Question

1-tan^2x+tan^4x-tan^6x+...

Answer 1

its alternationg Geometric series is'nt it?

Answer 2

so what do u want sum of the series?

Answer 3

@baaymaax

Answer 4

\[f(x)=1-\tan^2x+\tan^4x-\tan^6x+.....\\first~term=a=1\\common~ratio=r=\frac{ -\tan^2x }{ 1 }=-\tan^2x\]
so sum of the series\[S=\frac{ a }{ 1-r }=\frac{ 1 }{ 1-(-\tan^2x) }=\frac{ 1 }{ 1+\tan^2x }=\frac{ 1 }{ \sec^2x }=\cos^2x\]