Question

If $\theta =3\alpha$ and $\sin \theta =\frac{a}{\sqrt{a^2+b^2}}$. The value of the expression, $a\text{cosec}\alpha -b\sec \alpha$ is:

• A. $\frac{1}{\sqrt{a^2+b^2}}$
• B. $2\sqrt{a^2+b^2}$
• C. $a+b$
• D. None of the above

Answer 1

Answer: (B)

$2\sqrt{a^2+b^2}$

Given $\theta =3\alpha$

$\sin \theta =\frac{a}{\sqrt{a^2+b^2}}$

$\cos \theta =\frac{b}{\sqrt{a^2+b^2}}$

$\theta -\alpha =2\alpha$

$\Rightarrow \sin (\theta -\alpha )=\sin 2\alpha$

$\Rightarrow \sin \theta \cos \alpha -\cos \theta \sin \alpha =2\sin \alpha \cos \alpha$

$\Rightarrow \frac{a\cos \alpha -b\sin \alpha }{\sqrt{a^2+b^2}}=2\sin \alpha \cos \alpha$

$\Rightarrow \frac{a\cos \alpha -b\sin \alpha }{\sin \alpha \cos \alpha }=2\sqrt{a^2+b^2}$

$\Rightarrow a\csc \alpha =b\sec \alpha =2\sqrt{a^2+b^2}$