\left(\frac{1 - 1/25}{1 + 1/25}ight) = \cos^{-1}\left(\frac{24/25}{26/25}ight) = \cos^{-1}\left(\frac{12}{13}ight)$.હવે,પદાવલિ $\cos^{-1}\left(\frac{15}{17}ight) + \cos^{-1}\left(\frac{12}{13}ight)$ બને છે.સૂત્ર $\cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}ight)$ નો ઉપયોગ કરતા:$\cos^{-1}\left(\frac{15}{17} 	imes \frac{12}{13} - \sqrt{1 - \left(\frac{15}{17}ight)^2} \sqrt{1 - \left(\frac{12}{13}ight)^2}ight)$$= \cos^{-1}\left(\frac{180}{221} - \sqrt{1 - \frac{225}{289}} \sqrt{1 - \frac{144}{169}}ight)$$= \cos^{-1}\left(\frac{180}{221} - \sqrt{\frac{64}{289}} \sqrt{\frac{25}{169}}ight)$$= \cos^{-1}\left(\frac{180}{221} - \frac{8}{17} 	imes \frac{5}{13}ight)$$= \cos^{-1}\left(\frac{180}{221} - \frac{40}{221}ight) = \cos^{-1}\left(\frac{140}{221}ight)$.આમ,$\cos^{-1}\left(\frac{140}{221}ight)$ વિકલ્પોમાં આપેલ નથી,તેથી સાચો વિકલ્પ $(d)$ છે.

$\cos^{-1}\left(\frac{15}{17}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) = $

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