Question

Prove that $\dfrac{\cos A+\sin A}{\cos A-\sin A}+\dfrac{\cos A-\sin A}{\cos A+\sin A}=2\sec 2A$ ?

Answer 1

Hint: Here we have to prove whether the given equation is true or not. So we will take the $LHS$ value and simplify it to get our $RHS$ value. We will simplify the $LHS$ 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 $RHS$ or not and prove the equation.

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

Now we will use the algebraic identity:
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
$\Rightarrow \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Using above identity in equation (1) where $a=\cos A$ and $b=\sin A$ we get,
$\Rightarrow \dfrac{\left( {{\cos }^{2}}A+{{\sin }^{2}}A+2\cos A\sin A \right)+\left( {{\cos }^{2}}A+{{\sin }^{2}}A-2\cos A\sin A \right)}{{{\cos }^{2}}A-{{\sin }^{2}}A}$
$\Rightarrow \dfrac{2{{\cos }^{2}}A+2{{\sin }^{2}}A}{{{\cos }^{2}}A-{{\sin }^{2}}A}$

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

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

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.