Question

Solve :$\frac{\tan \left(\frac{\pi }{4}+\theta \right)+\tan \left(\frac{\pi }{4}-\theta \right)}{\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)}=$

• A. $\sin 2\theta$
• B. $\cos 2\theta$
• C. $\sec 2\theta$
• D. cosec $2\theta$

Answer 1

The correct option is D cosec $2\theta$

$\frac{tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta ))}{tan(\frac{\pi }{4}+\theta ))-tan\frac{\pi }{4}-\theta )}$

$=\frac{\left(\frac{1+tan\theta }{1-tan\theta }\right)+\left(\frac{1-tan\theta }{1+tan\theta }\right)}{\left(\frac{1+tan\theta }{1-tan\theta }\right)-\left(\frac{1-tan\theta }{1+tan\theta }\right)}$

$=\frac{(1+tan\theta {)}^2+(1-tan\theta {)}^2}{(1+tan\theta {)}^2-(1-tan\theta {)}^2}$

$=\frac{1+ta{n}^2\theta +2tan\theta +1+ta{n}^2\theta -2tan\theta }{1+ta{n}^2\theta +2tan\theta -1-ta{n}^2\theta +2tan\theta }$

$\Rightarrow \frac{2(1+ta{n}^2\theta )}{4tan\theta }\Rightarrow \frac{1\left(se{c}^2\theta \right)}{2\left(sin\theta /cos\theta \right)}\Rightarrow \frac{cos\theta }{\left(2\right)sin\theta co{s}^2\theta }$

$\Rightarrow \frac{1}{2sin\theta cos\theta }=\frac{1}{sin2\theta }=cosec2\theta$

$\therefore \frac{tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )}{tan(\frac{\pi }{4}+\theta )-tan(\frac{\pi }{4}-\theta )}=cosec2\theta$