Question

prove sin theta tan theta = 1-cos squared theta/cos theta

Answer 1

\[\sin \theta \tan \theta = \frac{1-\cos^2 \theta}{\cos \theta}\]
\[LHS=\sin \theta \tan \theta =\sin \theta \times \frac{\sin \theta} {\cos \theta} =\frac{\sin^2 \theta} {\cos \theta}\]
\[=\frac{1-\cos^2 \theta} {\cos \theta} =RHS\]
Hence LHS=RHS
i.e. \[\sin \theta \tan \theta = \frac{1-\cos^2 \theta}{\cos \theta}\]
Thus proved.

@shawnray

Answer 2

Reasons:

\[\tan \theta = \frac{\sin \theta}{\cos \theta} \]

\[\sin^2 \theta = 1- \cos^2 \theta\]

Answer 3

@shawnray

Answer 4

thank you