1yaar please question thik se likho
why u wrote just sin ^n + cosec ^n shuld n't it have been sin^(n)x + cosec^(n)x
4 Answers
11here's ur solution
answer is 2
see given is
→sinx + cosecx = 2
→sin2x + 1 = 2sinx
→(sinx - 1)2 = 0
→ sinx = 1 → x = \frac{\pi }{2 }
hence
sin nx + cosecnx = 1n + 1n = 2
1when u get eqn of the form a+1/a=2,a=1/a
from AM>=GM,
a+1/a>=2√(a*1/a)
implies a+1/a>=2
here equality holds when a=1/a,which happens here
so a=1(note a=-1 doesnt satisfy the above eqn)
so sin x=cosec x=1
and thus sinnx+cosecnx=2