Question

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

Answer 1

First convert the given trigonometric expression in terms of sine and cosine. Next, take the L.C.M. of the denominators. Now, multiply and divide by 2 in both numerator and denominator of the obtained fraction. Use angle sum property of sine to obtain the value of the given expression.

Complete step-by-step answer:
The given trigonometric expression $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$ is solved as shown below.

$$\,\,\,\,\,\sqrt 3 \cos ec20^\circ - \sec 20^\circ \\\ \Rightarrow \dfrac{{\sqrt 3 }}{{\sin 20^\circ }} - \dfrac{1}{{\cos 20^\circ }} \\\$$

Now, take the L.C.M. of the denominator as shown below.
$\dfrac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }}$
Multiply and divide by 2 on both the numerator and denominator of the above expression.

$$\,\,\,\,\,\dfrac{{\dfrac{2}{2} \times \left( {\sqrt 3 \cos 20^\circ - \sin 20^\circ } \right)}}{{\dfrac{2}{2} \times \left( {\sin 20^\circ \cos 20^\circ } \right)}} \\\ \Rightarrow \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2} \cdot \cos 20^\circ - \dfrac{1}{2} \cdot \sin 20^\circ } \right)}}{{\dfrac{1}{2}\left( {2\sin 20^\circ \cos 20^\circ } \right)}} \\\ \Rightarrow \dfrac{{2\left( {\sin 60^\circ \cdot \cos 20^\circ - \cos 60^\circ \cdot \sin 20^\circ } \right)}}{{\dfrac{1}{2}\left( {\sin 40^\circ } \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {2\sin A\cos A = \sin 2A} \right) \\\ \Rightarrow \dfrac{{4\sin \left( {60^\circ - 20^\circ } \right)}}{{\left( {\sin 40^\circ } \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sin A\cos B - \cos A\sin B = sin\left( {A - B} \right)} \right) \\\$$

Further, simplify the above trigonometric expression.

$$\,\,\,\,\,\,\,\dfrac{{4\sin 40^\circ }}{{\left( {\sin 40^\circ } \right)}} \\\ \Rightarrow 4 \\\$$

Thus, the value of the trigonometric expression $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$ is 4.
Hence, option (A) is the correct answer.

Note: Try to remember all trigonometric formulas. Use the following trigonometric identities while solving the given problem.

$$\sin A\cos B - \cos A\sin B = sin\left( {A - B} \right) \\\ 2\sin A\cos A = \sin 2A \\\$$