Question

If $a{\sin }^2\theta +b{\cos }^2\theta =a{\cos }^2\varphi +b{\sin }^2\varphi =1$ and $a\tan \theta =b\tan \varphi$, then choose the correct option.

• A. $a+b=2ab$
• B. $a-b=2ab$
• C. $a-b+2ab=0$
• D. $a+b+2ab=0$

Answer 1

The correct option is A $a+b=2ab$

Given: $a{\sin }^2\theta +b{\cos }^2\theta =1$ .... $\left(i\right)$ and $a{\cos }^2\varphi +b{\sin }^2\varphi =1$ .... $\left(ii\right)$

Dividing equation $\left(i\right)$ by ${\cos }^2\theta$, we get

$a{\tan }^2\theta +b={\sec }^2\theta$

$\Rightarrow a{\tan }^2\theta +b=1+{\tan }^2\theta$

$\Rightarrow {\tan }^2\theta =\frac{b-1}{1-a}=\tan \theta =\sqrt{\frac{b-1}{1-a}}$

Similarly in the equation $\left(ii\right)$, on dividing by ${\cos }^2\varphi$ we get,

$b{\tan }^2\varphi +a={\sec }^2\varphi$

$\Rightarrow b{\tan }^2\varphi +a=1+{\tan }^2\varphi$

$\Rightarrow {\tan }^2\varphi =\frac{1-a}{b-1}=\tan \varphi =\sqrt{\frac{1-a}{b-1}}$

Now, it is given that

$a\tan \theta =b\tan \varphi$

$\Rightarrow a\sqrt{\frac{b-1}{1-a}}=b\sqrt{\frac{1-a}{b-1}}$

$\Rightarrow \frac{a}{b}=\frac{1-a}{b-1}$

$\Rightarrow ab-a=b-ab$

$\Rightarrow a+b=2ab$