Question

Prove that $\dfrac{{\tan ({{45}^ \circ } + A) - \tan ({{45}^ \circ } - A)}}{{\tan ({{45}^ \circ } + A) + \tan ({{45}^ \circ } - A)}} = \sin 2A$.

Answer 1

The question requires the application of trigonometric identities for ${45^ \circ }$angle. Trigonometric identities are equations that connect various trigonometric functions and hold for any value of the variable in the domain. An identity is a mathematical expression that holds true for all values of the variable(s) it contains.

Complete step-by-step answer:
In the given question, we have to use the following trigonometric identity:
$\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$ and
$\tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$
Here we know that $x = {45^ \circ }$and $y = A$
Now substituting the value in the above formula, we get,
$\tan ({45^ \circ } + A) = \dfrac{{\tan {{45}^ \circ } + \tan A}}{{1 - \tan {{45}^ \circ }\tan A}}$
$\tan ({45^ \circ } - A) = \dfrac{{\tan {{45}^ \circ } - \tan A}}{{1 + \tan {{45}^ \circ }\tan A}}$
Using trigonometric ratio table, we get $\tan {45^ \circ } = 1$, and comparing with above formula, we get,
$\tan ({45^ \circ } + A) = \dfrac{{(1 + \tan A)}}{{(1 - \tan A)}}$
$\tan ({45^ \circ } - A) = \dfrac{{(1 - \tan A)}}{{(1 + \tan A)}}$
Substituting these values in left-hand side of the equation, we get,
$= \dfrac{{\dfrac{{(1 + \tan A)}}{{(1 - \tan A)}} - \dfrac{{(1 - \tan A)}}{{(1 + \tan A)}}}}{{\dfrac{{(1 + \tan A)}}{{(1 - \tan A)}} + \dfrac{{(1 - \tan A)}}{{(1 + \tan A)}}}}$
Cross-multiplying the denominators, we get,
$= \dfrac{{\dfrac{{{{(1 + \tan A)}^2} - {{(1 - \tan A)}^2}}}{{(1 - \tan A)}}}}{{\dfrac{{{{(1 + \tan A)}^2} + {{(1 - \tan A)}^2}}}{{(1 - \tan A)}}}}$
Dividing by $(1 - \tan A)$, we get,
$= \dfrac{{{{(1 + \tan A)}^2} - {{(1 - \tan A)}^2}}}{{{{(1 + \tan A)}^2} + {{(1 - \tan A)}^2}}}$
Now solving the brackets by factoring using following equation:
${(a + b)^2} = {a^2} + 2ab + {b^2}$and
${(a - b)^2} = {a^2} - 2ab + {b^2}$
Comparing with the above formula, we get,
$= \dfrac{{(1 + 2\tan A + {{\tan }^2}A) - (1 - 2\tan A + {{\tan }^2}A)}}{{(1 + 2\tan A + {{\tan }^2}A) + (1 - 2\tan A + {{\tan }^2}A)}}$
Opening the brackets, we get,
$= \dfrac{{1 + 2\tan A + {{\tan }^2}A - 1 + 2\tan A - {{\tan }^2}A}}{{1 + 2\tan A + {{\tan }^2}A + 1 - 2\tan A + {{\tan }^2}A}}$
Adding and subtracting the variables, we get,
$= \dfrac{{4\tan A}}{{2 + 2{{\tan }^2}A}}$
Dividing by $2$, we get,
$= \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$
Now we will use the following trigonometric identities to solve the question:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and
$1 + {\tan ^2}\theta = {\sec ^2}\theta$and
${\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}$

Comparing with above formula and substituting the value, we get,
$= \dfrac{{2\dfrac{{\sin A}}{{\cos A}}}}{{{{\sec }^2}A}}$
$= \dfrac{{2\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{1}{{{{\cos }^2}A}}}}$
$= 2\dfrac{{\sin A}}{{\cos A}} \times {\cos ^2}A$
$= 2\sin A \times \cos A$
Now according to formula:
$2\sin \theta \cos \theta = \sin 2\theta$

Comparing with above formula, we get,
$= \sin 2A$
Hence LHS=RHS. Therefore, it is proved as per above solution that
$\dfrac{{tan({{45}^ \circ } + A) - tan({{45}^ \circ } - A)}}{{tan({{45}^ \circ } + A) + tan({{45}^ \circ } - A)}} = \sin 2A$

Note: Meaning of tangent and sine is given below for better understanding:
Tangent: The ratio of side opposite to given angle and its adjacent side is called tangent. It is denoted as $\tan \theta$.
$\tan \theta = \dfrac{{Opposite{\text{ }}side\,to\,given\,angle}}{{{\rm A}djacent{\text{ }}side\,to\,given\,angle}}$
Sine: The ratio of side opposite to given angle and hypotenuse is called sine. It is denoted as $\sin \theta$.
$\sin \theta = \dfrac{{Opposite{\text{ }}side\,to\,given\,angle}}{{Hypotenuse}}$