Question

If acos2$\theta$ + bsin2$\theta$ = C has $\alpha$ and $\beta$ as its solution, then values of tan $\alpha$+ tan$\beta$,tan $\alpha$.tan$\beta$ are respectivly equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

a cos2$\theta$ + b sin2$\theta$ = C

⇒ $a(co{s}^2\theta -si{n}^2\theta )$ + $2bsin\theta cos\theta$ = C

⇒ $a\left(1ta{n}^2\theta \right)$ + $2btan\theta$ = $cse{c}^2\theta$ = $C(1+ta{n}^2\theta )$

⇒ $ta{n}^2\theta (c+a)-2btan\theta$ + c - a = 0 $\to$ 1

The equation 1 has tan$\alpha$ and tan$\beta$ as its roots

⇒ tan$\alpha$ + tan$\beta$ = $\frac{2b}{c+a}$, tan$\alpha$tan$\beta$ = $\frac{c-a}{c+a}$