Question

Can someone help me please?
Verify the identity.
(Sinx - tanx)^2 = (tan^2x)(cosx - 1)^2

Answer 1

I figures starting with the left side would be easier, so after u expand it to sin^2x - 2sinxtanx + tan^2x

Answer 2

Since the right side is 1 term, we probably want the left side to be 1 term as well.

Answer 3

\[(\sin x-\frac{\sin x}{\cos x})^2=(\frac{\sin x \cos x-\sin x}{\cos x})^2=\]
\[\frac{\sin ^2 \cos ^2x-2\sin ^2x \cos x+\sin ^2x}{\cos ^2x}=\]
\[\frac{\sin ^2x(\cos ^2x-2\cos x+1)}{\cos ^2x}=\frac{\sin ^2x}{\cos ^2x}(\cos x-1)^2\]

Answer 4

Oh that makes sense. Thank you so much.

Answer 5

yw