ત્રિકોણમિતીય પ્રતિવિધેયો — ગણિત ધોરણ 12 સાયન્સ — Question

$\cot\left[\frac{\pi}{4}-2\cos^{-1}3\right]=\tan\left[\frac{\pi}{2}-\left(\frac{\pi}{4}-2\cot^{-1}3\right)\right]$

$=\tan\left[\frac{\pi}{4}+2\cot^{-1}3\right]$

$=\tan\left[\frac{\pi}{4}+2\tan^{-1}\frac{1}{3}\right]$

$=\tan\left[\frac{\pi}{4}+\tan^{-1}\left(\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{9}\right)}\right)\right]$

$(\because$ જો $0 < x < 1,$ તો $2tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2})$

$=\tan\left[\frac{\pi}{4}+\tan^{-1}\frac{2(3)}{9-1}\right]$

$=\tan\left[\frac{\pi}{4}+\tan^{-1}\frac{6}{8}\right]$

$=\tan\left[\frac{\pi}{4}+\tan^{-1}\frac{3}{4}\right]$

$=\frac{\tan\frac{\pi}{4}+\tan\left(\tan^{-1}\frac{3}{4}\right)}{1-\tan\frac{\pi}{4}\tan\left(\tan^{-1}\frac{3}{4}\right)}$

$=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}=\frac{4+3}{4-3}=7$