Question

Help? Verify the identity..

Answer 1

\[\frac{ \cot^2 x }{ \csc x + 1 } = \frac{ 1 - \sin x }{ \sin x }\]

Answer 2

@surjithayer ?

Answer 3

\[\csc ^2x-\cot ^2 x=1,\csc ^2x-1=\cot ^2x\]
\[L.H.S.=\frac{ \csc ^2x-1 }{ \csc x+1 }=\frac{ \left( \csc x+1 \right)\left( \csc x-1 \right) }{ \left( \csc x+1 \right) }\]
\[=\csc x-1=\frac{ 1 }{ \sin x }-1=\frac{ 1-\sin x }{ \sin x }=R.H.S.\]

Answer 4

Only one question, just so I fully understand. How did you get from (1/sin x) -1 to (1-sin x)/sin x?

Answer 5

@surjithayer

Answer 6

\[\frac{ 1-\sin x }{ \sin x }=\frac{ 1 }{\sin x }-\frac{ \sin x }{ \sin x }=\frac{ 1 }{\sin x }-1\]

Answer 7

Oh... Okay... I think I got it now! :D Thank you so much! God bless!