Prove that the following is an identity

Question

$\frac{\sin(x)}{\cot(x)} + \cos(x) = \sec(x)$

Note that this involves identities, so you can't treat it like an equation and multiply both sides by a number. ( when I multiplied both sides by $\cos(x)$, it turned out I could easily use the pathagorean identity, but the solution obtained was WRONG)

Answer 1

$$\frac{\sin}{\cot}+\cos =\frac{\sin}{\frac{\cos}{\sin}}+\cos = \frac{\sin^2}{\cos} + \cos = \frac{\overbrace{\sin^2+\cos^2}^{=1}}{\cos} = \frac{1}{\cos} = \sec$$

Answer 2

Guessing you mean "equivalent to a well-known identity", here's a helpful hint:

Hint: $$\dfrac{\sin x}{\cot x}+\cos x=\sec x\\ \implies \dfrac{\sin^2 x}{\cos x}+\cos x=\dfrac{1}{\cos x}\\ \implies \dfrac{\sin^2 x}{\cos x}+\dfrac{\cos^2 x}{\cos x}=\dfrac{1}{\cos x}$$

Answer 3

Working from the expression on the left, we have:

$$\frac{\sin x}{\cot x}+\cos x=\frac{\sin x+\cos x\cot x}{\cot x} =\frac{\sin x+\cos x\cdot \displaystyle\frac{\cos x}{\sin x}}{\cot x}=$$ $$=\frac{\displaystyle\frac{\sin^2x+\cos^2x }{\sin x}}{\cot x}=\frac{1}{\sin x}\cdot \frac{1}{\cot x}=\frac{1}{\sin x}\cdot \frac{\sin x}{\cos x}=\frac{1}{\cos x}=\sec x. $$

Then, the equality follow, i.e., $$ \frac{\sin x}{\cot x}+\cos x=\sec x.$$