Trigonometry identity $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$ 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.
 A: $RHS = \dfrac{\cos^3 x}{\sin^2 x}+ \cos x=\dfrac{\cos^3 x+\cos x \sin^2x}{\sin^2 x}=\dfrac{\cos x (\sin^2x+\cos^2 x)}{\sin^2 x}=\dfrac{\cos x}{\sin^2x}$ BECAUSE $\sin^2 x + \cos^2 x =1$...
Now, $\dfrac{\cos x}{\sin^2x}=\dfrac{\cos x}{\sin x} \dfrac{1}{\sin x}=\cot x \csc x.$
A: $$\frac{\cos ^3x}{\sin^2 x}+\cos x$$
$$\cos x\left(\cot^2x+1\right)=\cos x\left(\csc^2x\right)= \frac{\cos x}{\sin x}\left(\csc\right)=\cot x \csc x$$
A: HINT:
$$\csc x\cot x-\cos x=\cos x\left(\dfrac1{\cos^2x}-1\right)$$
$$\dfrac1{\cos^2x}-1=\dfrac{1-\cos^2x}{\cos^2x}=\dfrac{\sin^2x}{\cos^2x}$$
A: Notice, we have to prove $$\csc x\cot x=\frac{\cos^3x}{\sin^2x}+\cos x$$ Taking RHS as follows $$RHS=\frac{\cos^3x}{\sin^2x}+\cos x$$ $$=\cos x\left(\frac{\cos x}{\sin x}\right)^2+\cos x$$ $$=\cos x\cot^2x+\cos x$$ $$=\cos x(1+\cot^2x)$$ $$=\cos x(\csc^2x)$$ $$=\csc x\frac{\cos x}{\sin x}$$$$=\csc x\cot x=\color{blue}{LHS}$$
A: $$\csc(x)\cot(x)=\frac{\cos^3(x)}{\sin^2(x)}+\cos(x)\Longleftrightarrow$$
$$\csc(x)\cot(x)=\frac{\cos^3(x)+\cos(x)\sin^2(x)}{\sin^2(x)}\Longleftrightarrow$$
$$\csc(x)\cot(x)\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\frac{1}{\sin(x)}\cdot\frac{\cos(x)}{\sin(x)}\cdot\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\cos(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\cos(x)=\cos(x)(\cos^2(x)+\sin^2(x))\Longleftrightarrow$$
$$1=\cos^2(x)+\sin^2(x)\Longleftrightarrow$$
$$1=1$$
The left hand side and right hand side are identical -> identity has been verified
