Question

Find the value of $\tan {22^0}30'$.

Answer 1

Hint:First express tan into the form of sin and cos. Then use the twice angle formulas for both sin and cos to express the previous relation into trigonometric ratios whose values are known generally. Hence find the desired value.

Complete step-by-step answer:
Given in the problem we need to find the value of $\tan {22^0}30'$.
In a right-angle triangle, the trigonometric ratio tan is equal to the ratio of perpendicular to the base.
Since we do not know the value of $\tan {22^0}30'$directly, we need to express the same into a transformation such that the value of the trigonometric ratios on the RHS of the equations are known.
We could also express tan in the form of sin and cos as,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Replacing $\theta$ by $\dfrac{\theta }{2}$ in the above, we get
$\tan \dfrac{\theta }{2} = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}{\text{ (1)}}$
Multiplying both numerator and denominator of the equation (1) by $2\cos \dfrac{\theta }{2}$, we get
$\Rightarrow \tan \dfrac{\theta }{2} = \dfrac{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2{{\cos }^2}\dfrac{\theta }{2}}}$
We know that $\sin 2\theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$. Using the same in above equation, we get
$\Rightarrow \tan \dfrac{\theta }{2} = \dfrac{{\sin \theta }}{{2{{\cos }^2}\dfrac{\theta }{2}}}{\text{ (2)}}$
Also, we know that,
$\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1 \\\ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = 1 + \cos \theta \\\$
Using the above result in equation (2), we get
$\Rightarrow \tan \dfrac{\theta }{2} = \dfrac{{\sin \theta }}{{1 + \cos \theta }}{\text{ (3)}}$

Since we need to find the value of $\tan {22^0}30'$
We can use equation (3) to find the value of $\tan {22^0}30'$as the double of ${22^0}30'$is ${45^0}$whose values for the trigonometric ratios are known.
Let us assume $\theta = {45^0}$.
$\dfrac{\theta }{2} = \dfrac{{{{45}^0}}}{2} = {22.5^0} = {22^0}30'$
Using the above in equation (3), we get
$\Rightarrow \tan \left( {{{22}^0}30'} \right) = \dfrac{{\sin {{45}^0}}}{{1 + \cos {{45}^0}}}$
Using $\sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }}$ in above, we get
$\Rightarrow \tan \left( {{{22}^0}30'} \right) = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}}} = \dfrac{1}{{\sqrt 2 + 1}}$
Rationalising the denominator in the above value by multiplying both numerator and denominator by $\sqrt 2 - 1$, we get
$\Rightarrow \tan \left( {{{22}^0}30'} \right) = \dfrac{1}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} = \dfrac{{\sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}}$
Using the identity, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator above, we get

$$\Rightarrow \tan \left( {{{22}^0}30'} \right) = \dfrac{{\sqrt 2 - 1}}{{2 - 1}} = \sqrt 2 - 1 \\\ \Rightarrow \tan \left( {{{22}^0}30'} \right) = \sqrt 2 - 1 \\\$$

Hence, the value of $\tan {22^0}30'$is $\sqrt 2 - 1$.

Note:The relation between trigonometric ratios tan, sin and cos and the twice angle formulas used in the above solution should be kept in mind while solving problems like above. Values of trigonometric ratios for general angles should also be kept in mind while solving problems like above. A degree consists of sixty minutes and a minute consists of 60 seconds.