Question

If $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \dfrac{x}{y}$, then the value of $\tan A$ is:
(A) $\dfrac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}$
(B) $\dfrac{{2xy}}{{{x^2} + {y^2}}}$
(C) $\dfrac{{2xy}}{{{x^2} - {y^2}}}$
(D) $\dfrac{{2xy}}{{{y^2} - {x^2}}}$

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 half angle formula for cosine and tangent. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in 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 value of $\tan A$ when we are provided with the relation $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \dfrac{x}{y}$.
Now, we simplify the left side of the equation given to us by using the different forms of half angle formulae of cosine $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2} = 2{\cos ^2}\dfrac{x}{2} - 1$. So, we get,
$ \Rightarrow \sqrt {\dfrac{{1 + \left( {2{{\cos }^2}\dfrac{A}{2} - 1} \right)}}{{1 - \left( {1 - 2{{\sin }^2}\dfrac{A}{2}} \right)}}} = \dfrac{x}{y}$
Opening the brackets,
$ \Rightarrow \sqrt {\dfrac{{1 + 2{{\cos }^2}\dfrac{A}{2} - 1}}{{1 - 1 + 2{{\sin }^2}\dfrac{A}{2}}}} = \dfrac{x}{y}$
Cancelling the like terms with opposite signs, we get,
$ \Rightarrow \sqrt {\dfrac{{2{{\cos }^2}\dfrac{A}{2}}}{{2{{\sin }^2}\dfrac{A}{2}}}} = \dfrac{x}{y}$
Cancelling the common terms in numerator and denominator, we get,
$ \Rightarrow \sqrt {\dfrac{{{{\cos }^2}\dfrac{A}{2}}}{{{{\sin }^2}\dfrac{A}{2}}}} = \dfrac{x}{y}$
Computing the square root,
$ \Rightarrow \cot \dfrac{A}{2} = \dfrac{x}{y}$
But we have to calculate the value of $\tan A$ for which the value of $\tan \dfrac{A}{2}$ must be known. We know that tangent and cotangent are reciprocal functions of each other. Taking reciprocal on both sides,
\[ \Rightarrow \dfrac{1}{{\cot \dfrac{A}{2}}} = \dfrac{y}{x}\]
\[ \Rightarrow \tan \dfrac{A}{2} = \dfrac{y}{x}\]
Now, we use the trigonometric formula $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$. So, we get value of $\tan A$ as,
$\tan A = \dfrac{{2\tan \left( {\dfrac{A}{2}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{A}{2}} \right)}}$
Substituting the value of $\tan \left( {\dfrac{A}{2}} \right)$, we get,
$ \Rightarrow \tan A = \dfrac{{2\left( {\dfrac{y}{x}} \right)}}{{1 - {{\left( {\dfrac{y}{x}} \right)}^2}}}$
Simplifying the equation, we get,
\[ \Rightarrow \tan A = \dfrac{{\dfrac{{2y}}{x}}}{{\dfrac{{{x^2} - {y^2}}}{{{x^2}}}}}\]
\[ \Rightarrow \tan A = \dfrac{{2xy}}{{{x^2} - {y^2}}}\]
So, we get the value of $\tan A$ as \[\dfrac{{2xy}}{{{x^2} - {y^2}}}\]. Hence, option (C) is the correct answer.
So, the correct answer is “Option B”.

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.