Question

Prove the given trigonometric expression: $2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=6$

Answer 1

Hint: First consider the left hand side of the equation then put the standard values which should be used such as $\cos 45=\dfrac{1}{\sqrt{2}},\tan 60=\sqrt{3},\sin 45=\dfrac{1}{\sqrt{2}}$ and $\tan 30=\dfrac{1}{\sqrt{3}}$ then square it and simplify and prove that its value is ‘6’.

Complete step by step answer:

In the question we are asked to prove that $2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=6$

Let’s consider only the left hand side of equation which is,
$2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)$
By the use of standard values of standard angles we can say that,
$\cos 45=\dfrac{1}{\sqrt{2}},\tan 60=\sqrt{3},\sin 45=\dfrac{1}{\sqrt{2}},\tan 30=\dfrac{1}{\sqrt{3}}$
So their square will be,
${{\cos }^{2}}45=\dfrac{1}{2},{{\tan }^{2}}60=3,{{\sin }^{2}}45=\dfrac{1}{2},{{\tan }^{2}}30=\dfrac{1}{3}$
So by substituting value of ${{\cos }^{2}}45=\dfrac{1}{2},{{\tan }^{2}}60=3,{{\sin }^{2}}45=\dfrac{1}{2}$ and ${{\tan }^{2}}30=\dfrac{1}{3}$
In left hand side of equation we get,
$2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=2\left( \dfrac{1}{2}+3 \right)-6\left( \dfrac{1}{2}-\dfrac{1}{3} \right)$
On taking the LCM and solving, we get
$\begin{align}
  & 2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=2\left( \dfrac{1+6}{2} \right)-6\left( \dfrac{3-2}{6} \right) \\
 & \Rightarrow 2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=2\left( \dfrac{7}{2} \right)-6\left( \dfrac{1}{6} \right) \\
\end{align}$
Cancelling the like terms, we get
$2\left( {{\cos }^{2}}45+{{\tan }^{2}}60 \right)-6\left( {{\sin }^{2}}45-{{\tan }^{2}}30 \right)=7-1=6$

Hence the value of the LHS is 6.

So, LHS=RHS

Hence proved

Note: Students always forget to learn the value of standard angles of sin, cos and tan ratios. So for this kind of problem to solve they should learn it to do these problems easily.
Another approach is to substitute $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, then solve the expression accordingly.