Question

The smallest positive value of $x$ in degrees for which
\[\tan x = \dfrac{{\cos {5^ \circ }\cos {{20}^ \circ } + \cos {{35}^ \circ }\cos {{50}^ \circ } - \sin {{50}^ \circ }\sin {{20}^ \circ } - \sin {{35}^ \circ }\sin {{50}^ \circ }}}{{\sin {5^ \circ }\cos {{20}^ \circ } - \sin {{35}^ \circ }\cos {{50}^ \circ } + \cos {5^ \circ }\sin {{20}^ \circ } - \cos {{35}^ \circ }\sin {{50}^ \circ }}}\] is equal to:
$\left( a \right){\text{ 3}}{{\text{0}}^ \circ }$
$\left( b \right){\text{ 6}}{{\text{0}}^ \circ }$
$\left( c \right){\text{ 7}}{{\text{5}}^ \circ }$
$\left( d \right){\text{ 12}}{{\text{0}}^ \circ }$

Answer 1

Hint:
In this question, we will use some trigonometric formula to reduce the equation and will then put the values and solve it. But for this, we need first to reduce it. So we will use it here $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$and also we will use it$\sin \left( {A + B} \right) = \sin A\cos B - \cos A\sin B$. By using these formulas we will solve this problem easily.

Formula used:
Formula related to sums and differences
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$\sin \left( {A + B} \right) = \sin A\cos B - \cos A\sin B$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$

Complete step by step solution:
 We have the equation as
\[ \Rightarrow \tan x = \dfrac{{\cos {5^ \circ }\cos {{20}^ \circ } + \cos {{35}^ \circ }\cos {{50}^ \circ } - \sin {{50}^ \circ }\sin {{20}^ \circ } - \sin {{35}^ \circ }\sin {{50}^ \circ }}}{{\sin {5^ \circ }\cos {{20}^ \circ } - \sin {{35}^ \circ }\cos {{50}^ \circ } + \cos {5^ \circ }\sin {{20}^ \circ } - \cos {{35}^ \circ }\sin {{50}^ \circ }}}\]
By using the above formula, on applying it we get
$ \Rightarrow \tan x = \dfrac{{\cos \left( {{5^ \circ } + {{20}^0}} \right) + \cos \left( {{{35}^ \circ } + {{50}^ \circ }} \right)}}{{\sin \left( {{5^ \circ } + {{20}^0}} \right) - \sin \left( {{{35}^ \circ } + {{50}^ \circ }} \right)}}$
So on solving the above equation can be written as
$ \Rightarrow \tan x = \dfrac{{\cos {{25}^0} + \cos {{85}^0}}}{{\sin {{25}^0} - \sin {{85}^0}}}$
And as we know
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
And also we know that $\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$
So on applying the formula in the equation, we get
$ \Rightarrow \tan x = \dfrac{{2\cos {{55}^0}.\cos {{30}^0}}}{{ - 2\cos {{55}^0}.\sin {{30}^0}}}$
Therefore, on solving the above equation we get
$ \Rightarrow \tan x = \dfrac{{\cos {{30}^0}}}{{ - .\sin {{30}^0}}}$
Now as we know,
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$, and $\sin {30^ \circ } = \dfrac{1}{2}$
So putting these values in the above equation, we get
$ \Rightarrow \tan x = - \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}$
And on solving it, we get
$ \Rightarrow \tan x = - \sqrt 3 $
Hence the value for $\tan x = - \sqrt 3 $ and it will be equal to ${120^ \circ }$.

Therefore, the option $\left( d \right)$ is correct.

Note:
We have seen initially how big the question was looking but when it comes to solving it becomes very easy to solve by using some trigonometric formula. So for solving such a type of problem we have to memorize the important formula and by practice only we can do it. So we must be clear about using the trigonometric identities and somewhere we have to make the identity in the given order. It makes the question simpler and hence helps you to reach a correct solution. In this way, we can solve this type of problem very easily.