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How to prove that $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$?

I tried manupulating the left hand side but ended up in $\frac{\cos x}{\sin^2 x}$.

Can someone show me? Thanks in advance.

Mathxx
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5 Answers5

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$RHS = \dfrac{\cos^3 x}{\sin^2 x}+ \cos x=\dfrac{\cos^3 x+\cos x \sin^2x}{\sin^2 x}=\dfrac{\cos x (\sin^2x+\cos^2 x)}{\sin^2 x}=\dfrac{\cos x}{\sin^2x}$ BECAUSE

$\sin^2 x + \cos^2 x =1$...

Now, $\dfrac{\cos x}{\sin^2x}=\dfrac{\cos x}{\sin x} \dfrac{1}{\sin x}=\cot x \csc x.$

Chiranjeev_Kumar
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curious_mind
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$$\frac{\cos ^3x}{\sin^2 x}+\cos x$$ $$\cos x\left(\cot^2x+1\right)=\cos x\left(\csc^2x\right)= \frac{\cos x}{\sin x}\left(\csc\right)=\cot x \csc x$$

Panglossian Oporopolist
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HINT:

$$\csc x\cot x-\cos x=\cos x\left(\dfrac1{\cos^2x}-1\right)$$

$$\dfrac1{\cos^2x}-1=\dfrac{1-\cos^2x}{\cos^2x}=\dfrac{\sin^2x}{\cos^2x}$$

lab bhattacharjee
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Notice, we have to prove $$\csc x\cot x=\frac{\cos^3x}{\sin^2x}+\cos x$$ Taking RHS as follows $$RHS=\frac{\cos^3x}{\sin^2x}+\cos x$$ $$=\cos x\left(\frac{\cos x}{\sin x}\right)^2+\cos x$$ $$=\cos x\cot^2x+\cos x$$ $$=\cos x(1+\cot^2x)$$ $$=\cos x(\csc^2x)$$ $$=\csc x\frac{\cos x}{\sin x}$$$$=\csc x\cot x=\color{blue}{LHS}$$

Harish Chandra Rajpoot
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$$\csc(x)\cot(x)=\frac{\cos^3(x)}{\sin^2(x)}+\cos(x)\Longleftrightarrow$$ $$\csc(x)\cot(x)=\frac{\cos^3(x)+\cos(x)\sin^2(x)}{\sin^2(x)}\Longleftrightarrow$$ $$\csc(x)\cot(x)\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$ $$\frac{1}{\sin(x)}\cdot\frac{\cos(x)}{\sin(x)}\cdot\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$ $$\cos(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$ $$\cos(x)=\cos(x)(\cos^2(x)+\sin^2(x))\Longleftrightarrow$$ $$1=\cos^2(x)+\sin^2(x)\Longleftrightarrow$$ $$1=1$$

The left hand side and right hand side are identical -> identity has been verified

Jan Eerland
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