Question

lf $\alpha$ and $\beta$ are two different solutions lying between $\frac{-\pi }{2}$ and $\frac{\pi }{2}$ of the equation $2tan\theta +Sec\theta =2$ then $tan\alpha +tan\beta$

• A. 0
• B. 1
• C. 4/3
• D. 8/3

Answer 1

The correct option is D 8/3

The given equation is $2\tan \theta +\sec \theta =2$

$\Rightarrow \sec \theta =2-2\tan \theta$

Squaring on both sides, we get

${\sec }^2\theta =4{(1-\tan \theta )}^2$

${\sec }^2\theta =4(1+{\tan }^2\theta -2\tan \theta )$

$1+{\tan }^2\theta =4(1+{\tan }^2\theta )-8\tan \theta$

$3{\tan }^2\theta -8\tan \theta +3=0$

Since $\alpha$ and $\beta$ are the roots of the equation,

$\therefore \tan \alpha +\tan \beta =\frac{8}{3}$