Question

If $tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )=\lambda sec2\theta$, then

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

The correct option is D

2

Given:

$tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )=\lambda sec2\theta \Rightarrow \frac{tan\frac{\pi }{4}+tan\theta }{1-tan\frac{\pi }{4}\times tan\theta }+\frac{tan\frac{\pi }{4}-tan\theta }{1+tan\frac{\pi }{4}\times tan\theta }=\lambda sec2\theta \Rightarrow \frac{1+tan\theta }{1-tan\theta }+\frac{1-tan\theta }{1+tan\theta }=\lambda sec2\theta \Rightarrow \frac{(1+tan\theta {)}^2(1-tan\theta {)}^2}{(1-tan\theta )+(1+tan\theta )}=\lambda sec2\theta$

$\Rightarrow \frac{2(1+ta{n}^2\theta )}{1-ta{n}^2\theta }=\lambda sec2\theta \Rightarrow \frac{2se{c}^2\theta }{1-ta{n}^2\theta }=\lambda sec2\theta \Rightarrow \frac{2}{co{s}^2\theta (1-ta{n}^2\theta )}=\lambda sec2\theta \Rightarrow \frac{2}{co{s}^2\theta (1-\frac{si{n}^2\theta }{co{s}^2\theta })}=\lambda sec2\theta \Rightarrow \frac{2}{co{s}^2\theta -si{n}^2\theta }=\lambda sec2\theta$

$\Rightarrow \frac{2}{cos2\theta }=\lambda sec2\theta \Rightarrow 2sec2\theta =\lambda sec2\theta \Rightarrow 2=\lambda \therefore \lambda =2$