Question

The value of the expression $\dfrac{{1 - 4\sin {{10}^0}\sin {{70}^0}}}{{2\sin {{10}^0}}}$ is
(A)$\dfrac{1}{2}$
(B)1
(C) 2
(D) None of these

Answer 1

Hint- To solve this question we will use trigonometric identities such as $\sin (90 - \theta ) = \cos \theta {\text{ and cosA - cosB = 2sin(}}\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2})$

Complete step-by-step solution -
Given expression is $\dfrac{{1 - 4\sin {{10}^0}\sin {{70}^0}}}{{2\sin {{10}^0}}}............................(1)$
As we know that
$\sin (90 - \theta ) = \cos \theta {\text{ }} \\\ {\text{cosA - cosB = 2sin(}}\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2}) \\\$
From equation (1) write the angles of sin as a sum or difference of two angles such as ${70^0} = \dfrac{{{{60}^0} + {{80}^0}}}{2}{\text{ and 1}}{{\text{0}}^0} = \dfrac{{{{80}^0} - {{60}^0}}}{2}$ , we get
$= \dfrac{{1 - 4\sin \left( {\dfrac{{{{60}^0} + {{80}^0}}}{2}{\text{ }}} \right)\sin \left( {\dfrac{{{{80}^0} - {{60}^0}}}{2}} \right)}}{{2\sin ({{90}^0} - {{10}^0})}}$
Now, using the formulas mentioned above, we get
$= \dfrac{{1 - 2[\cos \left( {{\text{6}}{{\text{0}}^0}} \right) - \cos \left( {{{80}^0}} \right)]}}{{2\cos ({{80}^0})}}$
As we know that $\cos {60^0} = \dfrac{1}{2}$ substituting this value in the above equation, we get
$= \dfrac{{1 - 2 \times \dfrac{1}{2} + 2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\\ = \dfrac{{1 - 1 + 2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\\ = \dfrac{{2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\\ = 1 \\\$
So, the value of the given expression is 1.
Hence, the correct answer is option B.

Note- To solve this question, we used the trigonometric identities and some manipulation. Whenever we have an unknown or random angle in the problem, whose trigonometric values are unknown, try to manipulate some angle by using trigonometric identities in order to cancel that term or to bring the angle in some known value. Remember the trigonometric identities.