Question

Sin⁴theta/ p + cos⁴theta/q = 1/p+m

PT: sin^2m+2 theta/ p^{​​​​​​m} + cos^{​​​​2m+2}theta/ q^{​​​​​​m} = 1/(p+q)^m

Answer 1

the question has been entered incorrectly, I have made the necessary changes while answering.

cos^4x=(cos^2x)^2=(1-sin^2x)^2=1+ sin^4x-2sin^2x

=>sin^4x / p +cos^4x / q=1/p+q


=>sin^4x/p + (1+sin^4x-2sin^2x)/q = 1/(p+q)

[q*sin^4x + p(sin^4x-2sin^2x+1)] /pq = 1/(p+q)

=>[(p+q)sin^4x-2p sin^2x+p]/pq = 1/p+q

=>(p+q)^2 sin^4x - 2p(p+q)sin^2x + p(p+q) =pq

=>(p+q)^2 sin^4x - 2p(p+q)sin^2x + p^2


=> [(p+q)sin^2x-p]^2 = 0
=>(p+q)sin^2x - p = 0

sin^2x=p/(p+q).........(1)

(take m power of both side)
=>sin^2mx=p^m/(p+q)^m

(divide by p^m and multiplying by sin^2x on both side )
=>sin^(2m+2)x/p^m=sin^2x/(p+q)^m. ........(2)

=>cos^2x=1 - sin^2x=1-p/(p+q)=q/(p+q). (from eq 1 substituting value of sin^2x)

=>cos^2x=q/(p+q)...........(3)
(take m power of both side)

=>cos^2mx=q^m/(p+q)^m

(divide by q^m both side and multiply by cos^2x)
=>cos^(2m+2)x/q^m =cos^2x/(p+q)^m......(4)

(adding eq 2@4)
=>sin^(2m+2)x/p^m + cos^(2m+2)x/q^m=cos^2x/(p+q)^m + sin^2x/(p+q)^m =1/(p+q)^m ........proved

Hope this helps