Question

How do you prove that $- \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}$?

Answer 1

We directly use the formula of $\tan 2x$ after we convert the cotangent function in left hand side of the equation to tangent function using the formula $\cot x = \dfrac{1}{{\tan x}}$. Substitute the value of $\tan 2x$ in the denominator and multiply negative sign wherever required.

• $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$

Complete step-by-step solution:
We have to prove that $- \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}$
We take left hand side of the equation and convert the cotangent function into tangent function
Left hand side of the equation is $- \cot 2x$
We use the conversion of $\cot x = \dfrac{1}{{\tan x}}$ to write the left hand side of the equation in terms of tangent
$\Rightarrow - \cot 2x = \dfrac{{ - 1}}{{\tan 2x}}$
Now we know the formula for $\tan 2x = \dfrac{{1 - {{\tan }^2}x}}{{2\tan x}}$
$\Rightarrow - \cot 2x = \dfrac{{ - (1 - {{\tan }^2}x)}}{{2\tan x}}$
Multiply negative sign outside the bracket with terms inside the bracket
$\Rightarrow - \cot 2x = \dfrac{{ - 1 - ( - {{\tan }^2}x)}}{{2\tan x}}$
Use the concept that when a negative number is multiplied by a negative number, we get the result as a positive number.
$\Rightarrow - \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}$
Since the term on right hand side of the equation is same as right hand side of the equation, we can write LHS $=$ RHS
Hence Proved

Note: Alternate method:
We can convert the terms given on the right hand side of the equation in terms of sine and cosine using the formula $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Later use the formula ${\cos ^2}x - {\sin ^2}x = \cos 2x$ and $2\sin x\cos = \sin 2x$ to convert the angle from x to 2x.
Right hand side of the equation is $\dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}$
Substitute the value of $\tan x = \dfrac{{\sin x}}{{\cos x}}$ in both numerator and denominator of the equation on right hand side
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1}}{{\dfrac{{2\sin x}}{{\cos x}}}}$
Take LCM of the terms in numerator of the equation
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}}}}{{\dfrac{{2\sin x}}{{\cos x}}}}$
Write the fraction in simpler form
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}} \times \dfrac{{\cos x}}{{2\sin x}}$
Cancel same factors from numerator and denominator of the fraction
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\cos x}} \times \dfrac{1}{{2\sin x}}$
Now we can write the product of two fractions as one fraction
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{2\sin x\cos x}}$
Take -1 common from numerator of the fraction
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{ - ({{\cos }^2}x - {{\sin }^2}x)}}{{2\sin x\cos x}}$
Now we know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$ and $2\sin x\cos = \sin 2x$.
Substitute the value of ${\cos ^2}x - {\sin ^2}x = \cos 2x$ in numerator of the fraction and $2\sin x\cos = \sin 2x$ in the denominator of the fraction.
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{ - \cos 2x}}{{\sin 2x}}$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$ then changing the angle from x to 2x we can write $\cot 2x = \dfrac{{\cos 2x}}{{\sin 2x}}$
$\Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = - \cot 2x$
Since the term on right hand side of the equation is same as left hand side of the equation, we can write LHS $=$ RHS
Hence Proved