Question

Prove the following trigonometric relation-
$\dfrac{{\cot {\text{A + cscA - 1}}}}{{\cot {\text{A - cscA + 1}}}} = \dfrac{{{\text{1 + cosA}}}}{{{\text{sinA}}}}$

Answer 1

Hint: In order to solve this question, proceed by substitute 1 by a trigonometric identity and using the difference of two squares formula in the LHS i.e. $1{\text{ = cs}}{{\text{c}}^2}{\text{A - co}}{{\text{t}}^2}{\text{A }}$ Then simplify the expression. We will get our RHS.

Complete step-by-step answer:
To prove - $\dfrac{{\cot {\text{A + cscA - 1}}}}{{\cot {\text{A - cscA + 1}}}} = \dfrac{{{\text{1 + cosA}}}}{{{\text{sinA}}}}$
Consider the left hand side,
$\dfrac{{\cot {\text{A + cscA - 1}}}}{{\cot {\text{A - cscA + 1}}}}$ …………… (1)

We know that, $1 + {\text{co}}{{\text{t}}^2}{\text{A = cs}}{{\text{c}}^2}{\text{A}}$
This can be written as, $1{\text{ = cs}}{{\text{c}}^2}{\text{A - co}}{{\text{t}}^2}{\text{A }}$
first put the value of 1 in the numerator of equation (1) in terms of $1{\text{ = cs}}{{\text{c}}^2}{\text{A - co}}{{\text{t}}^2}{\text{A }}$
so, Substituting this in the numerator of...............(1)
$\dfrac{{\cot {\text{A + cscA - }}\left( {{\text{cs}}{{\text{c}}^2}{\text{A - }}{{\cot }^2}{\text{A}}} \right)}}{{\cot {\text{A - cscA + 1}}}}$

In algebra, there is a formula known as the Difference of two squares:$({{\text{m}}^2}{\text{ - }}{{\text{n}}^2}) = ({\text{m + n}})({\text{m - n}})$ ………..(2)
Here, ${\text{m = }}({\text{cscA) and n = }}({\text{cotA)}}$
So using expression (2)
$\Rightarrow {(\csc {\text{A)}}^2} - {(\cot {\text{A)}}^2} = \left( {\csc {\text{A + }}\cot {\text{A}}} \right)\left( {\csc {\text{A - }}\cot {\text{A}}} \right)$ …. (3)
Now on putting the value of ${(\csc {\text{A)}}^2} - {(\cot {\text{A)}}^2}$in equation (1) from equation (3)
$\Rightarrow \dfrac{{\cot {\text{A + cscA - }}\left( {{\text{cscA + }}\cot {\text{A}}} \right)\left( {{\text{cscA - }}\cot {\text{A}}} \right)}}{{\cot {\text{A - cscA + 1}}}}$
$\Rightarrow \dfrac{{(\cot {\text{A + cscA)}}\left( {{\text{1 - }}\left( {{\text{cscA - }}\cot {\text{A}}} \right)} \right)}}{{\cot {\text{A - cscA + 1}}}}$
This can be written as
$\Rightarrow \dfrac{{(\cot {\text{A + cscA)}}\left( {{\text{1 - cscA + }}\cot {\text{A}}} \right)}}{{\cot {\text{A - cscA + 1}}}}$
On cancelling $\left( {{\text{1 - cscA + }}\cot {\text{A}}} \right)$ from numerator and denominator both
$\Rightarrow (\cot {\text{A + cscA)}}$ ……. (4)
We now that $\cot {\text{A = }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}\& {\text{ cscA = }}\dfrac{1}{{\sin {\text{A}}}}$, so on substituting these results in equation (4)
$\Rightarrow \dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}{\text{ + }}\dfrac{1}{{\sin {\text{A}}}}$
Here denominator of both terms are same so can be taken as common
$\Rightarrow \dfrac{{1 + \cos {\text{A}}}}{{\sin {\text{A}}}} = {\text{RHS}}$
Therefore, $\dfrac{{\cot {\text{A + cscA - 1}}}}{{\cot {\text{A - cscA + 1}}}}{\text{ (LHS)}} = \dfrac{{{\text{1 + cosA}}}}{{{\text{sinA}}}}{\text{ (RHS)}}$
Hence proved.

Note- Whenever we face such types of problems the key concept is that we always try to think of some short substitution so that some terms get cancelled and expression can be reduced. And whenever you find a trigonometric expression having even power we should use the identity “difference of two squares” which is stated above.