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

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

If u_(n)=sin(n theta)sec^(n)theta,v_(n)=cos(n theta)sec^(n)thetan in Nn!=1 then (v_(n)-v_(n-1))/(u_(n-1))+(1)/(n)((u_(n))/(v_(n)))=

If u_(n)=sin(n theta) sec^(n) theta, v_(n)=cos(n theta) sec^(n) theta , n in N, n ne 1," then " (v_(n)-v_(n-1))/(u_(n-1))+(1)/(n)(u_(n))/(v_(n)) =

If U_(n) = sin n theta*sec^(n)theta, V_(n) = cosntheta*sec^(n)theta for n= 0,1,2,… then V_(n) -V_(n-1)+U_(n-1)tantheta =

If U_(n) = sin n theta.sec^(n)theta, V_(n)=cosnthetasec^(n)theta for n=0,1,2,……. Then V_(n)-V_(n-1) + U_(n-1) tan theta = ?

If costheta + sectheta = (-2) then cos^ntheta + sec^ntheta = (If n is even)

If a_n=sin^n theta+cosec^ntheta and a_1=2 prove that, a_n=2 .

If S_n=cos^ntheta+sin^n theta then find teh value of 3S_4-2S_6