Question
Let $$ - \frac{\pi }{6} < \theta < - \frac{\pi }{{12}}.$$ Suppose $${\alpha _1}$$ and $${\beta _1}$$ are the roots of the equation $${x^2} - 2x\sec \alpha + 1 = 0$$ and $$\,{\alpha _2}$$ and $${\beta _2}$$ are the roots of the equation $${x^2} + 2x\tan \theta - 1 = 0.\,$$ If $${\alpha _1}\, > {\beta _1}$$ and $${\alpha _2}\, > {\beta _2}$$ then $$\,{\alpha _1}\, + {\beta _2}$$ equals
A.
$$2\left( {\sec \theta - \tan \theta } \right)$$
B.
$$2\sec \theta $$
C.
$$ - 2\tan \theta $$
D.
0
Answer :
$$ - 2\tan \theta $$
Solution :
$$\eqalign{
& {x^2} - 2x\sec \theta + 1 = 0 \cr
& \Rightarrow x = \sec \,\,\theta \pm \,\tan \theta \cr
& {\text{and }}{x^2} + 2x\tan \theta - 1 = 0 \cr
& \Rightarrow \,\,x = - \tan \theta \pm \sec \theta \cr
& \because - \frac{\pi }{6} < \theta < \, - \frac{\pi }{{12}} \cr
& \Rightarrow \sec \frac{\pi }{6} > \sec \theta > \sec \frac{\pi }{{12}} \cr
& {\text{and}}\,\, - \tan \frac{\pi }{6} < \tan \theta < - \frac{{\tan \pi }}{{12}} \cr
& {\text{also}}\,\,\tan \frac{\pi }{{12}} < - \tan \theta < \tan \frac{\pi }{6} \cr
& {\alpha _1},{\beta _1}\,{\text{are}}\,{\text{roots}}\,{\text{of }}{x^2} - 2x\,\sec \theta \, + 1 = 0 \cr
& {\text{and}}\,{\alpha _1} > {\beta _1} \cr
& \therefore {\alpha _1} = \sec \theta - \tan \theta \,\,{\text{and }}{\beta _1} = \sec \theta + \tan \theta \cr
& {\alpha _2},{\beta _2}\,{\text{are}}\,{\text{roots of }}{x^2} + 2x\,\tan {\text{ }}\theta - 1 = 0\,\,{\text{and}}\,{\alpha _2} > {\beta _2} \cr
& \therefore {\alpha _2} = - \tan \,\theta + \sec \theta ,{\beta _2} = - \tan \,\theta - \sec \theta \cr
& \therefore {\alpha _1} + {\beta _2} = \sec \theta \, - \tan \theta - \tan \theta - \sec \theta = - 2\tan \theta \cr} $$