Question

If $U_n=\sin \left(n\theta \right){\sec }^n\theta ,V_n=\cos \left(n\theta \right){\sec }^n\theta$ for $n=0,1,2$,

then $V_n-V_{n-1}+U_{n-1}\tan \theta =$?

• A. $0$
• B. $1$
• C. $\sin \theta$
• D. $\cos \theta$

Answer 1

The correct option is A $0$

Given $U_n=\sin \left(n\theta \right).{\sec }^n\theta =\cos \left(n\theta \right).{\sec }^n\theta$

$V_n-V_{n-1}+U_{n-1}\tan \theta$

$V_n=\cos \left(n\theta \right).{\sec }^n\theta$

$V_n=\cos \left(\right(n-1)\theta +\theta ).{\sec }^{n-1}\theta .\sec \theta$

$V_n={\sec }^{n-1}\theta \left[\cos \right(n-1)\theta .\cos \theta -\sin (n-1)\theta .\sin \theta ].\sec \theta$

$V_n={\sec }^{n-1}\theta \left[\cos \right(n-1)\theta .\cos \theta .\sec \theta -\sin (n-1)\theta .\sin \theta .\sec \theta ]$

$V_n={\sec }^{n-1}\theta \left[\cos \right(n-1)\theta -\sin (n-1)\theta .\frac{\sin \theta }{\cos \theta }]$

$V_n={\sec }^{n-1}\theta \left[\cos \right(n-1\left)\theta \right]-{\sec }^{n-1}\theta \left[\sin \right(n-1)\theta .\tan \theta ]$

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

$V_n=V_{n-1}-\left[\sin \right(n-1)\theta .{\sec }^{n-1}\theta ].\tan \theta$

$V_n=V_{n-1}-\left[U_{n-1}\right]\tan \theta$

$V_n-V_{n-1}+U_{n-1}.\tan \theta =0$