Question

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

Answer 1

In this question we solve the left side of the equation and equating to the right side by using basic trigonometric formulas and identities such as $\dfrac{{\cos x}}{{\sin x}} = \cot x$. We use the basic trigonometric identities such as $\sin 2x = 2\sin x\cos x$ ,$\cos 2x = 2{\cos ^2}x - 1$ and $\cos 2x = 1 - 2{\sin ^2}x$.

Complete step by step answer:
We have $\dfrac{{1 + \sin 2A + \cos 2A}}{{1 + \sin 2A - \cos 2A}} = \cot A$.
Taking Left side of the equation
$\dfrac{{1 + \sin 2A + \cos 2A}}{{1 + \sin 2A - \cos 2A}}$
We can write $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 2{\cos ^2}A - 1$ in numerator and in denominator $\cos 2A$ to be written as $\cos 2A = 1 - 2{\sin ^2}A$. We get,
$\Rightarrow \dfrac{{1 + 2\sin A\cos A + 2{{\cos }^2}A - 1}}{{1 + 2\sin A\cos A - (1 - 2{{\sin }^2}A)}}$

Simplifying the denominator of the given equation. We get,
$\Rightarrow \dfrac{{1 + 2\sin A\cos A + 2{{\cos }^2}A - 1}}{{1 + 2\sin A\cos A - 1 + 2{{\sin }^2}A}}$
Cancelling out $1$ from both the numerator and denominator. We get,
$\Rightarrow \dfrac{{2\sin A\cos A + 2{{\cos }^2}A}}{{2\sin A\cos A + 2{{\sin }^2}A}}$
Taking $2$ as common from both the numerator and denominator. We get,
$\Rightarrow \dfrac{{2(\sin A\cos A + {{\cos }^2}A)}}{{2(\sin A\cos A + {{\sin }^2}A)}}$
Cancelling out $2$. We get,
$\Rightarrow \dfrac{{\sin A\cos A + {{\cos }^2}A}}{{\sin A\cos A + {{\sin }^2}A}}$

Taking common $\cos A$ from the equation in numerator and $\sin A$ from the equation in the denominator. We get,
$\Rightarrow \dfrac{{\cos A(\sin A + \cos A)}}{{\sin A(\cos A + \sin A)}}$
Cancelling out the equal values in the above equation. We get,
$\Rightarrow \dfrac{{\cos A}}{{\sin A}}$
We know that $\dfrac{{\cos A}}{{\sin A}} = \cot A$
So, $\dfrac{{1 + \sin 2A + \cos 2A}}{{1 + \sin 2A - \cos 2A}} = \cot A$
Hence, the left side of the equation is equal to the right side of the equation.
Hence, proved.

Note: We use the trigonometric identities to substitute the term in the equation according to the result we need as we have use two trigonometric identities for $\cos 2A$ as $\cos 2A = 2{\cos ^2}A - 1$ and $\cos 2A = 1 - 2{\sin ^2}A$ as we want $\cot 2A$ as a result and $\dfrac{{\cos A}}{{\sin A}} = \cot A$. So, we have to use both identities. If we use only one we can’t get the result.