Question

Find the value of $\dfrac{{\cos {{20}^ \circ } + 8\sin {{70}^ \circ }\sin {{50}^ \circ }\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$.
(A) $1$
(B) $2$
(C) $\dfrac{3}{4}$
(D) None of these

Answer 1

According to this question, we need to know the trigonometric formulas which include the sum and difference identities. These include formulas for $\sin x + \sin y$, $\sin x - \sin y$, $\cos x + \cos y$, $\cos x - \cos y$. By applying these formulas, we can easily solve this question.

Complete step-by-step solution:
We need to solve the given term. The term is:
$\dfrac{{\cos {{20}^ \circ } + 8\sin {{70}^ \circ }\sin {{50}^ \circ }\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
First, we will use the trigonometric formula which is:
$\cos A - \cos B = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$
Now, we will try to solve and make the question, according to the given formula, and we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4(2\sin {{70}^ \circ }\sin {{50}^ \circ })\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
According to the formula, we can change the term $2\sin {70^ \circ }\sin {50^ \circ } = 2\sin \dfrac{{20 + 120}}{2}\sin \dfrac{{120 - 20}}{2}$, and we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4\left( {2\sin \dfrac{{20 + 120}}{2}\sin \dfrac{{120 - 20}}{2}} \right)\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Now, when we apply the formula $\cos A - \cos B = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$here, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4(\cos {{20}^ \circ } - \cos {{120}^ \circ })\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Now, we will open the bracket, and we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4\cos {{20}^ \circ }\sin {{10}^ \circ } - 4\cos {{120}^ \circ }\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Now, from the trigonometric table, we know that the value of $\cos {120^ \circ }$is $- \dfrac{1}{2}$. When we put the value of $\cos {120^ \circ }$here, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4\cos {{20}^ \circ }\sin {{10}^ \circ } - 4\left( { - \dfrac{1}{2}} \right)\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
After cancelling the divisible terms and after simplifying the term, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 4\cos {{20}^ \circ }\sin {{10}^ \circ } + 2\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
We can modify the term as:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2(2\cos {{20}^ \circ }\sin {{10}^ \circ }) + 2\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Now, we will use the trigonometric formula which is:
$\sin A - \sin B = 2\cos \dfrac{{x + y}}{2}\sin \dfrac{{x - y}}{2}$
We will try to make the solving term be in this form, and we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2\left( {2\cos \left( {\dfrac{{10 + 30}}{2}} \right)\sin \left( {\dfrac{{10 - 30}}{2}} \right)} \right) + 2\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
By applying the formula, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2(\sin {{30}^ \circ } - \sin {{10}^ \circ }) + 2\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
When we open the bracket, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2\sin {{30}^ \circ } - 2\sin {{10}^ \circ } + 2\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Here, similar terms are getting cancel out, and we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2\sin {{30}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}}$
Now, we know the value of $\sin 30$is $\dfrac{1}{2}$. When we put this value, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 2\left( {\dfrac{1}{2}} \right)}}{{{{\sin }^2}{{80}^ \circ }}}$
We can rewrite the denominator as:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 1}}{{{{\sin }^2}({{90}^ \circ } - {{10}^ \circ })}}$
We will apply the co-function identities for the denominator part. The formula is:
$\sin ({90^ \circ } - x) = \cos x$
After applying this formula in the denominator part, we get:
$\Rightarrow \dfrac{{\cos {{20}^ \circ } + 1}}{{{{\cos }^2}{{10}^ \circ }}}$
We know that $\cos 2\theta = 2{\cos ^2}\theta - 1$. When we rewrite this formula, we get:
$\cos 2\theta + 1 = 2{\cos ^2}\theta$
By applying this formula in our numerator part, we will assume $\theta = 10$, and we get:
$\Rightarrow \dfrac{{2{{\cos }^2}{{10}^ \circ }}}{{{{\cos }^2}{{10}^ \circ }}}$
The similar terms get cancelled out, and we get:
$= 2$

Therefore, $\dfrac{{\cos {{20}^ \circ } + 8\sin {{70}^ \circ }\sin {{50}^ \circ }\sin {{10}^ \circ }}}{{{{\sin }^2}{{80}^ \circ }}} = 2$. So, option (B) is correct.

Note: We need to always use the correct formula at the correct place. There are many places where we can use more than one formula, but we have to see which formula is more suitable. Choosing other formulas may also lead to an answer, but they may not be the desired answers.