Question

How do you verify the identity \[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\] ?

Answer 1

Hint: In this question, we need to prove that \[\dfrac{2\tan x}{1 + \tan^{2}x}\] is equal to \[sin2x\] . Sine , cosine and tangent are the basic trigonometric functions .Sine is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle . Similarly tangent is nothing but a ratio of the opposite side of a right angle to the adjacent side of the right angle. With the help of the Trigonometric functions and ratios , we can prove that \[\dfrac{2\tan x}{1 + \tan^{2}x}\] is equal to \[sin2x\]
Identity used :
1.\[1 + \tan^{2}\theta = \sec^{2}\theta\]
Formula used :
1.\[\tan \theta\ = \dfrac{{\sin \theta}}{{\cos \theta}}\]
2.\[\sec \theta\ = \dfrac{1}{{\cos \theta}}\]

Complete step by step solution:
To prove,
\[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\]
First we can consider the left part of the given expression.
\[\Rightarrow\dfrac{2\tan x}{1 + \tan^{2}x}\]
We know that \[\tan \theta\ = \dfrac{{\sin \theta}}{{\cos \theta}}\]
By replacing \[x\] in the place of \[\theta\], we get, \[{\tan }x = \dfrac{\sin x}{\cos x}\]
\[\Rightarrow \dfrac{2\tan x}{1 + \tan^{2}x} = \dfrac{2\left( \dfrac{\sin x}{\cos x} \right)}{1 + \tan^{2}x}\]
From the trigonometric identity \[1 + \tan^{2}\theta = sec^{2}\theta\] ,
We get,
\[= \dfrac{2\left( \dfrac{\sin x}{\cos x} \right)}{\sec^{2}x}\]
We know that \[\sec \theta = \dfrac{1}{{\cos \theta}}\]
\[= \dfrac{2\left( \dfrac{{\sin x}}{{\cos x}} \right)}{\dfrac{1}{\cos^{2}x}}\]
\[= 2\left( \dfrac{{\sin x}}{{\cos x}} \right) \times \dfrac{\cos^{2}x}{1}\]
By simplifying,
We get,
\[= 2(\sin x \times cosx)\]
From the trigonometry formula, \[2\sin x \cos x = \sin 2x\]
We get,
\[= sin2x\]
Thus we get the left part of the expression.
We have proved \[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\]
Hence proved.
Final answer :
We have proved the identity \[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\]

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions and ratios . Trigonometric functions are also known as circular functions or geometrical functions.
Alternative solution :
We can also prove this by considering the right part of the given expression first.
To prove,
\[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\]
First we can consider the right part of the given expression.
\[\Rightarrow sin2x\]
We can rewrite \[2x\] as \[x + x\] ,
\[\Rightarrow sin(x + x)\]
We know that
\[\sin \left( a + b \right) = \sin a \cos b + \cos a \sin b\]
Here \[a = b = x\]
Thus we get,
\[\Rightarrow \sin x\ cosx\ + \ cosx\ \sin x\]
By adding,
We get ,
\[\Rightarrow 2\sin x\ cosx\]
On dividing the term by \[\cos^{2}x + \sin^{2}x\] , since we know
That the value of \[\sin^{2}\theta + \cos^{2}\theta = 1\]
We get ,
\[\Rightarrow \dfrac{2\sin x \cos x}{\cos^{2}x + \sin^{2}x}\]
On dividing each and every terms in the numerator and denominator by \[\cos^{2}x\ \]
\[\Rightarrow \dfrac{\dfrac{2\sin x \cos x}{\cos^{2}x}}{\left\{ \left( \dfrac{\cos^{2}x}{\cos^{2}x} \right) + \left( \dfrac{\sin^{2}x}{\cos^{2}x} \right) \right\}}\]
By simplifying,
We get,
\[\Rightarrow \dfrac{2\dfrac{{\sin x}}{{\cos x}}}{1 + \left( \dfrac{\sin^{2}x}{\cos^{2}x} \right)}\]
We know that \[\tan \theta\ = \dfrac{{\sin \theta}}{{\cos \theta}}\]
\[\Rightarrow \dfrac{2{\tan x}}{1 + \\tan^{2}x}\]
Thus we get the left part of the expression.
We have proved \[\dfrac{2\tan x}{1 + \tan^{2}x} = sin2x\]