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

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

\csc x − \sin x=(\cot x)(\cos x) \csc x − \sin x=(\cot x)(\cos x)

So starting with LHS:
\frac{1}{\sin x} − \sin x &= \frac{1 − (\sin x)^2}{ \sin x } \\
&= \frac{(\cos x)^2 }{ \sin x }
\frac{1}{\sin x} − \sin x &= \frac{1 − (\sin x)^2}{ \sin x } \\
&= \frac{(\cos x)^2 }{ \sin x }

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



3 Answers


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




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



Please accept an answer if it is helpful
– Blue
2 days ago

I’ll solve it specifically

Yes, start with LHS-
cos\,x−sin\,x=\dfrac{1}{sin\,x}-sin\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because cos\,x=\dfrac{1}{sin\,x}\}cos\,x−sin\,x=\dfrac{1}{sin\,x}-sin\,x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{\because cos\,x=\dfrac{1}{sin\,x}\}
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.