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saveliy_v [14]
3 years ago
14

Prove that sinxtanx=1/cosx - cosx

Mathematics
1 answer:
maks197457 [2]3 years ago
5 0

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

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