Question

Prove that $\dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}{\text{ }}{\text{.}}$

Answer 1

Hint: In this question, we are asked to prove a trigonometric equality. To solve this question standard trigonometric identities and formulae are required. Firstly, we will take L.H.S. which is $\dfrac{{\cot A - \cos A}}{{\cot A + \cos A}}$ and then try to prove it equal to R.H.S. by simplifying it with the use of standard formulae. The basic trigonometric conversion identities are: $\left( 1 \right)\sin x = \dfrac{1}{{\cos ecx}}$ i.e. sine and cosecant function are inverse of each other. $\left( 2 \right)\cos x = \dfrac{1}{{\sec x}}$ i.e. cosine and secant functions are opposite of each other. $\left( 3 \right)\tan x = \dfrac{1}{{\cot x}}$ i.e. tangent and cotangent functions are inverse of each other.

Complete step-by-step answer:
The given expression is:
$ \Rightarrow \dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}$
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{\cot A - \cos A}}{{\cot A + \cos A}}$
By the knowledge of basic trigonometric identities we know that ;
$ \Rightarrow \cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}$
Taking L.H.S. and trying to simplify it , we get ;
Replacing cotangent in terms of sine and cosine in the L.H.S. expression , we get ;
$ \Rightarrow \dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \cos A}}{{\dfrac{{\cos A}}{{\sin A}} + \cos A}}$
Taking L.C.M. the expression becomes ;
$ \Rightarrow \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \cos A}}{{\dfrac{{\cos A}}{{\sin A}} + \cos A}} = \dfrac{{\dfrac{{\cos A - \sin A\cos A}}{{\sin A}}}}{{\dfrac{{\cos A + \sin A\cos A}}{{\sin A}}}}$
Further simplifying the above expression ;
$ \Rightarrow \dfrac{{\cos A - \cos A\sin A}}{{\cos A + \cos A\sin A}}$
Taking $\cos A$ outside in both numerator and denominator ;
$ \Rightarrow \dfrac{{\cos A\left( {1 - \sin A} \right)}}{{\cos A\left( {1 + \sin A} \right)}}$
On further simplification of the above equation, we get ;
$ = \dfrac{{1 - \sin A}}{{1 + \sin A}}{\text{ }}......\left( 1 \right)$
Since the given R.H.S. is in terms of cosecant function, therefore we have to proceed in such a way that we get the L.H.S. function also in terms of cosecant function.
By the standard trigonometric identities we know that;
$ \Rightarrow \cos ecx = \dfrac{1}{{\sin x}}$
So, dividing both numerator and denominator by $\sin A$ in equation $1$ , we get ;
$ \Rightarrow \dfrac{{1 - \sin A}}{{1 + \sin A}} = \dfrac{{\dfrac{{1 - \sin A}}{{\sin A}}}}{{\dfrac{{1 + \sin A}}{{\sin A}}}}$
Rearranging the above expression ;
$ \Rightarrow \dfrac{{\dfrac{{1 - \sin A}}{{\sin A}}}}{{\dfrac{{1 + \sin A}}{{\sin A}}}} = \dfrac{{\dfrac{1}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}}}}{{\dfrac{1}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}}}}$
Further simplifying the above expression, we get ;
$ = \dfrac{{\dfrac{1}{{\sin A}} - 1}}{{\dfrac{1}{{\sin A}} + 1}}$
Using the trigonometric identity stated above, i.e. $\sin x = \dfrac{1}{{\cos ecx}}$ , we can conclude that ;
$ \Rightarrow \dfrac{{\dfrac{1}{{\sin A}} - 1}}{{\dfrac{1}{{\sin A}} + 1}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$
$ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$
Therefore, $\dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}{\text{ }}{\text{.}}$
Hence proved.

Note: Trigonometric identities are of great importance to solve these kinds of questions. There are three important identities : $\left( 1 \right){\sin ^2}x + {\cos ^2}x = 1$ , from this identity we can also deduce the relations: $\left( {\text{i}} \right)1 - {\sin ^2}x = {\cos ^2}x$ and $\left( {{\text{ii}}} \right)1 - {\cos ^2}x = {\sin ^2}x$ . $\left( 2 \right)1 + {\tan ^2}x = {\sec ^2}x$ , from this identity we can also deduce the relations: $\left( {\text{i}} \right){\sec ^2}x - {\tan ^2}x = 1$ and $\left( {{\text{ii}}} \right){\sec ^2}x - 1 = {\tan ^2}x$ . $\left( 3 \right)1 + {\cot ^2}x = \cos e{c^2}x$ , from this identity we can also deduce the relations: $\left( {\text{i}} \right)\cos e{c^2}x - 1 = {\cot ^2}x$ and $\left( {{\text{ii}}} \right)\cos e{c^2}x - {\cot ^2}x = 1$ .
These identities are called Pythagorean identities. These identities are true for any value of x in the universe.