Question

if you need to verify a trig function and are given cot^2x - cos^2x = cot^2xcos^2x how do you solver

Answer 1

\[\cot^2 \theta -\cos^2 \theta = \cot^2 \theta \times \cos^2 \theta\]

Answer 2

expand so it is all sines and cosines

Answer 3

then work with one side and try to get the other side

Answer 4

can you do it from here?

Answer 5

\[\large RHS=\cot^2 \theta \times \cos^2 \theta\]
\[\large =\frac{\cos^2 \theta}{\tan^2 \theta}\]
\[\large =\cos^2 \theta \times \frac{\cos^2 \theta}{\sin^2 \theta}\]
\[\large =(1-\sin^2 \theta)\times \frac{\cos^2 \theta}{\sin^2 \theta}\]
\[\large =\frac{\cos^2 \theta}{\sin^2 \theta}-\cos^2 \theta\]
\[\large =\cot^2 \theta -\cos^2 \theta\]
\[\large =LHS\]

Answer 6

Nice work @Azteck!