Question

Evaluate $\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\tan 32-\dfrac{5}{3}\tan 13.\tan 37.\tan 45.\tan 53.\tan 77$. All the angles are in degrees.

Answer 1

Hint: Since we have the given trigonometric expression:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\tan 32-\dfrac{5}{3}\tan 13.\tan 37.\tan 45.\tan 53.\tan 77$
Try to find the relation between the given tangent of angles and then we can convert $\tan \theta $ to \[\cot \theta \] using $\tan \left( 90-\theta \right)=\cot \theta $. Later, after simplifying the terms, use the identity of cosecant and cotangent as: $\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1$ and solve the expression.

Complete step by step answer:
We have the given expression:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\tan 32-\dfrac{5}{3}\tan 13.\tan 37.\tan 45.\tan 53.\tan 77......(1)$
Now, we need to find a relation between the tangent of angles.
We get:
$\begin{align}
  & \tan 13=\tan \left( 90-77 \right)......(3) \\
 & \tan 37=\tan \left( 90-53 \right)......(4) \\
 & \tan 32=\tan \left( 90-58 \right)......(5) \\
\end{align}$
Substitute all the values in equation (1), we get:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\tan \left( 90-58 \right)-\dfrac{5}{3}\tan \left( 90-77 \right).\tan \left( 90-53 \right).\tan 45.\tan 53.\tan 77......(6)$
As we know that, $\tan \left( 90-\theta \right)=\cot \theta $
So, we can write equation (6) as:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\cot 58-\dfrac{5}{3}\cot 77.\cot 53.\tan 45.\tan 53.\tan 77......(7)$
Also, we know that: $\tan \theta .\cot \theta =1$
So, we have:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}{{\cot }^{2}}58-\dfrac{5}{3}\tan 45......(8)$
Since, $\tan 45=1$
By substituting the value $\tan 45=1$ in equation (8), we get:
$\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}{{\cot }^{2}}58-\dfrac{5}{3}......(9)$
We can rewrite equation (9) as:
$\dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}58-{{\cot }^{2}}58 \right)-\dfrac{5}{3}......(10)$
Since, we know that: $\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1$
So, we can write equation (10) as:
$\begin{align}
  & \Rightarrow \dfrac{2}{3}\left( 1 \right)-\dfrac{5}{3} \\
 & \Rightarrow \dfrac{2-5}{3} \\
 & \Rightarrow -1 \\
\end{align}$

So, the solution of the expression $\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}58-\dfrac{2}{3}\cot 58.\tan 32-\dfrac{5}{3}\tan 13.\tan 37.\tan 45.\tan 53.\tan 77$ is -1

Note: Always remember if we are given a trigonometric expression containing tangent, cotangent, and cosecant of different angles, to simplify the equation try to find if the tangent of angles is related to the cotangent of given angles. If so, then cancel out the terms and then get a simple equation. Since, we have cosecant and cotangent of angles in the equation formed, see if the equation formed resembles with the trigonometric identity $\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1$or use algebraic function to form this relation. This simplifies the equation further and you get a finite answer.
Also, if you are not good with trigonometric identities, it is always useful to convert all the trigonometric functions into sine and cosine of given angles and simplify the equation formed.