Question #e1c7e

1 Answer
Jul 27, 2016

See the Proof given below in Explanation.

Explanation:

Let costheta=x, rArr sectheta=1/xcosθ=x,secθ=1x. hence, given that,

costheta+sectheta=2rArr x+1/x=2, or, x^2+1=2xcosθ+secθ=2x+1x=2,or,x2+1=2x

:. x^2-2x+1=0 rArr (x-1)^2=0 rArr x=1

rArr costheta=1, sectheta=1

Hence, cos^ntheta+sec^ntheta=1^n+1^n=2.

Hence, the Proof.

Here, we have proved that, cos^ntheta+sec^ntheta=2.

But, cos^mtheta+sec^ntheta=2, where, m.n in RR is also true!

On the same lines, the following problems can be solved :

(1) : sintheta+csctheta=2rArrsin^mtheta+csc^ntheta=2, m,n in RR

(2) : tanx+cotx=2rarrtan^2016x+cot^2017x=2, etc.

Enjoy Maths!