Question

The value of $\cos ec{15^0} + \sec {15^0} = $
A.$2\sqrt 3 $
B.$\sqrt 6 $
C.$2\sqrt 6 $
D.$\sqrt 6 + \sqrt 2 $

Answer 1

Hint: We have given a trigonometric expression in cosecant and secant. With angle${15^0}$. We have to calculate the value of trigonometric expression. Firstly we write the angle in $A + B{\text{ or }}A - B$ form. The function \[cosecant{\text{ and }}secant\] will be converted in $\cos ec(A + B)$ form and $\sec (A + B)$ form. Then we apply trigonometric formula of $\cos ec(A + B){\text{ and secant}}(A + B)$ or we can convert it in $\sin (A + B){\text{ and cos}}(A + B)$ . After expanding this we will put the values of angles and solve it.

Complete step-by-step answer:
We have given a trigonometric expression. $\cos ec{15^0} + \sec {15^0}$. Angle ${15^0}$ can be written in the difference of two angles ${60^0}$ and ${45^0}$
So ${15^0} = {60^0} - {45^0}$
Therefore \[\cos ec{15^0} + \sec {15^0} = \cos ec({60^0} - {45^0}) + \sec ({60^0} - {45^0})\]
Also we know that $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$
So $\cos ec{15^0} + \sec {15^0} = \dfrac{1}{{\sin ({{60}^0} - {{45}^0})}} + \dfrac{1}{{\cos ({{60}^0} - {{45}^0})}}{\text{ - - - - - - - - - (i)}}$
We first solve $\sin \left( {{{60}^0} - {{45}^0}} \right)$
$\sin \left( {{{60}^0} - {{45}^0}} \right)$is in the form $\sin \left( {A - B} \right)$
Also we know that
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
So \[\sin \left( {{{60}^0} - {{45}^0}} \right) = \sin {60^0}\cos {45^0} - \cos {60^0}\sin {45^0}\]
Value of \[\sin {60^0} = \dfrac{{\sqrt 3 }}{2},{\text{ }}\cos {45^0} = \dfrac{1}{{\sqrt 2 }}\]
\[\cos {60^0} = \dfrac{1}{a},{\text{ }}\sin {45^0} = \dfrac{1}{{\sqrt 2 }}\]
Therefore \[\sin \left( {{{60}^0} - {{45}^0}} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\]
Now we calculate $\cos \left( {60{}^0 - {{45}^0}} \right)$
We know that $\cos \left( {A - B} \right)$
$ = \cos A - \cos B + \sin A\sin B$
So $\cos \left( {{{60}^0} - {{45}^0}} \right) = \cos {60^0}\cos {45^0} + \sin {60^0}\sin {45^0}$
$ = \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}$
\[ \Rightarrow \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }} = + \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\]
Putting these values in equation (i)
$\cos ec{15^0} - \sec {15^0} = \dfrac{1}{{\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}}} + \dfrac{1}{{\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}}$
$ \Rightarrow \dfrac{{2\sqrt 2 }}{{\sqrt 2 - 1}} + \dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}}$
Taking L.C.H. and solving
$\cos ec{15^0} - \sec {15^0} = \dfrac{{2\sqrt 2 \left( {\sqrt 3 + 1} \right) + 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}$
$\cos ec{15^0} - \sec {15^0} = \dfrac{{2\sqrt 6 + 2\sqrt 2 + 2\sqrt 6 - 2\sqrt 2 }}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}$
$ \Rightarrow \dfrac{{456}}{{31}} \Rightarrow {\text{ }}\dfrac{{4\sqrt 6 }}{2}{\text{ }} \Rightarrow {\text{ }}2\sqrt 6 $
So value of $\cos ec{15^0} + \sec {15^0} = 2\sqrt 6 $
Option (C) is correct .

Note: Trigonometry is the branch of mathematics that studies the relationship between side lengths and angles of the triangle. Trigonometry has six trigonometric functions. Which are ${\text{sin, cos, tan, cosec, sec and cot}}$. Trigonometric functions are the real functions which relate an angle of right angle triangles to the ratio of two sides of a triangle.
Trigonometric functions are also called circular functions. With the help of these trigonometric functions we can drive lots of trigonometric formulas.