Solveeit Logo
Login

Question

Question: \(\tan\left\lbrack \frac{\pi}{4} + \frac{1}{2}\cos^{- 1}\frac{a}{b} \right\rbrack + \tan\left\lbrack...

tan⁡[π4+12cos⁡−1ab]+tan⁡[π4−12cos⁡−1ab]\tan\left\lbrack \frac{\pi}{4} + \frac{1}{2}\cos^{- 1}\frac{a}{b} \right\rbrack + \tan\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\frac{a}{b} \right\rbrackequal to

A

2ab\frac{2a}{b}

B

2ba\frac{2b}{a}

C

ab\frac{a}{b}

D

ba\frac{b}{a}

Answer

2ba\frac{2b}{a}

Explanation

Solution

Let cos⁡−1ab=θ⇒cos⁡θ=ab\cos^{- 1}\frac{a}{b} = \theta \Rightarrow \cos\theta = \frac{a}{b}

tan⁡[π4+12cos⁡−1ab]+tan⁡[π4−12cos⁡−1ab]\tan\left\lbrack \frac{\pi}{4} + \frac{1}{2}\cos^{- 1}\frac{a}{b} \right\rbrack + \tan\left\lbrack \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}\frac{a}{b} \right\rbrack=1+t1−t+1−t1+t\frac{1 + t}{1 - t} + \frac{1 - t}{1 + t} ,

where t=tan⁡θ2t = \tan\frac{\theta}{2} =2(1+t2)1−t2=2cos⁡θ=2ba= 2\frac{(1 + t^{2})}{1 - t^{2}} = \frac{2}{\cos\theta} = \frac{2b}{a}