Question

Evaluate the following:
\[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

Answer 1

Hint: We will use the trigonometric ratio table and find out the values of trigonometric ratios to evaluate the given expression \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]. Then we will substitute the values of those trigonometric ratios in the equation given in the equation and simplify to get the answer.

Complete step-by-step answer:
In the question, we have been given the expression as \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\].
All the angles are standard angles whose values can be computed easily. We can refer to the following table for finding the value of trigonometric functions at standard angles.

Ratio/Angle(\[\theta \])	\[{{0}^{\circ }}\]	\[{{30}^{\circ }}\]	\[{{45}^{\circ }}\]	\[{{60}^{\circ }}\]	\[{{90}^{\circ }}\]
\[\sin \theta \]	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	1
\[\cos \theta \]	1	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	0
\[tan\theta \]	0	\[\dfrac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	Not defined
\[cosec\theta \]	Not defined	2	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	1
\[\sec \theta \]	1	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	Not defined
\[\cot \theta \]	Not defined	\[\sqrt{3}\]	1	\[\dfrac{1}{\sqrt{3}}\]	0

Referring to the table above, we can see that \[\tan {{45}^{\circ }}=1,\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\].
So by substituting these values in the equation given in the question and then simplifying it, we get as follows:
\[\begin{align}
  & \Rightarrow 2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }} \\
 & \Rightarrow 2{{\left( 1 \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\
 & \Rightarrow 2 \\
\end{align}\]
Therefore, the value of the given expression is equal to 2.
Note: Don’t get confused in the values of \[\cos {{30}^{\circ }}\] and \[\sin {{60}^{\circ }}\]. Sometimes by mistake we might use the values as \[\cos {{30}^{\circ }}=\dfrac{1}{2}\] and \[\sin {{60}^{\circ }}=\dfrac{1}{2}\] which is wrong and thus it might give us the wrong answer. So we must substitute the values of \[\cos {{30}^{\circ }}=\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\] in the expression carefully. Also we can solve the question by replacing \[{{\cos }^{2}}{{30}^{\circ }}\] with \[{{\left[ \sin \left( {{90}^{\circ }}-{{30}^{\circ }} \right) \right]}^{2}}\] since we know the relation that \[\cos \theta =\sin \left( {{90}^{\circ }}-\theta \right)\]. In these types of questions we must have to remember all the values of trigonometric ratios for the corresponding angles.