Question

If $U_n=\sin n\theta {\sec }^n\theta ,V_n=\cos n\theta {\sec }^n\theta \ne 1,$ then $\frac{V_n-V_{n-1}}{U_{n-1}}+\frac{1}{n}\frac{U_n}{V_n}$ is equal to

• A. $0$
• B. $\tan \theta$
• C. $-\tan \theta +\frac{\tan n\theta }{n}$
• D. $\tan \theta +\frac{\tan n\theta }{n}$

Answer 1

The correct option is C $-\tan \theta +\frac{\tan n\theta }{n}$

Let $U_n=\sin n\theta {\sec }^n\theta ,V_n=\cos n\theta {\sec }^n\theta \ne 1,$ .

Then $\frac{V_n-V_{n-1}}{U_{n-1}}+\frac{1}{n}\frac{U_n}{V_n}$

$=\frac{\cos n\theta .\sec \theta -\cos (n-1)\theta }{\sin (n-1)\theta }+\frac{1}{n}\tan n\theta$ (after simplification)

$=\frac{\cos (n-1+1)\theta .\sec \theta -\cos (n-1)\theta }{\sin (n-1)\theta }+\frac{1}{n}\tan n\theta$

$=\frac{\left\{\cos \right(n-1)\theta .\cos \theta -\sin (n-1)\theta .\sin \theta \}\sec \theta -\cos (n-1)\theta }{\sin (n-1)\theta }+\frac{1}{n}\tan n\theta$

$=-\sin \theta .\sec \theta +\frac{1}{n}\tan n\theta$

$=-\tan \theta +\frac{1}{n}\tan n\theta .$