Question

Evaluate the following: $\begin{array}{rcl}& & 2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)\end{array}$

Answer 1

Determine the value of $2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)$.

Trigonometry values for angles ${30}^{°},{60}^{°},{45}^{°}$ are as follows:

$\begin{array}{rcl}\cos \left({30}^{°}\right)& =& \frac{\sqrt{3}}{2}\\ \sin \left({60}^{°}\right)& =& \frac{\sqrt{3}}{2}\\ \tan \left({45}^{°}\right)& =& 1\end{array}$

Substitute the above values into the given expression.

$\begin{array}{rcl}2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)& =& 2{\left(1\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2\\ & \Rightarrow & 2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)=2+\frac{3}{4}-\frac{3}{4}\\ & \Rightarrow & 2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)=2+0\\ & \Rightarrow & 2{\tan }^2\left({45}^{°}\right)+{\cos }^2\left({30}^{°}\right)-{\sin }^2\left({60}^{°}\right)=2\end{array}$

Hence, the value is $2$.