Question

Compute the value of the following: $\text{cosec}{{15}^{\circ }}+\sec {{15}^{\circ }}$.

& A. 2\sqrt{2} \\\ & B.\sqrt{6} \\\ & C.2\sqrt{6} \\\ & D.\sqrt{6}+\sqrt{2} \\\ \end{aligned}$$

Answer 1

In order to find the value of $\text{cosec}{{15}^{\circ }}+\sec {{15}^{\circ }}$, firstly, we will be expressing $\text{cosec}{{15}^{\circ }}$ in terms of sine angle and then the angle would be expressed with the help of principle angle. And then we will be expressing $\sec {{15}^{\circ }}$ in terms of cos angle and then the angle would be expressed in terms of principal angle. After obtaining both of the values, upon adding them we will be obtaining the required solution.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between ${{0}^{\circ }}$ and ${{360}^{\circ }}$. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us start solving the problem.
Firstly let us consider $\text{cosec}{{15}^{\circ }}$. Upon expressing in terms of sine angle, we get
$\text{cosec}{{15}^{\circ }}=\dfrac{1}{\sin {{15}^{\circ }}}=\dfrac{1}{\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)}$
Now we will be applying the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and solve it. We get,
$\Rightarrow \dfrac{1}{\sin 45\cos 30-\cos 45\sin 30}=\dfrac{1}{\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}}=\dfrac{2\sqrt{2}}{\left( \sqrt{3}-1 \right)}$
Now let us consider $\sec {{15}^{\circ }}$. Upon expressing in terms of cos angle, we get
$sec{{15}^{\circ }}=\dfrac{1}{\cos {{15}^{\circ }}}=\dfrac{1}{\cos \left( {{45}^{\circ }}-{{30}^{\circ }} \right)}$
Now we will be applying the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and solve it, we get,
$\Rightarrow \dfrac{1}{\cos 45\cos 30+\sin 45\sin 30}=\dfrac{1}{\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}}=\dfrac{2\sqrt{2}}{\left( \sqrt{3}+1 \right)}$
Now upon adding both of the obtained values, we get
$\Rightarrow \dfrac{2\sqrt{2}}{\left( \sqrt{3}-1 \right)}+\dfrac{2\sqrt{2}}{\left( \sqrt{3}+1 \right)}=\dfrac{2\sqrt{6}+2\sqrt{2}+2\sqrt{6}-2\sqrt{2}}{3-1}=\dfrac{4\sqrt{6}}{2}=2\sqrt{6}$
$\therefore$ Option C is the correct answer.

Note: While solving such problems, we must always have a note that we must be converting the angles in terms of principal angles also the trigonometric ratios are to be converted accordingly. We must also check for the formulas of trigonometry to which it resembles for easy solving.