Question #e1c7e

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Answer 1 · 2016-07-27T12:40:55

See the Proof given below in Explanation.

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

#costheta+sectheta=2rArr x+1/x=2, or, x^2+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!