Question

Prove the following:
$\sec \left( \dfrac{\pi }{4}+\theta \right)\sec \left( \dfrac{\pi }{4}-\theta \right)=2\sec 2\theta$

Answer 1

Hint: First of all take, LHS of the given equation. Now use, $secx=\dfrac{1}{\cos x}$. Now use the formula of $\text{cos}\left( A\pm B \right)\text{ }=\cos A\cos B\mp \sin A\sin B$ and substitute the value of $\cos \dfrac{\pi }{4}\text{ and }\sin \dfrac{\pi }{4}$ from trigonometric table. Now simplify the equation and use ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x\text{ and }\cos x=\dfrac{1}{\sec x}$ to prove the desired result.

Complete step-by-step answer:

In this question, we have to prove that
$\sec \left( \dfrac{\pi }{4}+\theta \right)\sec \left( \dfrac{\pi }{4}-\theta \right)=2\sec 2\theta$
Let us consider the LHS of the equation given in the question.
$LHS=\sec \left( \dfrac{\pi }{4}+\theta \right)\sec \left( \dfrac{\pi }{4}-\theta \right)......\left( i \right)$
We know that $secx=\dfrac{1}{\cos x}$. By using this in the above equation, we get,
$LHS=\dfrac{1}{\cos \left( \dfrac{\pi }{4}+\theta \right)}.\dfrac{1}{\cos \left( \dfrac{\pi }{4}-\theta \right)}$
We know that cos (A + B) = cos A cos B - sin A sin B and cos (A – B) = cos A cos B + sin A sin B. So, by taking $A=\dfrac{\pi }{4}\text{ and }B=\theta$ and using these in the above equation, we get,
$LHS=\dfrac{1}{\left( \cos \dfrac{\pi }{4}\cos \theta -\sin \dfrac{\pi }{4}\sin \theta \right)}.\dfrac{1}{\left( \cos \dfrac{\pi }{4}\cos \theta +\sin \dfrac{\pi }{4}\sin \theta \right)}....\left( ii \right)$

From the trigonometric table, we know $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\text{ and }\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$. By substituting these in equation (ii), we get
$LHS=\dfrac{1}{\left( \dfrac{\cos \theta }{\sqrt{2}}-\dfrac{\sin \theta }{\sqrt{2}} \right)}.\dfrac{1}{\left( \dfrac{\cos \theta }{\sqrt{2}}+\dfrac{\sin \theta }{\sqrt{2}} \right)}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
By using this in the above equation, we get,
$LHS=\dfrac{1}{\left[ {{\left( \dfrac{\cos \theta }{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{\sin \theta }{\sqrt{2}} \right)}^{2}} \right]}$
$LHS=\dfrac{1}{\dfrac{{{\cos }^{2}}\theta }{2}-\dfrac{{{\sin }^{2}}\theta }{2}}$
$LHS=\dfrac{1}{\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{2}}$
$LHS=\dfrac{2}{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$
We know that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$. By using this in the above equation, we get,
$LHS=\dfrac{2}{\cos 2\theta }$
We know that $\cos x=\dfrac{1}{\sec x}$. By using this in the above equation, we get,
$LHS=\dfrac{2}{\dfrac{1}{\sec 2\theta }}$
$LHS=2\sec 2\theta$
LHS = RHS
Hence proved
So, we have proved that $\sec \left( \dfrac{\pi }{4}+\theta \right)\sec \left( \dfrac{\pi }{4}-\theta \right)=2\sec 2\theta$

Note: First of all, students should remember that in trigonometry, one formula can be transformed into multiple expressions as we have $\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x$. So, students are advised to remember all the transformations of common formulas to easily solve the question. Also, take special care while writing the angles.