Show that \[{\left( {{\text{cscA - cotA}}} \right)^{\text{2}}}\]= $\dfrac{{{\text{1 - cosA}}}}{{{\text{1 + cosA}}}}$

Hint- To solve this question, we need to apply the theory and various formulas of trigonometric functions. As we observe, cosec and cot are two trigonometry functions present in the left hand side (LHS) of the given expression. So, to show that LHS and RHS are equal, we will manipulate LHS using trigonometry identity or formula such that we will get RHS by solving LHS.

Complete step-by-step answer:

According to question, we have

\[{\left( {{\text{cscA - cotA}}} \right)^{\text{2}}}\]= $\dfrac{{{\text{1 - cosA}}}}{{{\text{1 + cosA}}}}$

Here, on the left hand side we have \[{\left( {{\text{cscA - cotA}}} \right)^{\text{2}}}\].

So, we need to manipulate LHS in such a manner that we will get RHS.

I.e. in this case RHS will be $\dfrac{{{\text{1 - cosA}}}}{{{\text{1 + cosA}}}}$

Now, LHS = \[{\left( {{\text{cscA - cotA}}} \right)^{\text{2}}}\]

= \[{\left( {\dfrac{{\text{1}}}{{{\text{sinA}}}}{\text{ - }}\dfrac{{{\text{cosA}}}}{{{\text{sinA}}}}} \right)^{\text{2}}}\]

= ${\left( {\dfrac{{{\text{1 - cosA}}}}{{{\text{sinA}}}}} \right)^{\text{2}}}$

= $\dfrac{{{{\left( {{\text{1 - cosA}}} \right)}^{\text{2}}}}}{{{{{\text{(sinA)}}}^{\text{2}}}}}$

= $\dfrac{{{{\left( {{\text{1 - cosA}}} \right)}^{\text{2}}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{A}}}}$

=$\dfrac{{{{\left( {{\text{1 - cosA}}} \right)}^{\text{2}}}}}{{{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{A}}}}$

=$\dfrac{{{{\left( {{\text{1 - cosA}}} \right)}^{\text{2}}}}}{{{{\text{1}}^2}{\text{ - co}}{{\text{s}}^{\text{2}}}{\text{A}}}}$

As we know, ${{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}{\text{ = (a + b)(a - b)}}$
= $\dfrac{{{{\left( {{\text{1 - cosA}}} \right)}^{\text{2}}}}}{{\left( {{\text{1 + cosA}}} \right)\left( {{\text{1 - cosA}}} \right)}}$

= $\dfrac{{\left( {{\text{1 - cosA}}} \right)}}{{\left( {{\text{1 + cosA}}} \right)}}$

= RHS

Since, LHS=RHS verified.

So, we say we show that \[{\left( {{\text{cscA - cotA}}} \right)^{\text{2}}}\]= $\dfrac{{{\text{1 - cosA}}}}{{{\text{1 + cosA}}}}$.

Note- We need to remember some basic formulas related to trigonometry. So that we easily understand the problem and apply these formulas. Some of them are mentioned below which we used in this question.

These Identities are given as-

sin$\theta $ = $\dfrac{{\text{1}}}{{{\text{cosec}}\theta }}$

cos$\theta $ = $\dfrac{{\text{1}}}{{{\text{sec}}\theta }}$

tan$\theta $ = $\dfrac{{\text{1}}}{{{\text{cot}}\theta }}$

\[{\text{cot}}\theta \]= $\dfrac{{{\text{cos}}\theta }}{{{\text{sin}}\theta }}$