Question

Prove the following.
\[\tan \dfrac{11\pi }{3}-2\sin \dfrac{4\pi }{6}-\dfrac{3}{4}{{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{17\pi }{6}=\dfrac{3-4\sqrt{3}}{2}\]

Answer 1

Hint: In this question, we need to prove the given trigonometric functions to be equal to \[\dfrac{3-4\sqrt{3}}{2}.\] For this, we first need to understand the signs of trigonometric functions in the different quadrants. Then we will use the following known values of trigonometric functions to get the final answer.
\[\left( i \right)\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
\[\left( ii \right)\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]
\[\left( iii \right)\tan \dfrac{\pi }{3}=\sqrt{3}\]
\[\left( iv \right)\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}\]
\[\left( v \right)\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}\]

Complete step-by-step answer:
Here, we need to prove \[\tan \dfrac{11\pi }{3}-2\sin \dfrac{4\pi }{6}-\dfrac{3}{4}{{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{17\pi }{6}=\dfrac{3-4\sqrt{3}}{2}.\] But first let us understand the effect on values of trigonometric function when we change the quadrant. When we add or subtract \[\pi \] from the angle of any trigonometric function, the trigonometric function remains the same but the sign varies as per quadrant. The different trigonometric functions possess different signs in different quadrants.

In the first quadrant, all trigonometric functions are positive. In the second quadrant, only sin and cosec functions are positive. In the third quadrant, only tan and cot functions are positive. In the fourth quadrant, only cos and sec functions are positive. Now, let us separate the terms individually and evaluate \[\tan \dfrac{11\pi }{3}.\] It can be written as \[\tan \left( 2\pi +\dfrac{5\pi }{3} \right).\] After the first revolution, it will remain in the same quadrant, hence it becomes equal to \[\tan \dfrac{5\pi }{3}.\] It can be written as \[\tan \left( \pi +\dfrac{2\pi }{3} \right).\] After adding \[\pi \] tan moves to the third quadrant where it is positive, so it becomes \[\tan \dfrac{2\pi }{3}.\] It can be written as \[\tan \left( \pi -\dfrac{\pi }{3} \right).\] Now, \[\pi -\dfrac{\pi }{3}\] will lie in the second quadrant, so tan becomes negative. Hence, we get, \[-\tan \dfrac{\pi }{3}.\] We know that \[\tan \dfrac{\pi }{3}=\sqrt{3}.\] Hence,
\[\tan \dfrac{11\pi }{3}=-\sqrt{3}......\left( i \right)\]
\[\left( ii \right)\sin \dfrac{4\pi }{6}\]
It can be written as \[\dfrac{2\pi }{3}\] which can be further written as \[\pi -\dfrac{\pi }{3}.\] So, we get, \[\sin \left( \pi -\dfrac{\pi }{3} \right).\] Now, it will lie in the second quadrant where it is positive, so, \[\sin \left( \pi -\dfrac{\pi }{3} \right)=\sin \dfrac{\pi }{3}.\] Since, \[\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}.\] So, \[\sin \dfrac{4\pi }{6}=\dfrac{\sqrt{3}}{2}.....\left( ii \right)\]
\[\left( iii \right){{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}\]
We know that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}},\] so \[{{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}=\dfrac{1}{{{\sin }^{2}}\dfrac{\pi }{4}}=\dfrac{1}{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}={{\left( \sqrt{2} \right)}^{2}}=2.\]
Hence, \[{{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}=2......\left( iii \right)\]
\[\left( iv \right){{\cos }^{2}}\dfrac{17\pi }{6}\]
It can be written as \[\cos \left( 2\pi +\dfrac{5\pi }{6} \right),\] the sign remains the same after the full revolution. So, we get, \[\cos \dfrac{5\pi }{6}.\] It can be written as \[\cos \left( \pi -\dfrac{\pi }{6} \right).\] Now, it will lie in the second quadrant, so we get \[\dfrac{-\cos \pi }{6}.\] Since, \[\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}.\] So,
\[\cos \dfrac{17\pi }{6}=\dfrac{-\sqrt{3}}{2}......\left( iv \right)\]
Now our equation was
\[\tan \dfrac{11\pi }{3}-2\sin \dfrac{4\pi }{6}-\dfrac{3}{4}{{\operatorname{cosec}}^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{\pi }{4}+4{{\cos }^{2}}\dfrac{17\pi }{6}=\dfrac{3-4\sqrt{3}}{2}\]
Putting values from (i), (ii), (iii) and (iv), we get,
\[\Rightarrow -3-2\left( \dfrac{\sqrt{3}}{2} \right)-\dfrac{3}{4}\left( 2 \right)+4{{\left( \dfrac{-\sqrt{3}}{2} \right)}^{2}}=\dfrac{3-4\sqrt{3}}{2}\]
\[\Rightarrow -\sqrt{3}-\sqrt{3}-\dfrac{3}{2}+4\left( \dfrac{3}{4} \right)=\dfrac{3-4\sqrt{3}}{2}\]
\[\Rightarrow -2\sqrt{3}-\dfrac{3}{2}+3=\dfrac{3-4\sqrt{3}}{2}\]
Taking 2 common, we get,
\[\Rightarrow \dfrac{-4\sqrt{3}-3+6}{2}=\dfrac{3-4\sqrt{3}}{2}\]
\[\Rightarrow \dfrac{3-4\sqrt{3}}{2}=\dfrac{3-4\sqrt{3}}{2}\]
Hence proved.

Note: Students should take care of the signs while evaluating all these values. If we take an angle as \[{{90}^{\circ }}+\theta \] or \[{{90}^{\circ }}-\theta ,\] then the trigonometric function also changes. For example, \[\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta .\] The \[\cos \theta \] changes to \[\sin \theta \] and vice versa and \[\tan \theta \] changes to \[\cot \theta \] and vice versa. We need to take care while adding or subtracting rational numbers with irrational numbers.