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Question: Prove that \(\dfrac{{1 - \sin A}}{{1 + \sin A}} = {\left( {\sec A - \tan A} \right)^2}\)...

Prove that 1−sin⁡A1+sin⁡A=(sec⁡A−tan⁡A)2\dfrac{{1 - \sin A}}{{1 + \sin A}} = {\left( {\sec A - \tan A} \right)^2}

Explanation

Solution

Hint : Here, we will take first of all the left hand side of the given equation and will replace the trigonometric functions in its equivalent value in terms of secant and tangent, since we require the answer on the right hand side of the equation in that form. Use the identity for cosine, sine, secant and tangent to simplify the intermediary steps.

Complete step-by-step answer :
LHS=1−sin⁡A1+sin⁡ALHS = \dfrac{{1 - \sin A}}{{1 + \sin A}}
Take the conjugate for the term in the denominator of the above expression and multiply and divide it with the given term.
LHS=1−sin⁡A1+sin⁡A×1−sin⁡A1−sin⁡ALHS = \dfrac{{1 - \sin A}}{{1 + \sin A}} \times \dfrac{{1 - \sin A}}{{1 - \sin A}}
Simplify the above expression by using the concepts of the whole square for the difference of two terms and the difference of two squares.
LHS=(1−sin⁡A)212−sin⁡2ALHS = \dfrac{{{{\left( {1 - \sin A} \right)}^2}}}{{{1^2} - {{\sin }^2}A}}
Expand the numerator of the above expression by using (a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}also 1−sin⁡2A=cos⁡2A1 - {\sin ^2}A = {\cos ^2}A
LHS=1−2sin⁡A+sin⁡2Acos⁡2ALHS = \dfrac{{1 - 2\sin A + {{\sin }^2}A}}{{{{\cos }^2}A}}
Now, numerators can be split by giving the denominator upon all the terms –
LHS=1cos⁡2A−2sin⁡Acos⁡2A+sin⁡2Acos⁡2ALHS = \dfrac{1}{{{{\cos }^2}A}} - \dfrac{{2\sin A}}{{{{\cos }^2}A}} + \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}
Now, place the equivalent values for all the reciprocal trigonometric functions. Such as sec⁡A=1cos⁡A\sec A = \dfrac{1}{{\cos A}}, sin⁡Acos⁡A=tan⁡A\dfrac{{\sin A}}{{\cos A}} = \tan A
LHS=sec2A−2tan⁡Asec⁡A+tan⁡2ALHS = se{c^2}A - 2\tan A\sec A + {\tan ^2}A
The above expression can be re-written as the whole square of the terms –
LHS=(secA−tan⁡A)2LHS = {(secA - \tan A)^2}
The above equation is the equation to the given right hand side of the equation.
LHS=RHSLHS = RHS
Hence, proved.

Note : Always remember the different identities and the correlation between all the six trigonometric functions. Don’t get confused between cosine and cosec functions. Always remember that any trigonometric is expressed in its equivalent functions by two ways, first reciprocal and the other is ratio of the other two functions. For example tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }} or tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}so be wise to choose the equivalent expression which is solely depended on the required resultant terms.