Question

How do you evaluate $\sec {15^\circ }$?

Answer 1

Hint : We have to find the value of the given trigonometric expression. We solve this question using the concept of the various formulas of trigonometric functions. We should also have the knowledge of the sine and cosine function for various angles. We should also have the knowledge of the relation between secant and cosine function. First, we will change the secant function in terms of cosine and then we will split the given angle such that we can substitute it in the formula of cosine of difference of two numbers. Then using the formula, we will expand the relation and then substituting the values in the expression we will get the value of the given function.

Complete step-by-step answer :
Given:
Evaluate $\sec {15^\circ }$
We know that the relation between cosine and secant is given as:
$\sec x = \dfrac{1}{{\cos x}}$
So, we can write the expression as:
$\sec {15^\circ } = \dfrac{1}{{\cos {{15}^\circ }}}$
Now, splitting the angles we can write the expression as:
$\sec {15^\circ } = \dfrac{1}{{\cos {{\left( {45 - 30} \right)}^\circ }}}$
We also know that the formula for cosine of difference of two angles is given as:
$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$
Using the formula, we can write the expression as:
$\sec {15^\circ } = \dfrac{1}{{\cos {{45}^\circ }\cos {{30}^\circ } + \sin {{45}^\circ }\sin {{30}^\circ }}}$
We also know that the values of the trigonometric functions is given as:
$\sin {30^\circ } = \dfrac{1}{2}$, $\cos {45^\circ } = \dfrac{1}{{\sqrt 2 }}$,$\cos {30^\circ } = \dfrac{{\sqrt 3 }}{2}$ , $\sin {45^\circ } = \dfrac{1}{{\sqrt 2 }}$
Substituting the values, we can write the expression as:
$\sec {15^\circ } = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}}}$
On further simplifying, we can write the expression as:
$\sec {15^\circ } = \dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}}$
Rationalising the above expression by multiplying numerator and denominator by $\sqrt 3 - 1$, we get
$\sec {15^\circ } = \dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}$
We know that the formula of difference of square of two numbers is given as:
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Using the formula, we can write the expression as:
$\sec {15^\circ } = \dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{3 - 1}}$
$\sec {15^\circ } = \dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{2}$
Simplifying, we get the value as:
$\sec {15^\circ } = \sqrt 6 - \sqrt 2$
Hence, we get the value of the expression $\sec {15^\circ }$ as $\sqrt 6 - \sqrt 2$.
So, the correct answer is “$\sqrt 6 - \sqrt 2$”.

Note : While splitting the given angle, we will split the angle such that we know the exact value for that particular angle. As in the above question we split the given angle as shown as we knew the value of both the sine and cosine function for ${30^\circ }$ and ${45^\circ }$. That we have split the above angle as ${15^\circ } = {20^\circ } - {5^\circ }$ or any other such combination would be of no use, as we don’t know the value for these angles.