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 - ( \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 ((1)/(\cos t) - \cos t) = \sin^2 t`

`(\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`