Question

The Value of expression $\left( {\dfrac{1}{{\cos {{290}^ \circ }}}} \right) + \left( {\dfrac{1}{{\sqrt 3 \sin {{250}^ \circ }}}} \right)$is
A. $\dfrac{{\sqrt 3 }}{4}$
B. $\dfrac{4}{{\sqrt 3 }}$
C. $\dfrac{2}{{\sqrt 3 }}$
D. $\dfrac{{\sqrt 3 }}{2}$

Answer 1

Hint: Use trigonometric sign convention identities or formulas to decrease the angle and make the expression easier to solve. Then multiply and divide two to form the expression same and trigonometric formula and apply.

Formula Used:
Trigonometric formulas –
$\cos ({270^ \circ } + \theta ) = \sin \theta $
$\sin ({270^ \circ } - \theta ) = - \cos \theta $
$\sin (A + B) = \sin A\cos B - \cos A\sin B$
$\operatorname{Sin} 2\theta = 2\sin \theta \cos \theta $

Complete step by step solution:
Given Expression,
$\left( {\dfrac{1}{{\cos {{290}^ \circ }}}} \right) + \left( {\dfrac{1}{{\sqrt 3 \sin {{250}^ \circ }}}} \right) - - - - - (1)$
Here, $\operatorname{Cos} {290^ \circ } = \cos ({270^ \circ } + {20^ \circ }) = \sin {20^ \circ }$
$\sin {250^ \circ } = \sin ({270^ \circ } - {20^ \circ }) = - \cos {20^ \circ }$
Equation (1) will be,
$ = \left( {\dfrac{1}{{\sin {{20}^ \circ }}}} \right) - \left( {\dfrac{1}{{\sqrt 3 \cos {{20}^ \circ }}}} \right)$
$ = \dfrac{{\sqrt 3 \cos {{20}^ \circ } - \sin {{20}^ \circ }}}{{\sqrt 3 \sin {{20}^ \circ }\cos {{20}^ \circ }}}$
$ = \dfrac{{2 \times \left( {\dfrac{{\sqrt 3 }}{2}\cos {{20}^ \circ } - \dfrac{1}{2}\sin {{20}^ \circ }} \right)}}{{2 \times \dfrac{{\sqrt 3 }}{2}\sin {{20}^ \circ }\cos {{20}^ \circ }}}$
Multiply and divide numerator and denominator by $2$,
$ = \dfrac{{2 \times \left( {\sin {{60}^ \circ }\cos {{20}^ \circ } - \cos {{60}^ \circ }\sin {{20}^ \circ }} \right)}}{{\dfrac{{\sqrt 3 }}{2} \times 2\sin {{20}^ \circ }\cos {{20}^ \circ }}}$
$ = \dfrac{{2 \times \left( {\sin ({{60}^ \circ } - {{20}^ \circ })} \right)}}{{\dfrac{{\sqrt 3 }}{2} \times \sin {{40}^ \circ }}}$
$ = \dfrac{{2 \times \sin {{40}^ \circ }}}{{\dfrac{{\sqrt 3 }}{2} \times \sin {{40}^ \circ }}}$
$ = \dfrac{4}{{\sqrt 3 }}$
Hence, Value of expression $\left( {\dfrac{1}{{\cos {{290}^ \circ }}}} \right) + \left( {\dfrac{1}{{\sqrt 3 \sin {{250}^ \circ }}}} \right)$is equal to $\dfrac{4}{{\sqrt 3 }}$

Option ‘B’ is correct

Note: In such question, where trigonometric expression will be given to solve. Use the formula carefully. For Example - In $\operatorname{Sin} 2\theta = 2\sin \theta \cos \theta $, here the angle is getting half while opening $\operatorname{Sin} 2\theta $ in terms of $\sin $ and $\cos $. For Trigonometric value learn to make the table and remember all the values on tip.