Question

If $tan\theta =\frac{a}{b}$, then b $cos2\theta +asin2\theta$ is equal to

• A. a
• B. b
• C. $\frac{a}{b}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B

b

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

$\therefore bcos2\theta +asin2\theta ==b\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)+a\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)=\left(\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}}\right)+a\left(\frac{\frac{2a}{b}}{1+\frac{a^2}{b^2}}\right)=b\left(\frac{b^2-a^2}{b^2+a^2}\right)+\left(\frac{2a^{2}b}{a^2+b^2}\right)=\frac{b}{a^2+b^2}[b^2-a^2+2a^2]=\frac{(a^2+b^2)b}{(a^2+b^2)}$
= b