Question

Prove the following:
\[\tan {{720}^{o}}-\cos {{270}^{o}}-\sin {{150}^{o}}\cos {{120}^{o}}=\dfrac{1}{4}\]

Answer 1

Hint: First of all, take the LHS of the given equation and convert all angles between \[{{0}^{o}}\] to \[{{90}^{o}}\] by using \[\tan \left( 2n\pi \right)=0,\cos \left( {{180}^{o}}\pm \theta \right)=-\cos \theta ,\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \] and then use the trigonometric ratio table to prove the desired result.

Complete step-by-step answer:

In this question, we have to prove that \[\tan {{720}^{o}}-\cos {{270}^{o}}-\sin {{150}^{o}}\cos {{120}^{o}}=\dfrac{1}{4}\]. Let us consider the LHS of the equation given in the question.

\[LHS=\tan {{720}^{o}}-\cos {{270}^{o}}-\sin {{150}^{o}}\cos {{120}^{o}}\]

We know that,

\[{{720}^{o}}=2\times {{360}^{o}}\]

\[{{270}^{o}}={{180}^{o}}+{{90}^{o}}\]

\[{{150}^{o}}={{180}^{o}}-{{30}^{o}}\]

\[{{120}^{o}}={{180}^{o}}-{{60}^{o}}\]

By using these in the above equation, we get,

\[LHS=\tan \left( 2\times {{360}^{o}} \right)-\cos \left( {{180}^{o}}+{{90}^{o}} \right)-\sin \left( {{180}^{o}}-{{30}^{o}} \right).\cos \left( {{180}^{o}}-{{60}^{o}} \right).....\left( i \right)\]
We know that \[\tan \left( 2n\pi \right)=0\] where \[n\in W\]. So, by substituting n = 2, \[\pi ={{180}^{o}}\], we get,

\[\tan \left( {{2.360}^{o}} \right)=0\]

By using this in equation (i), we get,

\[LHS=0-\cos \left( {{180}^{o}}+{{90}^{o}} \right)-\sin \left( {{180}^{o}}-{{30}^{o}} \right).\cos \left( {{180}^{o}}-{{60}^{o}} \right)\]

We know that \[\cos \left( {{180}^{o}}+\theta \right)=-\cos \theta \]. By using this in the above equation, we get,

\[LHS=-\cos \left( {{90}^{o}} \right)-\sin \left( {{180}^{o}}-{{30}^{o}} \right).\cos \left( {{180}^{o}}-{{60}^{o}} \right)\]

We also know that \[\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \] and \[\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta \]. By using these in the above equation, we get,

\[LHS=-\cos \left( {{90}^{o}} \right)-\left( \sin {{30}^{o}} \right)\left( -\cos {{60}^{o}} \right)\]

\[LHS=-\cos {{90}^{o}}+\sin {{30}^{o}}\cos {{60}^{o}}....\left( ii \right)\]

Now, let us find the values of \[\sin {{30}^{o}},\cos {{60}^{o}}\] and \[\cos {{90}^{o}}\] from the trigonometric table for general angles.

	\[\sin \theta \]	\[\cos \theta \]	\[\tan \theta \]	\[\operatorname{cosec}\theta \]	\[\sec \theta \]	\[\cot \theta \]
0	0	1	0	-	1	-
\[\dfrac{\pi }{6}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{3}}\]	2	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{3}\]
\[\dfrac{\pi }{4}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]	1	\[\sqrt{2}\]	\[\sqrt{2}\]	1
\[\dfrac{\pi }{3}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]	\[\sqrt{3}\]	\[\dfrac{2}{\sqrt{3}}\]	2	\[\dfrac{1}{\sqrt{3}}\]
\[\dfrac{\pi }{2}\]	1	0	-	1	-	0

From the above table, we get

\[\cos {{90}^{o}}=0\]

\[\cos {{60}^{o}}=\dfrac{1}{2}\]

\[\sin {{30}^{o}}=\dfrac{1}{2}\]

By substituting these values in equation (ii), we get,

\[LHS=-0+\dfrac{1}{2}.\dfrac{1}{2}\]

\[LHS=\dfrac{1}{4}\]

So, we get, LHS = RHS

Hence proved.

Therefore, we have proved that

\[\tan {{720}^{o}}-\cos {{270}^{o}}-\sin {{150}^{o}}\cos {{120}^{o}}=\dfrac{1}{4}\]

Note: In these types of questions, students should take care while calculating and reducing the angles between 0 and \[{{90}^{o}}\]. Also, students can solve this question in terms of \[\pi \] by converting each angle \[{{0}^{o}}\] into radians by multiplying it by \[\dfrac{\pi }{{{180}^{o}}}\]. Students should also remember the trigonometric ratios of general angles like sin, cos, tan, etc. to easily solve the questions.