Question

Prove that: $\sqrt {\dfrac{{1 + \sin 2A}}{{1 - \sin 2A}}} = \tan \left( {\dfrac{\pi }{4} + A} \right)$

Answer 1

Here we will start resolving from the LHS of the equation. First, we will use the basic trigonometry property of the $\sin 2\theta$ in terms of tangent function to simplify the equation. Then we will solve the equation using the formula of tangent function of the sum of two angles. We will simplify the equation further to get the value same as RHS of the given equation.

Formula Used:
We will use the following formulas:
1.$\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$
2.$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$

Complete step-by-step answer:
Given equation is $\sqrt {\dfrac{{1 + \sin 2A}}{{1 - \sin 2A}}} = \tan \left( {\dfrac{\pi }{4} + A} \right)$.
First, we will consider the LHS of the equation and solve it.
$LHS = \sqrt {\dfrac{{1 + \sin 2A}}{{1 - \sin 2A}}}$
Now we will use the basic formula of the trigonometric for $\sin 2\theta$ in terms of the tan function.
Therefore, by using this formula $\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$ in the above equation, we get
$\Rightarrow LHS = \sqrt {\dfrac{{1 + \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}}}{{1 - \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}}}}$
Taking the LCM in the numerator and denominator, we get
$\Rightarrow LHS = \sqrt {\dfrac{{\dfrac{{1 + {{\tan }^2}A + 2\tan A}}{{1 + {{\tan }^2}A}}}}{{\dfrac{{1 + {{\tan }^2}A - 2\tan A}}{{1 + {{\tan }^2}A}}}}}$
Cancelling out the common terms, we get
$\Rightarrow LHS = \sqrt {\dfrac{{1 + {{\tan }^2}A + 2\tan A}}{{1 + {{\tan }^2}A - 2\tan A}}}$
Now using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we can write above equation as
$\Rightarrow LHS = \sqrt {\dfrac{{{{\left( {1 + \tan A} \right)}^2}}}{{{{\left( {1 - \tan A} \right)}^2}}}}$
Now in the above equation square of the terms gets canceled out by the square root of the equation. Therefore, we get
$\Rightarrow LHS = \dfrac{{1 + \tan A}}{{1 - \tan A}}$
We know that the value of the tan function at $\dfrac{\pi }{4}$ is equal to 1. So, we can write the above equation as
$\Rightarrow LHS = \dfrac{{\tan \dfrac{\pi }{4} + \tan A}}{{1 - \tan \dfrac{\pi }{4} \cdot \tan A}}$
We know the property $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$. Therefore, by using this property in the above equation, we get
$\Rightarrow LHS = \tan \left( {\dfrac{\pi }{4} + A} \right)$
The term in the above equation is equal to the RHS of the equation. Therefore
$\therefore LHS=RHS$
So, $\sqrt {\dfrac{{1 + \sin 2A}}{{1 - \sin 2A}}} = \tan \left( {\dfrac{\pi }{4} + A} \right)$
Hence, proved.

Note: Here, we have evaluated the left hands side of the equation because it is easy to transform sine function into tangent function using the formulas and the identities. If we try to resolve from the right hand side then we will also get cosine function and remove than and getting only the sine function is difficult. In order to solve the question we should keep in mind the trigonometric formulas and identities of at least sine and tangent function.