Question

The value of expression $\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$is
1)$$$$\dfrac{{\sqrt 3 }}{4}
2)$$$\dfrac{4}{{\sqrt 3 }}$ 3)$$$\dfrac{2}{{\sqrt 3 }}$
4)$$$$\dfrac{{\sqrt 3 }}{2}

Answer 1

We have to find the value of the given trigonometric expression $\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$ . We solve this using the formula of the double angle of sine function . We should also have the knowledge of values for various angles of trigonometric functions . Firstly we find the value of expression using the formula of sum of two angles of a sine function . Also , the conversion of an angle in terms of another trigonometric function.

Complete step-by-step solution:
Given :
We have to find the value of the expression $\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$
We know , that
$\cos290^\circ = \cos\left( {360 - 70} \right)^\circ$ The angle lies in the fourth quadrant and the value of cos function in the fourth is positive .
So ,
$\cos290^\circ = \cos70^\circ$
Similarly ,
$\sin250^\circ = \sin\left( {180 + 70} \right)^\circ$
The angle lies in the third quadrant and the value of sine function in the third quadrant is negative .
So ,$\;\sin250^\circ = - \sin70^\circ$
Now , the expression becomes
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \left( {\dfrac{1}{{\cos70^\circ }}} \right) - \left( {\dfrac{1}{{\sqrt 3 \times \sin70^\circ }}} \right)
Taking L.C.M. we get
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \dfrac{{\left[ {\sqrt 3 \times \sin70^\circ - \cos70^\circ } \right]}}{{\left[ {{\text{ }}\sqrt 3 \times \sin70^\circ \times \cos70^\circ } \right]}}
Multiplying numerator and denominator by $4$ , we get
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \dfrac{{4 \times \left[ {\sqrt 3 \times \sin70^\circ - \cos70^\circ } \right]}}{{4 \times \left[ {{\text{ }}\sqrt 3 \times \sin70^\circ \times \cos70^\circ } \right]}}
Simplifying , we get
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \dfrac{{\left[ {\left( {\dfrac{{\sqrt 3 }}{2}} \right) \times \sin70^\circ - \left( {\dfrac{1}{2}} \right) \times \cos70^\circ } \right]}}{{\left( {\dfrac{{\sqrt 3 }}{4}} \right) \times \left[ {2 \times \sin70^\circ \times \cos70^\circ } \right]}}
We know that $\sin2x = 2\sin x \times \cos x$
We also know that
$\cos30^\circ = \dfrac{{\sqrt 3 }}{2}$
$\sin30^\circ = \dfrac{1}{2}$
Using these values , we get
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \dfrac{{\left[ {\cos30^\circ \times \sin70^\circ - \sin30^\circ \times \cos70^\circ } \right]}}{{\left[ {\left( {\dfrac{{\sqrt 3 }}{4}} \right) \times \sin140^\circ } \right]}}
Also , the formula of difference of sine function is given as :
$\sin\left( {A - B} \right) = \sin A \times \cos B - \sin B \times \cos A$
Also , we can write
$\sin140^\circ = \sin\left( {180 - 40} \right)^\circ$
The angle lies in the second quadrant and the value of sine function in the second quadrant is positive .
So , $\sin140^\circ = \sin40^\circ$
U\sing these trigonometric formulas , we get
\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$$$$ = \dfrac{{\sin40^\circ }}{{\left[ {\left( {\dfrac{{\sqrt 3 }}{4}} \right) \times \sin40^\circ } \right]}}
Cancelling the terms , we get
$\left( {\dfrac{1}{{\cos290^\circ }}} \right) + \left( {\dfrac{1}{{\sqrt 3 \times \sin250^\circ }}} \right)$ $= \dfrac{4}{{\sqrt 3 }}$
Thus the value of the expression is $\dfrac{4}{{\sqrt 3 }}$
Hence , the correct option is$\left( 2 \right)$.

Note: The various trigonometric formulas are given as :
$\sin2x = 2 \times \sin x \times \cos x$
$\cos 2x = 2 \times {\cos^{2}}x - 1 = 1 - 2 \times {\sin^{2}}x$
$\tan 2x = \dfrac{{tan2x}}{{(1 - {\tan^{2}}x)}}$
$\sin 3x = 3\\\sin x - 4{\sin^{3}}x$
$\cos 3x = 4{\cos^{3}}x - 3\cos x$