# Proving the identity cscxâˆ′sinx=(cotx)(cosx)\csc xâˆ’\sin x = (\cot x)(\cos x)

We recently started Trigonometry and I was trying to solve this.

\csc x âˆ’ \sin x=(\cot x)(\cos x) \csc x âˆ’ \sin x=(\cot x)(\cos x)

So starting with LHS:
\begin{align}
\frac{1}{\sin x} âˆ’ \sin x &= \frac{1 âˆ’ (\sin x)^2}{ \sin x } \\
&= \frac{(\cos x)^2 }{ \sin x }
\end{align}\begin{align}
\frac{1}{\sin x} âˆ’ \sin x &= \frac{1 âˆ’ (\sin x)^2}{ \sin x } \\
&= \frac{(\cos x)^2 }{ \sin x }
\end{align}

I am stuck now and wanted to know how should I proceed. Is this much correct?

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Its done. Just split the numerator as:

\frac{\cos x . \cos x }{\sin x} \frac{\cos x . \cos x }{\sin x}

= \frac{\cos x }{\sin x} . \cos x = \frac{\cos x }{\sin x} . \cos x

= \cot x . \cos x = \cot x . \cos x

Proved!

Awesome. Got it now. Thanks a lot.
– user381255
2 days ago

– Blue
2 days ago

I’ll solve it specifically

cos\,xâˆ’sin\,x=(cot\,x)(cos\,x)cos\,xâˆ’sin\,x=(cot\,x)(cos\,x)
cos\,xâˆ’sin\,x=\dfrac{1}{sin\,x}-sin\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because cos\,x=\dfrac{1}{sin\,x}\}cos\,xâˆ’sin\,x=\dfrac{1}{sin\,x}-sin\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because cos\,x=\dfrac{1}{sin\,x}\}
=\dfrac{1-sin^2x}{sin\,x}=\dfrac{1-sin^2x}{sin\,x}=\dfrac{cos^2x}{sin\,x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because1-sin^2x=cos^2x\}=\dfrac{cos^2x}{sin\,x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because1-sin^2x=cos^2x\}
Write cos^2xcos^2x as-

cos\,x\cdot cos\,xcos\,x\cdot cos\,x
\dfrac{cos\,x\cdot cos\,x}{sin\,x}=\dfrac{cos\,x}{sin\,x}\cdot cos\,x\dfrac{cos\,x\cdot cos\,x}{sin\,x}=\dfrac{cos\,x}{sin\,x}\cdot cos\,x
cot\,x\cdot cos\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because\dfrac{cos\,x}{sin\,x}=cot\,x\}cot\,x\cdot cos\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because\dfrac{cos\,x}{sin\,x}=cot\,x\}

I loved these in my precalculus class. Let’s go.

You’re right but went one too far. Write:

\frac{\cos^2 x}{\sin x} = \cos x \cdot \frac{\cos x}{\sin x} \frac{\cos^2 x}{\sin x} = \cos x \cdot \frac{\cos x}{\sin x}

from whence the result follows.