# Prove that the following is an identity

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

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)

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

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The question seems a lot like homework. Please do not reveal the complete answer - rather, give a hint. – Sanath K. Devalapurkar Apr 3 '14 at 0:05
@SanathDevalapurkar, I side with "Lord Soth". If people are stuck on problems of this difficulty, it is likely that no hints will help. It is best to show a complete solution as they will likely have many additional problems to do. I would only refrain from doing so, if - and it would then be obvious - they are asking a lot of questions. – Chris K Apr 3 '14 at 0:57
How is equation involving trig functions different from a normal equation? – user11355 Apr 3 '14 at 1:57
@user139024 Sorry, I do not understand what you exactly mean by your question. Can you elaborate? A trigonometric equation is no different than a normal equation, equation is an equation :) – Lord Soth Apr 3 '14 at 3:01
Notice if I multiply both sides by cosx, it would also work. But this is a invalid approach, why? – user11355 Apr 3 '14 at 11:03

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

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

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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., $$\frac{\sin x}{\cot x}+\cos x=\sec x.$$

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