Question

Prove the following expression, $3\sin \left( \dfrac{\pi }{6} \right)\sec \left( \dfrac{\pi }{3} \right)-4\sin \left( \dfrac{5\pi }{6} \right)\cot \left( \dfrac{\pi }{4} \right)=1$.

Answer 1

Hint: By looking at the question, we can see that all the angles are in the standard form. We can write, $\sin \left( \dfrac{\pi }{6} \right)=\sin \left( \dfrac{180{}^\circ }{6} \right)=\sin 30{}^\circ =\dfrac{1}{2}$. We will also use $\sin \left( \pi -\theta \right)=\sin \theta $ to find the values of the angles to prove the given expression. Similarly, we can put the values of all the angles in the given expression in the left hand side or the LHS to prove that it is equal to 1.

Complete step-by-step answer:

It is given in the question that we have to prove that the value of the expression, $3\sin \left( \dfrac{\pi }{6} \right)\sec \left( \dfrac{\pi }{3} \right)-4\sin \left( \dfrac{5\pi }{6} \right)\cot \left( \dfrac{\pi }{4} \right)$ is 1. For that, we will first consider the LHS of the expression. Now we will solve each of the given angles of the expression individually.

We know that $\sin \left( \dfrac{\pi }{6} \right)=\sin \left( \dfrac{180{}^\circ }{6} \right)=\sin 30{}^\circ =\dfrac{1}{2}$ and, $\sec \left( \dfrac{\pi }{3} \right)=\dfrac{1}{\cos \left( \dfrac{\pi }{3} \right)}=\dfrac{1}{\cos \left( \dfrac{180{}^\circ }{3} \right)}=\dfrac{1}{\cos \left( 60{}^\circ \right)}=\dfrac{1}{\dfrac{1}{2}}=2$.

We can write $\sin \left( \dfrac{5\pi }{6} \right)=\sin \left( \dfrac{6\pi -\pi }{6} \right)=\sin \left( \pi -\dfrac{\pi }{6} \right)$ . We also know that, $\sin \left( \pi -\theta \right)=\sin \theta $, because sin is positive in the first and the second quadrant. So, applying it, we get,
$\sin \left( \pi -\dfrac{\pi }{6} \right)=\sin \left( \dfrac{\pi }{6} \right)=\sin \left( \dfrac{180{}^\circ }{6} \right)=\sin 30{}^\circ =\dfrac{1}{2}$

Similarly, we get,

$\cot \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\tan \left( \dfrac{\pi }{4} \right)}$

We know that the value of $\tan \left( \dfrac{\pi }{4} \right)=1$, so applying that in the above, we get,

$\dfrac{1}{\tan \left( \dfrac{\pi }{4} \right)}=\dfrac{1}{1}=1$

Now, we will put the obtained values of the given expression in the LHS and we get LHS as,

$\begin{align}

  & =3\sin \left( \dfrac{\pi }{6} \right)\sec \left( \dfrac{\pi }{3} \right)-4\sin \left( \dfrac{5\pi }{6} \right)\cot \left( \dfrac{\pi }{4} \right) \\

 & =3\times \dfrac{1}{2}\times 2-4\times \dfrac{1}{2}\times 1 \\

 & =\dfrac{6}{2}-2 \\

 & =3-2 \\

 & =1 \\

\end{align}$

Which is equal to the right hand side of the given expression, therefore, we can see that LHS = RHS.

Hence, we have proved that the value of the expression given in the question is 1.

Note: Generally, the possible mistakes that the students can make while solving this question is that they use the incorrect plus and minus signs for the values, while doing the conversions of the angles. So, they should be careful and try to solve step by step to avoid the chances of errors. They should also know the basic trigonometric standard forms of the angles given in the question and apply the conversions appropriately.