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Question: Prove that: \({\left( {\cot A - \cos ecA} \right)^2} = \dfrac{{1 - \cos A}}{{1 + \cos A}}\)....

Prove that:
(cot⁡A−cos⁡ecA)2=1−cos⁡A1+cos⁡A{\left( {\cot A - \cos ecA} \right)^2} = \dfrac{{1 - \cos A}}{{1 + \cos A}}.

Explanation

Solution

We have to prove (cot⁡A−cos⁡ecA)2=1−cos⁡A1+cos⁡A{\left( {\cot A - \cos ecA} \right)^2} = \dfrac{{1 - \cos A}}{{1 + \cos A}}. Now, solve the expression which is on the left hand side of the equation that is (cot⁡A−cos⁡ecA)2{\left( {\cot A - \cos ecA} \right)^2} and arrange the result similar to the expression on the right hand side of the equation. If both the right hand side and left hand side of the equation are equal then the given equation is the same then the given equality is proved.

Complete step-by-step solution:
It is given that prove (cot⁡A−cos⁡ecA)2=1−cos⁡A1+cos⁡A{\left( {\cot A - \cos ecA} \right)^2} = \dfrac{{1 - \cos A}}{{1 + \cos A}}.
Here, (LHS) left hand side of the equality is (cot⁡A−cos⁡ecA)2{\left( {\cot A - \cos ecA} \right)^2}.
Now, we have to solve (cot⁡A−cos⁡ecA)2{\left( {\cot A - \cos ecA} \right)^2}.
We have studied in trigonometry that cot⁡A=cos⁡Asin⁡A\cot A = \dfrac{{\cos A}}{{\sin A}} that is inverse of tan⁡A\tan A and cos⁡ecA\cos ecA is inverse of sin⁡A\sin Athat is cos⁡ecA=1sin⁡A\cos ecA = \dfrac{1}{{\sin A}}.
Now put cot⁡A=cos⁡Asin⁡A\cot A = \dfrac{{\cos A}}{{\sin A}} and cos⁡ecA=1sin⁡A\cos ecA = \dfrac{1}{{\sin A}} in the above expression (cot⁡A−cos⁡ecA)2{\left( {\cot A - \cos ecA} \right)^2}and we get,
=(cos⁡Asin⁡A−1sin⁡A)2= {\left( {\dfrac{{\cos A}}{{\sin A}} - \dfrac{1}{{\sin A}}} \right)^2}
Now, subtracting these two terms we get,
=(cos⁡A−1sin⁡A)2= {\left( {\dfrac{{\cos A - 1}}{{\sin A}}} \right)^2}
Taking (−1)\left( { - 1} \right) as common from the terms of numerator, we can write
=((−1)(1−cos⁡A)sin⁡A)2= {\left( {\dfrac{{\left( { - 1} \right)\left( {1 - \cos A} \right)}}{{\sin A}}} \right)^2}
Now, by squaring numerator and denominator we can write,
=(1−cos⁡A)2sin⁡2A= \dfrac{{{{\left( {1 - \cos A} \right)}^2}}}{{{{\sin }^2}A}}
We have remembered a trigonometrical formula sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1 so , we can write sin⁡2A=1−cos⁡2A{\sin ^2}A = 1 - {\cos ^2}A.
=(1−cos⁡A)21−cos⁡2A= \dfrac{{{{\left( {1 - \cos A} \right)}^2}}}{{1 - {{\cos }^2}A}}
We know that a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right), now, denominator can we written as (1)2−(cos⁡A)2{\left( 1 \right)^2} - {\left( {\cos A} \right)^2} which is ultimately expressed as (1−cos⁡A)(1+cos⁡A)\left( {1 - \cos A} \right)\left( {1 + \cos A} \right).
So, above expression can we written as
=(1−cos⁡A)(1−cos⁡A)(1−cos⁡A)(1+cos⁡A)= \dfrac{{\left( {1 - \cos A} \right)\left( {1 - \cos A} \right)}}{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}
Here, (1−cos⁡A)\left( {1 - \cos A} \right) is common in both numerator and denominator so, it is canceled. And the expression became
=1−cos⁡A1+cos⁡A= \dfrac{{1 - \cos A}}{{1 + \cos A}}.
Now, (RHS) right hand side of equality is (1−cos⁡A)(1+cos⁡A)\dfrac{{\left( {1 - \cos A} \right)}}{{\left( {1 + \cos A} \right)}}.
We get that RHS and LHS of equality are equal.

Thus, the given equality (cot⁡A−cos⁡ecA)2=1−cos⁡A1+cos⁡A{\left( {\cot A - \cos ecA} \right)^2} = \dfrac{{1 - \cos A}}{{1 + \cos A}} is proved.

Note: Sometimes while we are proving the equality we may have to solve the expression on the (RHS) right hand side of the equality, then only we get LHS=RHSLHS = RHS.
This equality can also be proved by solving the RHS expression and then equating with LHS. For this multiply by (1−cos⁡A)\left( {1 - \cos A} \right) in both numerator and denominator, then proceed further and apply some trigonometrical formula to reach the result.