Question

If $(a+b)tan(\theta -\varphi )=(a-b)tan(\theta +\varphi )$, then $\frac{sin\left(2\theta \right)}{sin\left(2\varphi \right)}$ is equal to

• A. $ab$
• B. $\frac{a}{b}$
• C. $\frac{b}{a}$
• D. $a^2{b}^2$

Answer 1

Answer: (B)

$\frac{a}{b}$

Given $(a+b)tan(\theta -\varphi )=(a-b)tan(\theta +\varphi )$

$\Rightarrow \frac{tan(\theta -\varphi )}{tan(\theta +\varphi )}=\frac{a-b}{a+b}$
$\Rightarrow \frac{sin(\theta -\varphi ).cos(\theta +\varphi )}{sin(\theta +\varphi ).cos(\theta -\varphi )}=\frac{a-b.cos(\theta -\varphi )}{a+b}$
$\Rightarrow \frac{2cos(\theta +\varphi ).sin(\theta -\varphi )}{2sin(\theta +\varphi ).cos(\theta -\varphi )}=\frac{a-b}{a+b}$
$\Rightarrow \frac{sin\left(2\theta \right)-sin\left(2\varphi \right)}{sin\left(2\theta \right)+sin\left(2\varphi \right)}=\frac{a-b}{a+b}$

Apply componendo and dividendo
$\Rightarrow \frac{sin2\theta }{sin\left(2\varphi \right)}=\frac{a+b+a-b}{a+b-(a-b)}=\frac{a}{b}$