Question

Evaluate the following trigonometric expression:
$3{{\cot }^{2}}{{60}^{\circ }}+{{\sec }^{2}}{{45}^{\circ }}$

Answer 1

Hint: To find the value of the given trigonometric expression we need the value of $\cot {{60}^{\circ }}$ and $\sec {{45}^{\circ }}$. We know that the value of $\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}$ and the value of $\sec {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ so take the square of these values and then substitute in the given expression and solve it further to get the value in the simplest form.

Complete step-by-step solution:
The trigonometric expression which is given in the above problem of which we have to find the value of is:
$3{{\cot }^{2}}{{60}^{\circ }}+{{\sec }^{2}}{{45}^{\circ }}$
If we know the value of $\cot {{60}^{\circ }}$ and $\sec {{45}^{\circ }}$ then the above problem is pretty simple because after that we just have to square these values and substitute in the given expression.
We know that the value of $\cot {{60}^{\circ }}$ is equal to:
$\dfrac{1}{\sqrt{3}}$
And the value of $\sec {{45}^{\circ }}$ is equal to:
$\sqrt{2}$
Now, substituting these values which we have just solved in the given expression we get,
$\begin{align}
  & 3{{\cot }^{2}}{{60}^{\circ }}+{{\sec }^{2}}{{45}^{\circ }} \\
 & =3{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}} \\
\end{align}$
$\begin{align}
  & =3\left( \dfrac{1}{3} \right)+\dfrac{1}{2} \\
 & =1+\dfrac{1}{2} \\
\end{align}$
Taking 2 as L.C.M in the above expression we get,
$\begin{align}
  & \dfrac{2+1}{2} \\
 & =\dfrac{3}{2} \\
\end{align}$
Hence, we have evaluated the given expression and the result of that evaluation is $\dfrac{3}{2}$.

Note: The alternative way of solving the above problem is as follows:
If you don’t remember the value of $\cot {{60}^{\circ }}$ and $\sec {{45}^{\circ }}$ then you can first of all convert them into $\dfrac{\cos {{60}^{\circ }}}{\sin {{60}^{\circ }}}\And \dfrac{1}{\cos {{45}^{\circ }}}$ respectively and then put the values of $\cos {{60}^{\circ }},\sin {{60}^{\circ }}\And \cos {{45}^{\circ }}$ and get their values.
Generally, we can only remember the value of sine and cosine angles so simplifying tan, cot, sec and cosec and then substitute sine and cosine angles will give the values of tan, cot, sec and cosec.
$\begin{align}
  & \cot {{60}^{\circ }}=\dfrac{\cos {{60}^{\circ }}}{\sin {{60}^{\circ }}} \\
 & \Rightarrow \cot {{60}^{\circ }}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \\
 & \Rightarrow \cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}} \\
\end{align}$
And similarly, we can find the value of $\sec {{45}^{\circ }}$.
$\sec {{45}^{\circ }}=\dfrac{1}{\cos {{45}^{\circ }}}=\dfrac{1}{\dfrac{1}{\sqrt{2}}}=\sqrt{2}$
This is how, we can find the values of $\cot {{60}^{\circ }}\And \sec {{45}^{\circ }}$ if we forgot their values.