Question

If a tan $\theta$ = b, then a cos 2$\theta$ + b sin 2$\theta$ =

• A. a
• B. b
• C. -a
• D. -b

Answer 1

The correct option is A

a

Given that tan $\theta$ = $\frac{b}{a}$.

Now, a cos 2$\theta$ + b sin 2 $\theta$ = a$\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)$ + b$\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

Putting $tan\theta$ = $\frac{b}{a}$, we get

= a$\left(\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}\right)$ + b$\left(\frac{2\frac{b}{a}}{1+\frac{b^2}{a^2}}\right)$ = a$\left(\frac{a^2-b^2}{a^2+b^2}\right)$ + b$\left(\frac{2ba}{a^2+b^2}\right)$

= $\frac{1}{(a^2+b^2)}$ $a^3-a{b}^2+2a{b}^2$ = $\frac{a(a^2+b^2)}{a^2+b^2}$ = a.