Question

Evaluate $\dfrac{{\sec \theta \cos ec\left( {{{90}^ \circ } - \theta } \right) - \tan \theta \cot \left( {{{90}^ \circ } - \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\sin }^2}{{35}^ \circ }}}{{\tan {{10}^ \circ }\tan {{20}^ \circ }\tan {{60}^ \circ }\tan {{70}^ \circ }\tan {{80}^ \circ }}}$
A. $\dfrac{1}{{\sqrt 3 }}$
B. $\dfrac{2}{{\sqrt 3 }}$
C. $\sqrt 3$
D. 1

Answer 1

Here first of all we have to use trigonometry ratios for complementary angles then we will use some trigonometric identities i.e. Pythagorean Identity and Reciprocal Identities and we will get the required answer.

Complete step-by-step answer:
We have to evaluate so let I = $\dfrac{{\sec \theta \cos ec\left( {{{90}^ \circ } - \theta } \right) - \tan \theta \cot \left( {{{90}^ \circ } - \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\sin }^2}{{35}^ \circ }}}{{\tan {{10}^ \circ }\tan {{20}^ \circ }\tan {{60}^ \circ }\tan {{70}^ \circ }\tan {{80}^ \circ }}}$
Since we know that, $\cos ec\left( {{{90}^ \circ } - \theta } \right) = \sec \theta$, $\cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta$, $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta$ and $\cot \theta = \dfrac{1}{{\tan \theta }}$
Therefore, I = $\dfrac{{\sec \theta \left( {\sec \theta } \right) - \tan \theta \left( {\tan \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\cos }^2}{{55}^ \circ }}}{{\tan {{10}^ \circ }\tan \left( {{{90}^ \circ } - {{80}^ \circ }} \right)\tan {{20}^ \circ }\tan \left( {{{90}^ \circ } - {{70}^ \circ }} \right)\tan {{60}^ \circ }}}$
Now simplifying the above equation, we get
$\Rightarrow$I $= \dfrac{{\left( {\sec \theta \times \sec \theta } \right) - \left( {\tan \theta \times \tan \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\cos }^2}{{55}^ \circ }}}{{\tan {{10}^ \circ }\cot {{10}^ \circ }\tan {{20}^ \circ }\cot {{20}^ \circ }\tan {{60}^ \circ }}}$
As we know that $\cot \theta = \dfrac{1}{{\tan \theta }}$
So, I $= \dfrac{{\left( {\sec \theta \times \sec \theta } \right) - \left( {\tan \theta \times \tan \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\cos }^2}{{55}^ \circ }}}{{\tan {{10}^ \circ }\dfrac{1}{{\tan {{10}^ \circ }}}\tan {{20}^ \circ }\dfrac{1}{{\tan {{20}^ \circ }}}\tan {{60}^ \circ }}}$
Simplifying again the above equation we get
I $= \dfrac{{\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\cos }^2}{{55}^ \circ }}}{{\tan {{60}^ \circ }}}$
Now we know that $\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) = 1$, $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1$ and the value of $\tan {60^ \circ } = \sqrt 3$
After substituting the value in the above equation, we get
I $= \dfrac{{1 + 1}}{{\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }}$
Therefore, after evaluating $\dfrac{{\sec \theta \cos ec\left( {{{90}^ \circ } - \theta } \right) - \tan \theta \cot \left( {{{90}^ \circ } - \theta } \right) + {{\sin }^2}{{55}^ \circ } + {{\sin }^2}{{35}^ \circ }}}{{\tan {{10}^ \circ }\tan {{20}^ \circ }\tan {{60}^ \circ }\tan {{70}^ \circ }\tan {{80}^ \circ }}}$we got $\dfrac{2}{{\sqrt 3 }}$

So, the correct answer is “Option B”.

Note: In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.