Question

The sum of the series $\sin \theta .\sec 3\theta +\sin 3\theta .\sec 3^2+\sin 3^2\theta \sec 3^3\theta$+ .... up to n terms is

• A. $\frac{1}{2}(\tan 3^n\theta -\tan \theta )$
• B. $(\tan 3^n\theta -\tan \theta )$
• C. $\tan 3^n\theta -\tan 3^{n-1}\theta$
• D. $2(\tan 3^n\theta -\tan \theta )$

Answer 1

The correct option is A $\frac{1}{2}(\tan 3^n\theta -\tan \theta )$

Hence, Option $A$ is correct answer.