Question

simplify the left side to equal the right side: cos^2x(1+cot^2x)=cot^2x

Answer 1

help me please

Answer 2

I dont know trig!

Answer 3

ok

Answer 4

Change the trig terms

Answer 5

how?

Answer 6

\[\cos^2x(1+\cot^2x)\]
\[\cos^2x(\csc^2x)\]
\[\cos^2x(\frac{ 1 }{ \sin^2x) }\]

Answer 7

multiply it out and guess what you get? :P

Answer 8

if u remember the identity of this :
1+cot^2x = csc^2 (x)

so, the left side becomes
cos^2x * csc^2 (x)
= cos^2 x * 1/sin^2 x
= cos^2 x / sin^2 x
= (cosx/sinx)^2
= cot^2 x

QED

Answer 9

If you don't remember your identities (I never do), here it is the basic way:

\[\cos^2x(1+\cot^2x) = \cot^2x\]\[\cot x = 1/\tan x = \frac{1}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x}\]
\[\cos^2x(1+\frac{\cos^2x}{\sin^2x}) = \frac{\cos^2x}{\sin^2x}\]\[\cos^2x(\frac{\sin^2x}{\sin^2x} + \frac{\cos^2x}{\sin^2x}) = \frac{\cos^2x}{\sin^2x}\]\[\cos^2x(\frac{\sin^2x+\cos^2x}{\sin^2x}) = \frac{\cos^2x}{\sin^2x}\]but \(\sin^2x+\cos^2x=1\) so that becomes
\[\cos^2x(\frac{1}{\sin^2x}) = \frac{\cos^2x}{\sin^2x}\]and our work is done.