Question

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

Answer 1

Hint: This problem comes under trigonometry. Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. It is used to find the angles and missing sides with the help of trigonometric ratios. In general we solve trigonometry oriented sums with trigonometry ratios and trigonometric identities. Here we solve using some trigonometric identities of 2A angles.

Formula used: $\operatorname{Sin} 2x = 2\sin x\cos x$
$\cos 2x = 2{\cos ^2}x - 1$
\[\dfrac{{\sin x}}{{\cos x}} = \tan x\]

Complete step-by-step solution:
Let us consider the left hand side of equation
\[ \Rightarrow \dfrac{{\sin 2x}}{{1 + \cos 2x}}\]
Now apply the formula for required places mentioned in formula used we get,
\[ \Rightarrow \dfrac{{2\sin x\cos x}}{{1 + 2{{\cos }^2}x - 1}}\]
Now cancelling numerals on denominator, we get
\[ \Rightarrow \dfrac{{2\sin x\cos x}}{{2{{\cos }^2}x}}\]
\[ \Rightarrow \dfrac{{2\sin x\cos x}}{{2\cos x\cos x}}\]
Now cancelling 2 and \[\cos x\] on both numerator and denominator, we get
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}}\]
We have mentioned the formula mentioned in formula used, we get
\[ \Rightarrow \tan x\]
We obtain the right hand side of the equation which is the proof.
LHS = RHS.
Hence\[\dfrac{{\sin 2x}}{{1 + \cos 2x}} = \tan x\] is proved

Note: There are many trigonometric ratios and trigonometric angles and identities are there in the topic of trigonometric. There are many kinds of problems related to proving the left hand side of the equation to the right hand side. For that we need to solve using trigonometric ratios by taking any one side of the equation we need to simplify as we can if it is not enough to prove we need to solve simultaneously on both sides of the equations. Here we solved by taking LHS. With required formula and basic mathematical calculations.