Question

prove secθ sinθ/ tanθ+cotθ = sin^2θ

Answer 1

NOTE: \[\sec(\theta) = \frac{ 1 }{ \cos(\theta) }, \tan(\theta) = \frac{ \sin(\theta) }{ \cos(\theta) }, \cot(\theta) = \frac{ \cos(\theta) }{ \sin(\theta) }\]

Answer 2

Therefore, you have: \[\frac{\frac{ 1 }{ \cos(\theta)} \times \sin(\theta) }{ \frac{ \sin(\theta) }{ \cos(\theta) }\times \frac{ \cos(\theta) }{ \sin(\theta) } }\]

Answer 3

sorry, the denom should be + not x.

Answer 4

\[\frac{ \frac{ 1 }{ \cos(\theta) } \times \sin(\theta)}{ \frac{ \sin(\theta) }{ \cos(\theta) } + \frac{ \cos(\theta) }{ \sin(\theta) } }= \frac{ \frac{ \sin(\theta) }{ \cos(\theta) } }{ \frac{ \sin^2(\theta) }{ \cos(\theta)\sin(\theta) } + \frac{ \cos^2(\theta) }{ \cos(\theta)\sin(\theta) } }\]

Answer 5

I'm sure you can figure the rest out from here. I gave you more than I should have.

Answer 6

just for reference sin ^2 deita+ cos^2 (dita)=1

Answer 7

yes.