If `u_n=sin(n theta) sec^n theta,` `v_n=cos(ntheta)sec^ntheta ,` `n in N,` `n!=1` then `(v_n-v_(n-1))/(u_(n-1))+1/n (u_n/v_n)=`

Options

A. 0
B. `tan theta`
C. `-tan theta + (tan n theta) / n`
D. `tan theta + (tan n theta)/n`

Answer

C

Explanation

We have `U_n/V_n=tan n theta`
And `(V_n-V_(n-1))/U_(n-1) = (cos n theta sec^n theta - cos (n-1) theta sec^(n-1) theta)/(sin (n-1)theta sec^(n-1) theta)`
`=(cos n theta sec theta - cos (n-1)theta)/(sin (n-1) theta)=(cos n theta -cos (n-1) theta cos theta )/(cos theta sin (n-1) theta) " " because {cos n theta = cos {(n-1) theta + theta}}`
`=(cos (n-1) theta cos theta - sin (n-1) sin theta - cos (n-1) theta cos theta)/(cos theta sin (n-1) theta ) =-tan theta`
So that `(V_n -V_(n-1))/U_(n-1)+1/n U_n/V_n=-tan theta + (tan n theta)/n`