Question

If $\alpha$ and $\beta$ are two different solution lying between $\frac{-\pi }{2}$ and $\frac{\pi }{2}$ of the equation $2\mathrm{Tan}\theta +\mathrm{Sec}\theta -2$ then $\mathrm{Tan}\alpha +\mathrm{Tan}\beta$ is

• A. $0$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{8}{3}$

Answer 1

The correct option is A $0$

$2\tan \theta +\sec \theta -2=0\frac{2\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }-2=02\sin \theta -2\cos \theta +1=02(\sin \theta -\cos \theta )=-1\sin \theta -\cos \theta =\frac{-1}{2}{(\sin \theta -\cos \theta )}^2={\left(\frac{-1}{2}\right)}^2\sin {}^2\theta +\cos {}^2\theta -2\sin \theta \cos \theta =\frac{1}{4}1-2\sin \theta \cos \theta =\frac{1}{4}1-\sin 2\theta =\frac{1}{4}\Rightarrow \sin 2\theta =\frac{3}{4}\Rightarrow 2\theta ={\sin }^{-1}\frac{3}{4}$

It can be $\pi \&-\pi$

$\therefore \tan \pi +\tan (-\pi )=0$