Question

prove that:-
$\frac{ \sin(x) + \sin(2x) }{1 + \cos(x) + \cos(2x) } = \tan(x)$
​

Answer 1

Answer:-

To Prove:

$\sf \: \dfrac{ \sin \: x + \sin \: 2x}{1 + \cos \: x + \cos \: 2x} = \tan \: x$

Using sin 2x = 2 sin x cos x & cos 2x = 2cos² x - 1 we get,

$\sf \implies \dfrac{ \sin \: x + 2 \sin \: x .\cos \: x}{1 + 2 \cos ^{2} \: x \: - 1+ \cos \: x} = \tan \: x$

Taking sin x/cos x common in LHS we get,

$\sf \implies \: \dfrac{ \sin \: x(1 + 2 \cos \: x)}{ \cos \: x(1 + 2 \cos \: x)} = \tan \: x \\ \\ \sf \implies \: \dfrac{ \sin \: x}{ \cos \: x} = \tan \: x$

Now using sin x/cos x = tan x we get,

→ tan x = tan x

Hence,Proved.

Answer 2

$\underline{\underline{\red{To \: Prove:-}}}$

☞$\sf \: \dfrac{ \sin \: x + \sin \: 2x}{1 + \cos \: x + \cos \: 2x} = \tan \: x$

$\underline{\underline{\green{Proof}}}$

We know ,

• sin 2x = 2 sin x cos x
• cos 2x = 2cos² x - 1

Now,

$\begin{gathered}\sf :\implies \dfrac{ \sin \: x + 2 \sin \: x .\cos \: x}{1 + 2 \cos ^{2} \: x \: - 1+ \cos \: x} = \tan \: x \\ \\\end{gathered}$

$\begin{gathered}\sf :\implies \: \dfrac{ \sin \: x(1 + 2 \cos \: x)}{ \cos \: x(1 + 2 \cos \: x)} = \tan \: x \\ \\ \sf :\implies \: \dfrac{ \sin \: x}{ \cos \: x} = \tan \: x\end{gathered}$

Now ,

By sin x/cos x = tan x we get,

→ tan x = tan x

Hence proved.

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