Question

Prove that
$\dfrac{\cos {{22}^{\circ }}+\sin {{22}^{\circ }}}{\cos {{22}^{\circ }}-\sin {{22}^{\circ }}}=\tan {{67}^{\circ }}$

Answer 1

To prove the given equation we will use the relation between sine and cosine function. Firstly we will try to simplify the numerator and denominator by converting them into product value using the formula of trigonometry. Then we will simplify the obtained result such that we can get our desired answer.

Complete step by step answer:
We have to prove that:
$\dfrac{\cos {{22}^{\circ }}+\sin {{22}^{\circ }}}{\cos {{22}^{\circ }}-\sin {{22}^{\circ }}}=\tan {{67}^{\circ }}$……$\left( 1 \right)$
We will take left hand side value and simplify it to get the right hand side value as follows:
$\dfrac{\cos {{22}^{\circ }}+\sin {{22}^{\circ }}}{\cos {{22}^{\circ }}-\sin {{22}^{\circ }}}$…..$\left( 2 \right)$
We know we have a formula for changing the Sum or difference into products of the trigonometric functions.
Firstly we will convert the cosine value in numerator and denominator into sine by using the below property:
$\cos \left( 90-\theta \right)=\sin \theta$
Using the above property in equation (2) we get,
$\dfrac{\cos {{\left( 90-68 \right)}^{\circ }}+\sin {{22}^{\circ }}}{\cos {{\left( 90-68 \right)}^{\circ }}-\sin {{22}^{\circ }}}$
$\Rightarrow \dfrac{\sin {{68}^{\circ }}+\sin {{22}^{\circ }}}{\sin {{68}^{\circ }}-\sin {{22}^{\circ }}}$….$\left( 3 \right)$
Now we will use the Formula for changing the Sum or difference into Product given below:
$\begin{aligned} & \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) \\\ & \sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) \\\ \end{aligned}$
Using above formula in equation (3) and simplify it to get,
$\begin{aligned} & \Rightarrow \dfrac{2\sin {{\left( \dfrac{68+22}{2} \right)}^{\circ }}\cos {{\left( \dfrac{68-22}{2} \right)}^{\circ }}}{2\cos {{\left( \dfrac{68+22}{2} \right)}^{\circ }}\sin {{\left( \dfrac{68-22}{2} \right)}^{\circ }}} \\\ & \Rightarrow \dfrac{2\sin {{\left( \dfrac{90}{2} \right)}^{\circ }}\cos {{\left( \dfrac{46}{2} \right)}^{\circ }}}{2\cos {{\left( \dfrac{90}{2} \right)}^{\circ }}\sin {{\left( \dfrac{46}{2} \right)}^{\circ }}} \\\ & \Rightarrow \dfrac{2\sin {{45}^{\circ }}\cos {{23}^{\circ }}}{2\cos {{45}^{\circ }}\sin {{23}^{\circ }}} \\\ \end{aligned}$
Now as we know the value of below terms put in above value:
$\begin{aligned} & \sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\\ & \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
So we get,
$\begin{aligned} & \Rightarrow \dfrac{2\times \dfrac{1}{\sqrt{2}}\cos {{23}^{\circ }}}{2\times \dfrac{1}{\sqrt{2}}\sin {{23}^{\circ }}} \\\ & \Rightarrow \dfrac{\cos {{23}^{\circ }}}{\sin {{23}^{\circ }}} \\\ \end{aligned}$
Next we know the relation that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ using above we get,
$\Rightarrow \cot {{23}^{\circ }}$
As we want our answer in tangent function we will use the relation between cotangent and tangent above which is $\cot \left( 90-\theta \right)=\tan \theta$
$\begin{aligned} & \Rightarrow \cot {{\left( 90-67 \right)}^{\circ }} \\\ & \Rightarrow \tan {{67}^{\circ }} \\\ \end{aligned}$
So we got the right hand side value of equation (1).
Hence Proved that $\dfrac{\cos {{22}^{\circ }}+\sin {{22}^{\circ }}}{\cos {{22}^{\circ }}-\sin {{22}^{\circ }}}=\tan {{67}^{\circ }}$

Note: Trigonometric function is a very vast and important topic in mathematics. It studies the relation between the side lengths and angles of triangles. There are six trigonometric functions in all which are sine, cosine, tangent, cosecant, secant, cotangent and they all are related to each other in different ways. There are many formulas and properties of all the trigonometric functions which we use while solving the problems.