Question

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

Answer 1

Hint: 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 \[\sin \theta \].
\[\sin \theta = \dfrac{{Opposite{\text{ }}side\,to\,given\,angle}}{{Hypotenuse}}\]