Question

Simplify $\dfrac{{\cos 2A - 1}}{{\sin A}} + \dfrac{{\sin 2A \cdot \cos A}}{{\cos 2A + 1}}$.

Answer 1

Here, in the given question, we need to solve $\dfrac{{\cos 2A - 1}}{{\sin A}} + \dfrac{{\sin 2A \cdot \cos A}}{{\cos 2A + 1}}$. As we can see, we are given trigonometric ratios of angle $2A$ in the given trigonometric equation. At first, we will try to simplify the given trigonometric equation using identities of trigonometric ratios of angle $2A$. After this, we will cancel out the common terms to get our answer.

Formulas used:
$\cos 2A = 1 - 2{\sin ^2}A$
$\cos 2A = 2{\cos ^2}A - 1$
$\sin 2A = 2\sin A\cos A$
In the above written formulae it should be noted that the angle on the RHS is half of the angle on LHS.

Complete step by step answer:
We have,
$\dfrac{{\cos 2A - 1}}{{\sin A}} + \dfrac{{\sin 2A \cdot \cos A}}{{\cos 2A + 1}}$
As we know $\cos 2A = 1 - 2{\sin ^2}A$. Thus, we get $\cos 2A - 1 = - 2{\sin ^2}A$.
So, we will replace $\cos 2A - 1$ by $- 2{\sin ^2}A$.
As we know $\cos 2A = 2{\cos ^2}A - 1$. Thus, we get $\cos 2A + 1 = 2{\cos ^2}A$.
So, we will replace $\cos 2A + 1$ in the denominator by $2{\cos ^2}A$.
As we know $\sin 2A = 2\sin A\cos A$.
So, we will replace $\sin 2A$ by $2\sin A\cos A$.
So now, on substituting these values, we get
$\Rightarrow \dfrac{{ - 2{{\sin }^2}A}}{{\sin A}} + \dfrac{{2\sin A\cos A \cdot \cos A}}{{2{{\cos }^2}A}}$
This can also be written as,
$\Rightarrow \dfrac{{ - 2\sin A \cdot \sin A}}{{\sin A}} + \dfrac{{2\sin A\cos A \cdot \cos A}}{{2\cos A \cdot \cos A}}$
On canceling out common terms, we get
$\Rightarrow - 2\sin A + \sin A$
On subtraction of terms, we get
$\Rightarrow - \sin A$
Therefore, on solving $\dfrac{{\cos 2A - 1}}{{\sin A}} + \dfrac{{\sin 2A \cdot \cos A}}{{\cos 2A + 1}}$ we get $- \sin A$ as our answer.

Note:
These type of questions are completely based on substation of formulae. To solve this type of questions, one must know all the trigonometric formulae, especially trigonometric ratios formulae for half angle, double angle, triple angles, etc. We should take care of the calculations so as to be sure of our final answer.
Some formulas are written below:
Trigonometric ratios of angle $2A$ in terms of angle $A$.
1. $\sin 2A = 2\sin A\cos A$
2. $\cos 2A = {\cos ^2}A - {\sin ^2}A$
3. $\cos 2A = 2{\cos ^2}A - 1$ or, $1 + \cos 2A = 2{\cos ^2}A$
4. $\cos 2A = 1 - 2{\sin ^2}A$ or, $1 - \cos 2A = 2{\sin ^2}A$
5. $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
6. $\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$
7. $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
Trigonometric ratios of angle $3A$ in terms of angle $A$.
8. $\sin 3A = 3\sin A - 4{\sin ^3}A$
9. $\cos 3A = 4{\cos ^3}A - 3\cos A$
10 $\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$