Question

Can someone help me ? i have no clue how to figure this out

Answer 1

$\cos \: t ( \sec \: t - \cos \: t) = \sin^{2}t \\ \\ 1. \: 1 - \frac{ \cos \: t }{ \sec \: t } = \sin^{2} t \\ 2. \: 1 - \cos \: t \cos \: t = \sin^{2} t \\ 3. \: 1 - \cos^{2} t = \sin^{2} t \\ 4. \: 1 - \cos^{2} t - \sin^{2} t = 0 \\ 5. \: 1 - 1 = 0 \\ 6. \: 0 = 0 \\ 7. \: infinitely \: many \: solutions$

Answer 2

$\cos t (\sec t - \cos t) = \sin^2 t$

$\cos t (\dfrac{1}{\cos t} - \cos t) = \sin^2 t$

$\dfrac{\cos t}{\cos t} - \cos^2 t = \sin^2 t$

$1 - \cos^2 t = \sin^2 t$

Use the identity: $\sin^2 t + \cos^2 t = 1$ and solve for $\sin^2 t$. You get: $\sin^2 t = 1 - \cos^2 t$

Do the substitution on the left side to get:

$\sin^2 t = \sin ^2 t$