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Question: If \(m = \cos A - \sin A\) and \(n = \cos A + \sin A\) then, show that - \(\dfrac{{{m^2} + {n^2}}}{{...

If m=cos⁡A−sin⁡Am = \cos A - \sin A and n=cos⁡A+sin⁡An = \cos A + \sin A then, show that - m2+n2m2−n2=−12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}

Explanation

Solution

In this question, we are given a trigonometric equation and we have to prove that LHS = RHS. At first divide the question into two parts, namely - m2+n2m2−n2=−12sec⁡Acos⁡ecA\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{2}\sec A\cos ecA and −12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2 - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}. Prove both the parts individually by using certain very basic trigonometric identities. In both the parts, start by taking the LHS. Do your operations on LHS in such a way that you get RHS.

Complete step-by-step solution:
Let us first note down whatever we are given.
Given: If m=cos⁡A−sin⁡Am = \cos A - \sin A and n=cos⁡A+sin⁡An = \cos A + \sin A
To prove: m2+n2m2−n2=−12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}
We will divide the equation to be proved into two parts:
m2+n2m2−n2=−12sec⁡Acos⁡ecA\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{2}\sec A\cos ecA and −12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2 - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}.
Let us solve part 1 first.
Let us find the value of m2{m^2} and n2{n^2}. (Using cos⁡2A+sin⁡2A=1{\cos ^2}A + {\sin ^2}A = 1)
⇒m2=(cos⁡A−sin⁡A)2=cos⁡2A+sin⁡2A−2cos⁡Asin⁡A=1−2sin⁡Acos⁡A\Rightarrow {m^2} = {\left( {\cos A - \sin A} \right)^2} = {\cos ^2}A + {\sin ^2}A - 2\cos A\sin A = 1 - 2\sin A\cos A
⇒n2=(cos⁡A+sin⁡A)2=cos⁡2A+sin⁡2A+2sin⁡Acos⁡A=1+2sin⁡Acos⁡A\Rightarrow {n^2} = {\left( {\cos A + \sin A} \right)^2} = {\cos ^2}A + {\sin ^2}A + 2\sin A\cos A = 1 + 2\sin A\cos A
Now, we will add and subtract both.
⇒m2+n2=1−2sin⁡Acos⁡A+1+2sin⁡Acos⁡A=2\Rightarrow {m^2} + {n^2} = 1 - 2\sin A\cos A + 1 + 2\sin A\cos A = 2
⇒m2−n2=1−2sin⁡Acos⁡A−(1+2sin⁡Acos⁡A)=−4sin⁡Acos⁡A\Rightarrow {m^2} - {n^2} = 1 - 2\sin A\cos A - \left( {1 + 2\sin A\cos A} \right) = - 4\sin A\cos A
Putting both of them in m2+n2m2−n2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}},
⇒m2+n2m2−n2=2−4sin⁡Acos⁡A=−12sin⁡Acos⁡A\Rightarrow \dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = \dfrac{2}{{ - 4\sin A\cos A}} = - \dfrac{1}{{2\sin A\cos A}}
Now, we know that 1sin⁡A=cos⁡ecA\dfrac{1}{{\sin A}} = \cos ecA and 1cos⁡A=sec⁡A\dfrac{1}{{\cos A}} = \sec A. Using this, we get,
⇒m2+n2m2−n2=−12sin⁡Acos⁡A=−12cos⁡ecAsec⁡A\Rightarrow \dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{{2\sin A\cos A}} = - \dfrac{1}{2}\cos ecA\sec A
Hence, 1st part has been proved.
Moving towards 2nd part,
To prove: −12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2 - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}.
To prove this, we will use the formula - sec⁡2A=1+tan⁡2A{\sec ^2}A = 1 + {\tan ^2}A and cos⁡ec2A=1+cot⁡2A\cos e{c^2}A = 1 + {\cot ^2}A.
⇒sec⁡A=1+tan⁡2A\Rightarrow \sec A = \sqrt {1 + {{\tan }^2}A}, cos⁡ecA=1+cot⁡2A\cos ecA = \sqrt {1 + {{\cot }^2}A}
Putting in the LHS of the given equation,
LHS = −12(1+tan⁡2A)(1+cot⁡2A)- \dfrac{1}{2}\sqrt {\left( {1 + {{\tan }^2}A} \right)\left( {1 + {{\cot }^2}A} \right)}
Next, we will open the brackets by multiplying.
⇒−121+cot⁡2A+tan⁡2A+tan⁡2Acot⁡2A\Rightarrow - \dfrac{1}{2}\sqrt {1 + {{\cot }^2}A + {{\tan }^2}A + {{\tan }^2}A{{\cot }^2}A}
We know that tan⁡Acot⁡A=1\tan A\cot A = 1. Therefore, tan⁡2Acot⁡2A=1{\tan ^2}A{\cot ^2}A = 1. Putting this in the above equation,
⇒−121+cot⁡2A+tan⁡2A+1\Rightarrow - \dfrac{1}{2}\sqrt {1 + {{\cot }^2}A + {{\tan }^2}A + 1}
On simplifying we get,
⇒−122+cot⁡2A+tan⁡2A\Rightarrow - \dfrac{1}{2}\sqrt {2 + {{\cot }^2}A + {{\tan }^2}A}
Now, we can say that 2=2tan⁡Acot⁡A2 = 2\tan A\cot A as tan⁡Acot⁡A=1\tan A\cot A = 1.
Putting this in the above equation,
⇒−122tan⁡Acot⁡A+cot⁡2A+tan⁡2A\Rightarrow - \dfrac{1}{2}\sqrt {2\tan A\cot A + {{\cot }^2}A + {{\tan }^2}A}
Now, look closely and you will see that it is forming the identity of (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab.
⇒−12(tan⁡A+cot⁡A)2\Rightarrow - \dfrac{1}{2}\sqrt {{{\left( {\tan A + \cot A} \right)}^2}}
Simplifying the equation,
⇒−12(tan⁡A+cot⁡A)\Rightarrow - \dfrac{1}{2}\left( {\tan A + \cot A} \right) = RHS
Therefore, LHS = RHS.

Hence, m2+n2m2−n2=−12sec⁡Acos⁡ecA=−(cot⁡A+tan⁡A)2\dfrac{{{m^2} + {n^2}}}{{{m^2} - {n^2}}} = - \dfrac{1}{2}\sec A\cos ecA = - \dfrac{{\left( {\cot A + \tan A} \right)}}{2}.

Note: While proving the 2nd part, we used the identity - sec⁡2A=1+tan⁡2A{\sec ^2}A = 1 + {\tan ^2}A and cos⁡ec2A=1+cot⁡2A\cos e{c^2}A = 1 + {\cot ^2}A because the LHS part was in the terms of sec and cosec and he had to convert it into the terms of tan and cot. The only formula that we had were these two.
(While doing such questions, you need to understand the reason behind the steps that we are taking. Only then, you will be able to ask such questions yourself.)