Question

Prove that $\dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\cot A$ .

Answer 1

- Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like $\sin 2\theta =2\sin \theta \cos \theta$ , $\cos 2\theta +1=2{{\cos }^{2}}\theta$ and $1-\cos 2\theta =2{{\sin }^{2}}\theta$ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step solution -

Given:
We have to prove the following trigonometric equation:
$\dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\cot A$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following four formulas:
$\begin{aligned} & \sin 2\theta =2\sin \theta \cos \theta .................\left( 1 \right) \\\ & \cos 2\theta +1=2{{\cos }^{2}}\theta .................\left( 2 \right) \\\ & 1-\cos 2\theta =2{{\sin }^{2}}\theta ...................\left( 3 \right) \\\ & \dfrac{\cos \theta }{\sin \theta }=\cot \theta ............................\left( 4 \right) \\\ \end{aligned}$
Now, we will use the above four formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}$ .
Now, we will use the formula from the equation (1) to write $\sin 2A=2\sin A\cos A$ in the numerator and denominator of the term $\dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}$ . Then,
$\begin{aligned} & \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{2\sin A\cos A+1+\cos 2A}{2\sin A\cos A+1-\cos 2A} \\\ \end{aligned}$
Now, we will use the formula from the equation (2) to write $1+\cos 2A=2{{\cos }^{2}}A$ and formula from the equation (3) to write $1-\cos 2A=2{{\sin }^{2}}A$ in the above equation. Then,
$\begin{aligned} & \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{2\sin A\cos A+1+\cos 2A}{2\sin A\cos A+1-\cos 2A} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{2\sin A\cos A+2{{\cos }^{2}}A}{2\sin A\cos A+2{{\sin }^{2}}A} \\\ \end{aligned}$
Now, we will take $2\cos A$ common from the numerator, and $2\sin A$ common from the denominator of the term $\dfrac{2\sin A\cos A+2{{\cos }^{2}}A}{2\sin A\cos A+2{{\sin }^{2}}A}$ in the above equation. Then,
$\begin{aligned} & \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{2\sin A\cos A+2{{\cos }^{2}}A}{2\sin A\cos A+2{{\sin }^{2}}A} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{2\cos A\left( \sin A+\cos A \right)}{2\sin A\left( \cos A+\sin A \right)} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{\cos A\left( \sin A+\cos A \right)}{\sin A\left( \sin A+\cos A \right)} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{\cos A}{\sin A} \\\ \end{aligned}$
Now, we will use the formula from the equation (4) to write $\dfrac{\cos A}{\sin A}=\cot A$ in the above equation. Then,
$\begin{aligned} & \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\dfrac{\cos A}{\sin A} \\\ & \Rightarrow \dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\cot A \\\ \end{aligned}$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\dfrac{1+\sin 2A+\cos 2A}{1+\sin 2A-\cos 2A}=\cot A$ .
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result. After that, we should apply double angle trigonometric formulas of $\sin 2\theta$ and $\cos 2\theta$ correctly. Moreover, while simplifying we should be aware of the result and try to solve accurately without any calculation mistakes, so that we can prove the desired result easily.