Prove that \[\dfrac{{1 - \sin A}}{{1 + \sin A}} = {(\sec A - \tan A)^2}\]?

Hint: Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step solution:
 In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality \[\dfrac{{1 - \sin A}}{{1 + \sin A}} = {(\sec A - \tan A)^2}\]
Now we consider the LHS
 \[\dfrac{{1 - \sin A}}{{1 + \sin A}}\]
We can’t solve this directly, so now we multiply both numerator and denominator by the conjugate of the denominator term. The conjugate term of \[1 + \sin A\] is \[1 - \sin A\]
Now we will multiply the both numerator and the denominator terms by the \[1 - \sin A\], Now on multiplying we have
\[ \Rightarrow \dfrac{{1 - \sin A}}{{1 + \sin A}} \times \dfrac{{1 - \sin A}}{{1 - \sin A}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{(1 - \sin A)(1 - \sin A)}}{{(1 + \sin A)(1 - \sin A)}}\]
On further multiplying the terms we have
\[ \Rightarrow \dfrac{{{{(1 - \sin A)}^2}}}{{(1 + \sin A)(1 - \sin A)}}\]
The above inequality, in the denominator it is in the form of \[(a + b)(a - b)\], we have standard algebraic identity for this and it is given as \[(a + b)(a - b) = {a^2} - {b^2}\] Using this identity the above inequality is written as
\[ \Rightarrow \dfrac{{{{(1 - \sin A)}^2}}}{{1 - {{\sin }^2}A}}\]
From the trigonometry identities we have \[{\sin ^2}A + {\cos ^2}A = 1\]
In trigonometry identity taking \[{\sin ^2}A\] to RHS and it is written as \[1 - {\sin ^2}A = {\cos ^2}A\]
Therefore we have
\[ \Rightarrow \dfrac{{{{(1 - \sin A)}^2}}}{{{{\cos }^2}A}}\]
Since the both terms in the numerator and denominator we have square we can write the square for the whole term and so we have
\[ \Rightarrow {\left( {\dfrac{{1 - \sin A}}{{\cos A}}} \right)^2}\]
The denominator term is applicable to the both terms which are present in the numerator and so we have
 \[ \Rightarrow {\left( {\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}} \right)^2}\]
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosecant is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. From the reciprocal of the trigonometry ratios. The tangent trigonometry ratio is written in the form of ratio of sine to cosine. That is \[\tan A = \dfrac{{\sin A}}{{\cos A}}\]. Now the inequality is written as
\[ \Rightarrow {\left( {\sec A - \tan A} \right)^2}\]
\[ \Rightarrow RHS\]
Here we have proved LHS = RHS.

Note: The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.