Question

The value of $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$ is
A) 2
B) 3
C) 1
D) 4

Answer 1

Here we need to find the value of the given trigonometric expression. For that, we will use the basic trigonometric identities to simplify the expression. We will first use the reciprocal trigonometric function identities of trigonometry and then we will use the trigonometric difference identities to further simplify the expression. On further simplifying the terms, we will get the value of the given trigonometric expression.

Formula used:
We will use the following formulas:

1. $\cos ec \theta = \dfrac{1}{{\sin \theta }}$
2. $\sec \theta = \dfrac{1}{{\cos \theta }}$
3. $\sin A \cdot \cos B - \sin B \cdot \cos A = \sin \left( {A - B} \right)$

Complete Step by Step Solution:
The given trigonometric expression is $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$.
Now, we will use the reciprocal trigonometric identities $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$ in the given trigonometric expression. Therefore, we get
$\sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{\sqrt 3 }}{{\sin 20^\circ }} - \dfrac{1}{{\cos 20^\circ }}$
On further simplifying the terms, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cdot \cos 20^\circ }}$
On dividing and multiplying the terms of numerator by 2, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2} \times \cos 20^\circ - \dfrac{1}{2} \times \sin 20^\circ } \right)}}{{\sin 20^\circ \cdot \cos 20^\circ }}$
We know the values $\cos 60^\circ = \dfrac{1}{2}$ and $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$ .
Now, substituting these values in the above equation, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2\left( {\sin 60^\circ \times \cos 20^\circ - \cos 60^\circ \times \sin 20^\circ } \right)}}{{\sin 20^\circ \cdot \cos 20^\circ }}$
Now, using the trigonometric difference identities $\sin A \cdot \cos B - \sin B \cdot \cos A = \sin \left( {A - B} \right)$, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2 \times \sin \left( {60^\circ - 20^\circ } \right)}}{{\sin 20^\circ \cdot \cos 20^\circ }}$
On further simplifying the terms, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2 \times \sin 40^\circ }}{{\sin 20^\circ \cdot \cos 20^\circ }}$
Now, multiplying the numerator and denominator by 2, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2 \times \sin 40^\circ \times 2}}{{2 \times \sin 20^\circ \times \cos 20^\circ }}$
Now, using the double angle trigonometric identities $2\sin \theta \cos \theta = \sin 2\theta$, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = \dfrac{{2 \times \sin 40^\circ \times 2}}{{\sin 40^\circ }}$
On further simplifying the terms, we get
$\Rightarrow \sqrt 3 \cos ec20^\circ - \sec 20^\circ = 2 \times 2 = 4$
Therefore, the value of the given trigonometric expression is equal to 4.

Hence, the correct option is option D.

Note:
Trigonometric identities are defined as the equalities which involve the trigonometric functions and they are always true for every value of the occurring variables for which both sides of the equality are defined. We know that all the trigonometric identities are periodic in nature. They repeat their values after a certain interval.