Question

The value of $\tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right)$ is equal to:
(A) $\cos ecA + \cot A$
(B) $\cos ecA - \cot A$
(C) $\sec A + \tan A$
(D) $\sec A - \tan A$

Answer 1

Hint : The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as the compound formula for tangent function. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem. We must know the simplification rules to solve the problem with ease.

Complete step-by-step answer :
In the given problem, we have to find the value of the expression $\tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right)$.
So, we have, $\tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right)$
Now, we know the trigonometric compound angle formula for tangent as $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$. So, using the formula to find the value of the required expression, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{\tan {{45}^ \circ } - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + \tan {{45}^ \circ } \times \tan \left( {\dfrac{A}{2}} \right)}}$
Now, we know that the value of the tangent trigonometric function for the angle ${45^ \circ }$ is one. So, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{1 - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + 1 \times \tan \left( {\dfrac{A}{2}} \right)}}$
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Hence, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{1 - \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}{{1 + \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}$
Taking LCM of expressions, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}$
Cancelling common factors in numerator and denominator, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)}}$
Multiplying the numerator and denominator by $\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)$. So, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \left[ {\dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)}} \times \dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}} \right]$
Now, using the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ and ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{{{\left\\{ {\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)} \right\\}}^2}}}{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) - {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}
Now, using the trigonometric formula ${\cos ^2}x - {\sin ^2}x = \cos 2x$, we get,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) - 2\cos \dfrac{A}{2}\sin \dfrac{A}{2} + {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}{{\cos A}}$
Using the trigonometric identity ${\cos ^2}x + {\sin ^2}x = 1$ and the double angle formula for sine as $\sin 2x = 2\sin x\cos x$, we have,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{{1 - \sin A}}{{\cos A}}$
Separating the denominators,
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}$
Now, substituting $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$.
$\Rightarrow \tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right) = \sec A - \tan A$
So, we get the value of the expression $\tan \left( {{{45}^ \circ } - \dfrac{A}{2}} \right)$ is $\sec A - \tan A$.
Hence, Option (D) is the correct answer.
So, the correct answer is “Option D”.

Note : We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers but we have to mark the most appropriate option among the given choices.