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Question

Question: How do you prove \[\dfrac{{\sin 2x}}{{1 + \cos 2x}} = \tan x\]?...

How do you prove sin2x1+cos2x=tanx\dfrac{{\sin 2x}}{{1 + \cos 2x}} = \tan x?

Explanation

Solution

This question is related to the trigonometry, and we have to prove that the left hand side is equal to the right hand side of the expression, and this question can be solved by using trigonometric identities i.e.,sin2x=2sinxcosx\sin 2x = 2\sin x\cos x and cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1, and now by simplifying the expression that we will get by using the identities we will get the result which is on the right hand.

Complete step-by-step solution:
Given sin2x1+cos2x=tanx\dfrac{{\sin 2x}}{{1 + \cos 2x}} = \tan x,
Now we have to prove that left hand side is equal to the right hand side of the equation, now take the expression on the left hand side i.e.,
sin2x1+cos2x\Rightarrow \dfrac{{\sin 2x}}{{1 + \cos 2x}},
Now using the trigonometric identities i.e., sin2x=2sinxcosx\sin 2x = 2\sin x\cos x and cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1,
Now we have cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1,
By adding 1 to both sides of the identity, we get,
2cos2x1+1=1+cos2x\Rightarrow 2{\cos ^2}x - 1 + 1 = 1 + \cos 2x,
Now eliminating like terms we get,
2cos2x=1+cos2x\Rightarrow 2{\cos ^2}x = 1 + \cos 2x,
Now substituting the identities in the left hand side of the given expression we get,
sin2x1+cos2x=2sinxcosx2cos2x\Rightarrow \dfrac{{\sin 2x}}{{1 + \cos 2x}} = \dfrac{{2\sin x\cos x}}{{2{{\cos }^2}x}},
Now eliminating the like terms i.e., cosx\cos xand22 in both numerator and denominator, we get,
sin2x1+cos2x=sinxcosx\Rightarrow \dfrac{{\sin 2x}}{{1 + \cos 2x}} = \dfrac{{\sin x}}{{\cos x}},
We know from the trigonometric identity sinxcosx=tanx\dfrac{{\sin x}}{{\cos x}} = \tan x, we get,
sin2x1+cos2x=tanx\Rightarrow \dfrac{{\sin 2x}}{{1 + \cos 2x}} = \tan x,
Which is equal to the right hand side of the equation,
Hence proved.

\therefore By using identities we proved that expression sin2x1+cos2x\dfrac{{\sin 2x}}{{1 + \cos 2x}} is equal to tanx\tan x.

Note: An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
sin2x=1cos2x{\sin ^2}x = 1 - {\cos ^2}x,
cos2x+sin2x=1{\cos ^2}x + {\sin ^2}x = 1,
sec2xtan2x=1{\sec ^2}x - {\tan ^2}x = 1,
csc2x=1+cot2x{\csc ^2}x = 1 + {\cot ^2}x.
cos2xsin2x=12sin2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x,
cos2xsin2x=2cos2x1{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1,
sin2x=2sinxcosx\sin 2x = 2\sin x\cos x,
2cos2x=1+cos2x2{\cos ^2}x = 1 + \cos 2x,
tan2x=2tanx1tan2x\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}.