Question

If a tan$\theta$= b then a cos 2$\theta$ + b sin 2$\theta$ is equal to

• A. a
• B. b
• C. -a
• D. -b

Answer 1

The correct option is A

a

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

We need to find the value of a cos$2\theta$ + b sin$2\theta$ since we are given the value of tan$\theta$.Express sin$2\theta$

& cos$2\theta$ in terms of tan$\theta$

a.$\left(\frac{(1-ta{n}^2\theta )}{(1+ta{n}^2\theta )}\right)$ + b. $\left(\frac{2tan\theta }{(1-ta{n}^2\theta )}\right)$

Substituting the value of tan$\theta$.

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

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

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

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

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

Correct option is A.