Question & Answer
QUESTION

Without using trigonometric table evaluate:
$\dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{70}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{40}^{\circ }}}+2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\tan {{32}^{\circ }}-4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}$.

ANSWER Verified Verified
Hint: In this question, we will use the concept of trigonometric ratios of complementary angles and trigonometric identities, to find similar terms which can be cancelled.

Complete step-by-step answer:
In trigonometry, trigonometric ratios of complementary angles are given as below,
$\cos \left( 90-x \right)=\sin x\cdots \cdots \left( i \right)$,
$\cot \left( 90-x \right)=\tan x\cdots \cdots \left( ii \right)$,
$\tan \left( 90-x \right)=\cot x\cdots \cdots \left( iii \right)$.
Also, we know, trigonometric identities are as below,
${{\cos }^{2}}x+{{\sin }^{2}}x=1\cdots \cdots \left( iv \right)$,
${{\sec }^{2}}x-{{\tan }^{2}}x=1\cdots \cdots \left( v \right)$,
$\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1\cdots \cdots \left( vi \right)$.
Let us break the given question in three parts to make the solution easier.
Let, $A=\dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{70}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{40}^{\circ }}}$,
$\,B=2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\tan {{32}^{\circ }}$,
$C=4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}$.
Now, we can write $A$ as,
$\begin{align}
  & A=\dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{70}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{40}^{\circ }}} \\
 & =\dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{\left( 90-20 \right)}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{\left( 90-50 \right)}^{\circ }}} \\
\end{align}$
Applying equation $\left( i \right)$ in numerator and equation $\left( ii \right)$ in denominator of $A$, we get,
$A=\dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\sin }^{2}}{{20}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\tan }^{2}}{{50}^{\circ }}}$
Applying equation $\left( iv \right)$ in numerator and equation $\left( v \right)$ in denominator here, we get,
$A=\dfrac{1}{1}=1$
Now, we can write $B$ as,
$\,B=2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\tan {{\left( 90-58 \right)}^{\circ }}$
Applying equation $\left( iii \right)$here, we get,
 $\begin{align}
  & \,B=2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\cot {{58}^{\circ }} \\
 & \,=2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2{{\cot }^{2}}{{58}^{\circ }} \\
\end{align}$
Applying equation $\left( v \right)$ here, we get,
 $B=2$
Now, we can write $C$ as,
$\begin{align}
  & C=4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }} \\
 & =4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{\left( 90-37 \right)}^{\circ }}\tan {{\left( 90-13 \right)}^{\circ }} \\
\end{align}$
Applying equation $\left( iii \right)$ here, we get,
$\begin{align}
  & C=4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }} \\
 & =4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\cot {{37}^{\circ }}\cot {{13}^{\circ }} \\
\end{align}$
Now, we know that $\cot x=\dfrac{1}{\tan x}$. Using this in above equation, we get,
 $C=4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\dfrac{1}{\tan {{37}^{\circ }}}\dfrac{1}{\tan {{13}^{\circ }}}$
Cancelling tan terms from numerator and denominator, we get,
$C=\tan {{45}^{\circ }}$
As we know that $\tan x=\dfrac{\sin x}{\cos x}$, therefore,
$\begin{align}
  & C=\dfrac{\sin {{45}^{\circ }}}{\cos {{45}^{\circ }}} \\
 & =\dfrac{\sin {{45}^{\circ }}}{\cos {{\left( 90-45 \right)}^{\circ }}} \\
\end{align}$
Using equation $\left( i \right)$ here, we get,
$C=\dfrac{\sin {{45}^{\circ }}}{\sin {{45}^{\circ }}}$
cancelling the terms, we get,
$C=1$
Now, we can write given question as,
$\begin{align}
  & \dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{70}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{40}^{\circ }}}+2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\tan {{32}^{\circ }}-4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }} \\
 & =A+B-C \\
\end{align}$
Putting values of $A,\,B\,\text{and}\,C$ here, we get,
$\begin{align}
  & \dfrac{co{{s}^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{70}^{\circ }}}{{{\sec }^{2}}{{50}^{\circ }}-{{\cot }^{2}}{{40}^{\circ }}}+2\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-2\cot {{58}^{\circ }}\tan {{32}^{\circ }}-4\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }} \\
 & =1+2-1 \\
 & =2 \\
\end{align}$
Hence the value of the given expression is 2.

Note: This question can also be done by changing other trigonometric ratios. The objective is to express it in similar terms which can cancel each other or can together give numerical value