Question

trigs ques( see in comments)

Answer 1

\[\frac { \cos { \theta } }{ 1-\tan { \theta } } +\frac { \sin ^{ 2 }{ \theta } }{ \sin { \theta } -\cos { \theta } } =\sin { \theta } -\cos { \theta } \]

Answer 2

Multiply everything by \[
\sin\theta-\cos\theta
\]See what happens.

Answer 3

\[LHS=\quad \frac { \cos { \theta } }{ 1-\quad \tan { \theta } } +\quad \frac { \sin ^{ 2 }{ \theta } }{ \sin { \theta } -\cos { \theta } } \\ \quad \quad \quad =\quad \frac { \cos { \theta } }{ 1-\frac { \sin { \theta } }{ \cos { \theta } } } \quad +\quad \frac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } -\sin { \theta } } \\ =\frac { \cos ^{ 2 }{ \theta } }{ \cos { \theta } -\sin { \theta } } +\frac { \sin ^{ 2 }{ \theta } }{ \sin { \theta } -\cos { \theta } } =\frac { \cos ^{ 2 }{ \theta } }{ \cos { \theta } -\sin { \theta } } -\frac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } \sin { \theta } } \\ =\frac { \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } }{ \cos { \theta } -\sin { \theta } } =\frac { (\cos { \theta } \sin { \theta } )(\cos { \theta } \sin { \theta } ) }{ \cos { \theta } -\sin { \theta } } \\ =\cos { \theta } +\sin { \theta } =R.H.S.\]

I think you under stood!

Answer 4

Well I would prefer first to put \(\tan \theta = \cfrac{\sin \theta}{\cos \theta}\)

Answer 5

Thanks @Rohangrr

Answer 6

No problem!

Answer 7

Well @Rohangrr great work , but it would have been excellent if you have let @Dr.Professor do it on himself.

We need to concentrate more on the asker and our interaction rather than one post soln or answer.

But though, great work

Answer 8

There are mistakes in your work @Rohangrr

Answer 9

@Rohangrr

check out some..