Question

How to prove? $\tan A(1 + \sec 2A) = \tan 2A$

Answer 1

In this question we will use the double angle formulas and substitute them accordingly to simplify the expression so that we get the simplified expression which is equal to the right-hand side of the equation.

Formulae used:
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
${\cos ^2}\theta + {\sin ^2}\theta = 1$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta}}$
$\sin 2\theta = 2\sin \theta \cos \theta$

Complete step-by-step answer:
We have the expression as:
$\tan A(1 + \sec 2A) = \tan 2A$
On taking the left-hand side of the equation, we get:
$\Rightarrow \tan A(1 + \sec 2A)$
Now we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$therefore, we get:
$\Rightarrow \tan A\left( {1 + \dfrac{1}{{\cos 2A}}} \right)$
Now we know that $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$therefore, we get:
$\Rightarrow \tan A\left( {1 + \dfrac{1}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
on taking the lowest common multiple for the term, we get:
$\Rightarrow \tan A\left( {\dfrac{{1 + {{\cos }^2}A - {{\sin }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$therefore, on substituting this instead of $1$ in the numerator, we get:
$\Rightarrow \tan A\left( {\dfrac{{{{\sin }^2}A + {{\cos }^2}A + {{\cos }^2}A - {{\sin }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
on simplifying the numerator, we get:
$\Rightarrow \tan A\left( {\dfrac{{{{\cos }^2}A + {{\cos }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
On adding the terms in the numerator, we get:
$\Rightarrow \tan A\left( {\dfrac{{2{{\cos }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
Now we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$therefore, we get:
$\Rightarrow \left( {\dfrac{{\sin A}}{{\cos A}}} \right)\left( {\dfrac{{2{{\cos }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
On cancelling the terms, we get:
$\Rightarrow \sin A\left( {\dfrac{{2\cos A}}{{{{\cos }^2}A - {{\sin }^2}A}}} \right)$
Now on re-writing the value of ${\cos ^2}A - {\sin ^2}A$as $\cos 2A$, we get:
$\Rightarrow \sin A\left( {\dfrac{{2\cos A}}{{\cos 2A}}} \right)$
On rearranging the expression, we get:
$\Rightarrow \dfrac{{2\sin A\cos A}}{{\cos 2A}}$
Now we know that $\sin 2\theta = 2\sin \theta \cos \theta$ therefore, on using this, we get:
$\Rightarrow \dfrac{{\sin 2A}}{{\cos 2A}}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ therefore, the equation becomes:

$\Rightarrow \tan 2A$, which is the left-hand side of the expression, hence $\tan A(1 + \sec 2A) = \tan 2A$ is proved.

Note:
It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken.
The various trigonometric identities and formulae should be remembered while doing these types of sums. the various Pythagorean identities should also be remembered while doing these type of questions
To simplify any given equation, it is good practice to convert all the identities into $\sin$ and $\cos$ for simplifying.
If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.