Question

Prove the identity

Answer 1

\[\frac{ \sin \theta + \tan \theta }{ \cot \theta \csc \theta} = \sin \theta \tan \theta\]

Answer 2

that should read \[\cot \theta + \csc \theta \] in the denominator

Answer 3

Numerator: sin x + sin x/cos x = (sin x*cos x + sin x)/cos x =
= sin x (1 + cos x)/cos x
Denominator: cos x/sin x + 1/sin x = (cos x + 1)/sin x
Finally, after simplification by (cos x + 1)
y = sin x*sin x/cos x = sin x*tan x.

Answer 4

\[ \tan \theta=\frac{\sin\theta}{\cos\theta}\]\[\cot\theta=\frac{\cos\theta}{\sin\theta}\]\[\csc\theta=\frac{1}{\sin\theta}\]
So:
\[\frac{ \sin \theta + \tan \theta }{ \cot \theta +\csc \theta} =\large\frac{\sin\theta+\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\sin\theta}+\frac{1}{\sin\theta}} \\
\, \\

=\large\frac{\frac{\sin\theta\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta+1}{\sin\theta}} \\
\, \\
=\frac{\sin\theta\cos\theta+\sin\theta}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta+1}\\
\, \\
=\frac{\sin\theta(\cos\theta+1)}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta+1}\\
\, \\
=\frac{\sin\theta}{\cos\theta}\cdot\sin\theta\\
=\tan\theta\sin\theta\]

Answer 5

@kirbykirby what happened between \[\frac{ \sin \theta \cos \theta + \sin \theta }{ \cos \theta} * \frac{ \sin \theta }{ \cos \theta +1 }\] and \[\frac{ \sin \theta (\cos \theta +1) }{ \cos \theta } * \frac{ \sin \theta }{ \cos \theta + 1 }\]

Answer 6

in the first line, I factored out a \(\sin\theta\) term
If you think of "x" as \(\sin\theta\) and "y" as \(\cos\theta\), then what you have in the first fraction is \[\frac{xy+x}{y}=\frac{x(y+1)}{y}\]
Now rethink this by substituting x for sin, and y for cos.