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

Hint: 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 :
\[\sin2x = 2 \times \sin x \times \cos x\]
$\cos 2x = 2 \times {\cos^{2}}x - 1 = 1 - 2 \times {\sin^{2}}x$
$\tan 2x = \dfrac{{tan2x}}{{(1 - {\tan^{2}}x)}}$
$\sin 3x = 3\\sin x - 4{\sin^{3}}x$
$\cos 3x = 4{\cos^{3}}x - 3\cos x$