Question

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

Answer 1

Hint: We have to prove the given expression, and for that we need to use some trigonometric conversions and convert the terms in the expression into simpler terms. For example, we can convert as follows, $\tan \left( \dfrac{11\pi }{3} \right)=\tan \left( \dfrac{9\pi +2\pi }{3} \right)=\tan \left( 3\pi +\dfrac{2\pi }{3} \right)$. Using the relation, $\tan \left( n\pi +\theta \right)=\tan \theta $, we get the value as $\tan \left( \dfrac{2\pi }{3} \right)$. Similarly, we can find the values of the other terms and prove the given equation.

Complete step-by-step answer:

It is given in the question that we have to prove the expression, $\tan \dfrac{11\pi }{3}-2\sin \dfrac{4\pi }{6}-\dfrac{3}{4}{{\operatorname{cosec}}^{2}}\dfrac{\pi }{2}+4{{\cos }^{2}}\dfrac{17\pi }{6}=\dfrac{3-4\sqrt{3}}{2}$. To solve the question, we will first consider the left hand side or the LHS of the expression. We will solve each of the angles in the LHS individually. So, we get,

$\tan \left( \dfrac{11\pi }{3} \right)=\tan \left( \dfrac{9\pi +2\pi }{3} \right)=\tan \left( 3\pi +\dfrac{2\pi }{3} \right)$

We know that, $\tan \left( n\pi +\theta \right)=\tan \theta $, so by applying it in the above expression, we get the angle as,

$\tan \left( 3\pi +\dfrac{2\pi }{3} \right)=\tan \left( \dfrac{2\pi }{3} \right)$

Again, we can write $\tan \left( \dfrac{2\pi }{3} \right)=\tan \left( \pi -\dfrac{\pi }{3} \right)$. Now, using the formula, $\tan \left( n\pi -\theta \right)=-\tan \theta $, we get $\tan \left( \pi -\dfrac{\pi }{3} \right)=-\tan \left( \dfrac{\pi }{3} \right)$

Now, we can also write it in terms of angles. So, we get,

$-\tan \left( \dfrac{\pi }{3} \right)=-\tan \left( \dfrac{180{}^\circ }{3} \right)=-\tan 60{}^\circ =-\sqrt{3}$

Similarly, we get,

$\sin \left( \dfrac{4\pi }{6} \right)=\sin \left( \dfrac{2\pi }{3} \right)=\sin \left( \dfrac{2\times 180{}^\circ }{3} \right)=\sin \left( 120{}^\circ \right)$

We know that, $\sin \left( 90{}^\circ +\theta \right)=\cos \theta $, so applying that, we get,
$\sin \left( 120{}^\circ \right)=\sin \left( 90{}^\circ +30{}^\circ \right)=\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

We know that, $1+{{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta $, so by applying that in our next term, we get,

${{\operatorname{cosec}}^{2}}\left( \dfrac{\pi }{4} \right)=1+{{\cot }^{2}}\left( \dfrac{\pi }{4} \right)$

Also, we know that, $\cot \left( \dfrac{\pi }{4} \right)=1$, so by applying it, we get,

${{\operatorname{cosec}}^{2}}\left( \dfrac{\pi }{4} \right)=1+{{\left( 1 \right)}^{2}}=1+1=2$
Similarly, we get,

${{\cos }^{2}}\left( \dfrac{17\pi }{6} \right)={{\cos }^{2}}\left( \dfrac{18\pi -\pi }{6} \right)={{\cos }^{2}}\left( 3\pi -\dfrac{\pi }{6} \right)$

We know that, $\cos \left( \pi -\theta \right)=-\cos \theta $, so applying it, we get,

${{\cos }^{2}}\left( 3\pi -\dfrac{\pi }{6} \right)={{\left[ -\cos \left( \dfrac{\pi }{6} \right) \right]}^{2}}={{\left[ \cos \left( \dfrac{\pi }{6} \right) \right]}^{2}}={{\left[ \cos \left( \dfrac{180{}^\circ }{6} \right) \right]}^{2}}={{\left( \cos 30{}^\circ \right)}^{2}}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=\dfrac{3}{4}$

Now, we will put all the above obtained values in the expression and get LHS as,

$\begin{align}

  & =\left( -\sqrt{3} \right)-\left( 2\times \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{3}{4}\times 2 \right)+\left( 4\times \dfrac{3}{4} \right) \\

 & =\left( -\sqrt{3} \right)-\left( \sqrt{3} \right)-\left( \dfrac{3}{2} \right)+3 \\

 & =-2\sqrt{3}+\dfrac{3}{2} \\

 & =\dfrac{-4\sqrt{3}+3}{2} \\

 & =\dfrac{3-4\sqrt{3}}{2} \\

\end{align}$

Which is equal to the right hand side or RHS of the given expression, therefore LHS = RHS.

Hence, we have proved the given expression.


Note: The students usually make mistakes with the use of plus and minus signs while they are taking the value of the angles during the conversions. So, it is advisable to do the conversions carefully and step by step to avoid the chances of errors and getting the wrong result. They should also be thorough with the trigonometric relations and their appropriate usage.