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Question: Prove that \(\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}=2\sec 2A\) ?...

Prove that cos⁡A+sin⁡Acos⁡A−sin⁡A+cos⁡A−sin⁡Acos⁡A+sin⁡A=2sec⁡2A\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}=2\sec 2A ?

Explanation

Solution

Here we have to prove whether the given equation is true or not. So we will take the LHSLHS value and simplify it to get our RHSRHS value. We will simplify the LHSLHS value by taking the LCM of the two values given then we will use the square relation and double angle formula to simplify it further. Finally we will see whether the value obtained is our RHSRHS or not and prove the equation.

Complete step by step answer:
We have to prove the equation given below:
cos⁡A+sin⁡Acos⁡A−sin⁡A+cos⁡A−sin⁡Acos⁡A+sin⁡A=2sec⁡2A\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}=2\sec 2A
So we will take the LHSLHS value and simplify it to get our RHSRHS value.
So our LHSLHS (Left Hand Side) value is as follows:
cos⁡A+sin⁡Acos⁡A−sin⁡A+cos⁡A−sin⁡Acos⁡A+sin⁡A\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}
Taking LCM we get,
⇒(cos⁡A+sin⁡A)(cos⁡A+sin⁡A)+(cos⁡A−sin⁡A)(cos⁡A−sin⁡A)(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)\Rightarrow \dfrac{\left( \cos A+\sin A \right)\left( \cos A+\sin A \right)+\left( \cos A-\sin A \right)\left( \cos A-\sin A \right)}{\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)}
⇒(cos⁡A+sin⁡A)2+(cos⁡A−sin⁡A)2(cos⁡A+sin⁡A)(cos⁡A−sin⁡A)\Rightarrow \dfrac{{{\left( \cos A+\sin A \right)}^{2}}+{{\left( \cos A-\sin A \right)}^{2}}}{\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)}…..(1)\left( 1 \right)

Now we will use the algebraic identity:
(a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
⇒(a−b)2=a2+b2−2ab\Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab
⇒(a+b)(a−b)=a2−b2\Rightarrow \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}
Using above identity in equation (1) where a=cos⁡Aa=\cos A and b=sin⁡Ab=\sin A we get,
⇒(cos⁡2A+sin⁡2A+2cos⁡Asin⁡A)+(cos⁡2A+sin⁡2A−2cos⁡Asin⁡A)cos⁡2A−sin⁡2A\Rightarrow \dfrac{\left( {{\cos }^{2}}A+{{\sin }^{2}}A+2\cos A\sin A \right)+\left( {{\cos }^{2}}A+{{\sin }^{2}}A-2\cos A\sin A \right)}{{{\cos }^{2}}A-{{\sin }^{2}}A}
⇒2cos⁡2A+2sin⁡2Acos⁡2A−sin⁡2A\Rightarrow \dfrac{2{{\cos }^{2}}A+2{{\sin }^{2}}A}{{{\cos }^{2}}A-{{\sin }^{2}}A}

Take 22 common in the numerator,
⇒2(cos⁡2A+sin⁡2A)cos⁡2A−sin⁡2A\Rightarrow \dfrac{2\left( {{\cos }^{2}}A+{{\sin }^{2}}A \right)}{{{\cos }^{2}}A-{{\sin }^{2}}A}
Now we know the square relation is that cos⁡2A+sin⁡2A=1{{\cos }^{2}}A+{{\sin }^{2}}A=1 and double angle formula is that cos⁡2A−sin⁡2A=cos⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A substitute these values above,
⇒2×1cos⁡2A\Rightarrow \dfrac{2\times 1}{\cos 2A}
⇒2cos⁡2A\Rightarrow \dfrac{2}{\cos 2A}
Next we know that inverse cosine function is equal to secant function that is sec⁡A=1cos⁡A\sec A=\dfrac{1}{\cos A} so,
⇒2sec⁡2A\Rightarrow 2\sec 2A
This is our RHSRHS (Right Hand side).

Hence proved that cos⁡A+sin⁡Acos⁡A−sin⁡A+cos⁡A−sin⁡Acos⁡A+sin⁡A=2sec⁡2A\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}=2\sec 2A.

Note: As we have been given two fraction values our first step will mostly be to take the LCM as it will simplify our value further. We have to get the value as a secant function which means that we need the cosine function in the denominator so we have used the formula accordingly. In this type of question what should be our outcome is to be kept in mind while solving the value. Basic relations and formulas are very useful in these questions as they simplify the equation.