Question

Prove that sinxtanx=1/cosx - cosx
$\sin(x) \tan(x) = \frac{1}{ \cos(x) } - \cos(x)$
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Answer 1

Answer:

See below

Step-by-step explanation:

We want to prove that

$\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}$

Taking the RHS, note

$\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}$

Remember that

$\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)$

Therefore,

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

Once

$\dfrac{\sin(x)}{\cos(x)} = \tan(x)$

Then,

$\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)$

Hence, it is proved