Question

How do you prove $\dfrac{{2\tan x}}{{1 + {{\tan }^2}x}} = \sin 2x$?

Answer 1

Hint: In order to proof the above statement ,take the left hand side of the equation and put $\tan x = \dfrac{{\sin x}}{{\cos x}},{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}},$ now taking LCM and combining terms, you will get \[{\sin ^2}x + {\cos ^2}x\] in the numerator put it equal to 1 according to the identity then simplifying further will give your final result which is equal to right-hand side of the equation.

Complete step by step answer:
To prove: $\dfrac{{2\tan x}}{{1 + {{\tan }^2}x}} = \sin 2x$
Proof:
Taking Left-hand Side of the equation,
 $ \Rightarrow \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$
As we know that $\tan x$ is equal to the ratio of $\sin x$ to $\cos x$ and similarly $\cot x$is equal to the ratio of $\cos x$ to $\sin x$. In simple words, $\tan x = \dfrac{{\sin x}}{{\cos x}},$ and if we square on both sides of this rule we get${\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}},$
Putting these values in the above equation, we get
\[ \Rightarrow \dfrac{{2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}\]
Simplifying further by taking LCM as \[{\cos ^2}x\], we get
\[ \Rightarrow \dfrac{{2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}}} \right)}}\]
From the identity of trigonometry which states that sum of squares of sine and cosine is equal to 1 i.e. \[{\sin ^2}x + {\cos ^2}x = 1\]. Replacing in the equation, we get
\[
   \Rightarrow \dfrac{{2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}} \\
   \Rightarrow 2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)({\cos ^2}x) \\
   \Rightarrow 2\sin x\cos x \\
 \]
From the formula sine of double angle $\sin 2x = 2\sin x\cos x$, rewriting the equation we get
\[ \Rightarrow \sin 2x\]
$\therefore LHS = \sin 2x$
Taking Right-hand Side part of the equation
$RHS = \sin 2x$
$\therefore LHS = RHS$
Hence, proved.

Note: 1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Even Function – A function $f(x)$ is said to be an even function, if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function, if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, $\sin \theta $ and $\tan \theta $ and their reciprocals, $\cos ec\theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.