Question

Prove the Identity:
1+sinx-cos^2x/1+sinx=sinx

Answer 1

Use the identity \(\sin^2x+\cos^2x=1\).

Or, equivalently, \(\sin^2x=1-\cos^2x\).

\[\begin{align*}\frac{1+\sin x-\cos^2x}{1+\sin x}&=\frac{1-\cos^2x+\sin x}{1+\sin x}\\
&=\frac{\sin^2x+\sin x}{1+\sin x}
\end{align*}\]
Then factor out a power of \(\sin x\) from the numerator:
\[\begin{align*}\frac{1+\sin x-\cos^2x}{1+\sin x}&=\frac{1-\cos^2x+\sin x}{1+\sin x}\\
&=\frac{\sin^2x+\sin x}{1+\sin x}\\
&=\frac{\sin x(\sin x+1)}{1+\sin x}
\end{align*}\]
This is no different from what @Ahmad1 said the first time you asked this question, by the way.

Answer 2

From here, \(\dfrac{\sin x+1}{1+\sin x}=\dfrac{1+\sin x}{1+\sin x}=1\), so you're left with just \(\sin x\).

Answer 3

Wow........Muuuuuuuuch clearer. Thanks

Answer 4

you're welcome