Question

When $\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin \theta }{\cos 3\theta }=k(\tan 27\theta -\tan \theta )$ is defined, then $k=$

• A. $\frac{\pi }{2}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{1}{2}$

Given that,

$\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin \theta }{\cos 3\theta }=k$

Multiplying Numerator and denominator by $2\cos \theta$ the

$\begin{array}{c}\frac{\sin \theta }{\cos 3\theta }=\frac{2\sin \theta \cos \theta }{2\cos \theta \cos 3\theta }=\frac{\sin 2\theta }{2\cos \theta \cos \theta }\\ =\frac{\sin \left(3\theta -\theta \right)}{2\cos \theta \cos 3\theta }\end{array}$

$\begin{array}{c}=\frac{\sin 3\theta -\cos 3\theta \sin \theta }{2\cos \theta \cos 3\theta }\\ =\frac{1}{2}\left(\tan 3\theta .\tan \theta \right)..........\left(1\right)\end{array}$

Similarly,

$\begin{array}{c}\frac{\sin 3\theta }{\cos 9\theta }=\frac{1}{2}\left(\tan 9\theta -\tan 3\theta \right).......\left(2\right)\\ and\frac{\sin 9\theta }{\cos 27\theta }=\frac{1}{2}\left(\tan 27\theta -\tan 9\theta \right).......\left(3\right)\end{array}$

On Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$

We get,

$\begin{array}{c}=\frac{1}{2}\left(\tan 27\theta -\tan \theta \right)\\ k=\frac{1}{2}\end{array}$

Hence,

Option $C$ is correct answer.