Question

# Without using trigonometric tables, evaluate that:$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}$

Hint: Use the formulas of the trigonometric functions. These trigonometric formulas are $\tan x=\cot \left( 90-x \right)$, $\cot x=\dfrac{1}{\tan x}$ and $\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1$. Here, the angle x is in degree. Using these formulas, we can solve this question.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In trigonometry, we have some formulas that relate tan and cot functions. These formulas are,
$\tan x=\cot \left( 90-x \right)............\left( 1 \right)$
$\cot x=\dfrac{1}{\tan x}...............\left( 2 \right)$

Also, in trigonometry we have an identity,
$\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1.................\left( 3 \right)$

In all the above 3 formulas, the angle x should be in degree.
In the question, we are required to evaluate $\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}$
In formula $\left( 1 \right)$, substituting x = 53, we can write $\tan {{53}^{\circ }}=\cot \left( 90-53 \right)=\cot {{37}^{\circ }}$ and substituting x = 77, we can write $\tan {{77}^{\circ }}=\cot \left( 90-77 \right)=\cot {{13}^{\circ }}$. Substituting $\tan {{53}^{\circ }}$ and $\tan {{77}^{\circ }}$ in the above expression, we get,
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\cot {{37}^{\circ }}\cot {{13}^{\circ }}$
In formula $\left( 2 \right)$, substituting x = 13, we get$\cot {{13}^{\circ }}=\dfrac{1}{\tan {{13}^{\circ }}}$ and substituting x = 37, we get $\cot {{37}^{\circ }}=\dfrac{1}{\tan {{37}^{\circ }}}$. Substituting these in the above expression, we get,
\begin{align} & \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\dfrac{1}{\tan {{37}^{\circ }}}\dfrac{1}{\tan {{13}^{\circ }}} \\ & \Rightarrow \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{45}^{\circ }} \\ & \Rightarrow \dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\cot {{58}^{\circ }}\tan {{32}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\ \end{align}
Using formula $\left( 1 \right)$, substituting $\tan {{32}^{\circ }}=\cot \left( 90-32 \right)=\cot {{32}^{\circ }}$ in the above expression, we get,
$\dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }}$
From formula $\left( 3 \right)$, substituting x = 58, we get,
$\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1$
Substituting $\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1$ in the above expression, we get,
\begin{align} & \dfrac{2}{3}\left( 1 \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\ & \Rightarrow \dfrac{2}{3}-\dfrac{5}{3}\tan {{45}^{\circ }} \\ \end{align}
Also, from trigonometry, we have $\tan {{45}^{\circ }}=1$. Substituting $\tan {{45}^{\circ }}=1$ in the above equation, we get,
\begin{align} & \dfrac{2}{3}-\dfrac{5}{3}\left( 1 \right) \\ & \Rightarrow \dfrac{2}{3}-\dfrac{5}{3} \\ & \Rightarrow \dfrac{-3}{3} \\ & \Rightarrow -1 \\ \end{align}
Hence, the answer of the trigonometric expression is -1.

Note: There is a possibility that one may commit a mistake while using the identity $\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1$. It is a very common mistake that one may write this as $\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}+{{\cot }^{2}}{{58}^{\circ }}=1$ instead of $\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1$. To avoid this mistake, one should start from the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and divide both sides of the equation by ${{\sin }^{2}}x$. Then substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$. Substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$ in ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we will get the identity $\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1$.