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Question: Solve \[\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}}\]...

Solve (2tanx)(1+tan2x)\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}}

Explanation

Solution

since the denominator of the given problem contains squares of a trigonometric ratio, we can prove the given trigonometric identity by using the square relation or Pythagorean identity in the denominator. And in the numerator converting the trigonometric ratio tan in terms of sin and cos.

Complete step by step answer:
Let us consider the given problem (2tanx)(1+tan2x)\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}}
Since in the denominator we have a term containing square of a trigonometric ratio so let us use a square relation to convert 1+tan2x1 + {\tan ^2}x in terms of sec2x{\sec ^2}x then the above expression becomes
(2tanx)(1+tan2x)=(2tanx)(sec2x)\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}} = \dfrac{{(2\tan x)}}{{({{\sec }^2}x)}}
Now let us convert the numerator tanx\tan x in terms of sinx\sin x and cosx\cos x that is by using the quotient relation in the above equation we get
(2tanx)(1+tan2x)=(2sinxcosx)(sec2x)\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}} = \dfrac{{\left( {2\dfrac{{\sin x}}{{\cos x}}} \right)}}{{({{\sec }^2}x)}}
Since the numerator is in terms of sin and cos so let us convert denominator sec2x{\sec ^2}x in terms of cos then the above equation becomes
(2tanx)(1+tan2x)=(2sinxcosx)(1cos2x)\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}} = \dfrac{{\left( {2\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}}
On simplification the above equation becomes
(2tanx)(1+tan2x)=2sinxcosx×cos2x1\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}} = \dfrac{{2\sin x}}{{\cos x}} \times \dfrac{{{{\cos }^2}x}}{1}
Now on simplification
(2tanx)(1+tan2x)=2sinxcosx\dfrac{{(2\tan x)}}{{(1 + {{\tan }^2}x)}} = 2\sin x\cos x
Therefore, the value of the given expression is 2sinxcosx=sin2x2\sin x\cos x = \sin 2x.

Note: Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ.
The relation tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}} is called a quotient relation. In trigonometry, a trigonometric function can be divided by another trigonometric function and the quotient of them is also a trigonometric function. Hence, it is called the quotient identity of trigonometric functions.