Question

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

Answer 1

Hint: In this question we have to use the formula of $\cos 2A$ in terms of ${\sin ^2}A\,\,and\,\,{\cos ^2}A$ , then we have to do some simplification and cancel out common terms to get to the end result. There is a great importance of formulas in trigonometry. Use of formulae and bringing both numerator and denominator in the same terms is the most important thing.
Formula used: $\left( 1 \right)\,\,1 + \cos 2A = 2{\cos ^2}A$
$\left( 2 \right)\,\,1 - \cos 2A = 2{\sin ^2}A$
$\left( 3 \right)\,\,\sin 2A = 2\sin A\cos A$

Complete step-by-step answer:
In the given question, we have given a equation in which there is a use of formulas of $\cos 2A$ in terms of ${\sin ^2}A\,\,and\,\,{\cos ^2}A$ and also there is a use of formula of $\sin 2A$ in terms of $\sin A\,\,and\,\,\cos A$ .
Now, we have
$ \Rightarrow \dfrac{{1 + \cos 2A + \sin 2A}}{{1 - \cos 2A + \sin 2A}}$
Now we will use the formula $\,\,1 + \cos 2A = 2{\cos ^2}A$ in the numerator and $\,1 - \cos 2A = 2{\sin ^2}A$ in the denominator.
On applying the formulas, we get
$ \Rightarrow \dfrac{{2{{\cos }^2}A + \sin 2A}}{{2{{\sin }^2}A + \sin 2A}}$
Now use the formula $\,\sin 2A = 2\sin A\cos A$ in both numerator and denominator.
$ \Rightarrow \dfrac{{2{{\cos }^2}A + 2\sin A\cos A}}{{2{{\sin }^2}A + 2\sin A\cos A}}$
Taking common $2\cos A$ in the numerator and $2\sin A$ in the denominator.
$ \Rightarrow \dfrac{{2\cos A(\cos A + \sin A)}}{{2\sin A(\sin A + \cos A)}}$
  Now dividing the common term in numerator and denominator.
$ \Rightarrow \dfrac{{\cos A}}{{\sin A}}$
We know that $\dfrac{{\cos A}}{{\sin A}} = \cot A$.$\cot A$
So,
$ \Rightarrow \cot A$
Therefore, the value of $\dfrac{{1 + \cos 2A + \sin 2A}}{{1 - \cos 2A + \sin 2A}}$ is $\cot A$ .
So, the correct answer is “ $\cot A$”.

Note: The double angle trigonometric identities can be derived from the additional trigonometric identities. Basically, all you need to do is change all of the B’s to A’s.
\[ \Rightarrow sin{\text{ }}\left( {A + B} \right){\text{ }} = {\text{ }}sinAcosB{\text{ }} + {\text{ }}cosAsinB\]
Now changing the B’s to A’s you get:
\[ \Rightarrow sin\left( {A + A} \right){\text{ }} = {\text{ }}sinAcosA{\text{ }} + {\text{ }}cosAsinA\]
This simplifies down to: \[{\mathbf{sin}}\left( {{\mathbf{2A}}} \right){\text{ }} = {\text{ }}{\mathbf{2sinAcosA}}\]
Likewise, we can also find the double angle identities for other trigonometric functions also using existing identities.