Question

Prove that:
$2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\dfrac{{7\pi }}{6}{\cos ^2}\dfrac{\pi }{3} = \dfrac{3}{2}$

Answer 1

Note that, ${\text{since }}\dfrac{{7\pi }}{6}{\text{ is in third quadrant, therefore }}\operatorname{cosec} \dfrac{{7\pi }}{6}{\text{ is negative}}{\text{.}}$
Then in the left hand side, substitute the values of $\sin \dfrac{\pi }{6}$, $\operatorname{cosec} \dfrac{{7\pi }}{6}$ and $\cos \dfrac{\pi }{3}$.
And on solving we get the desired result.

Complete step-by-step answer:
Given to prove that $2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\dfrac{{7\pi }}{6}{\cos ^2}\dfrac{\pi }{3} = \dfrac{3}{2}$
Left hand side is given by,
$2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\dfrac{{7\pi }}{6}{\cos ^2}\dfrac{\pi }{3}$
The above expression can be written as,
$= 2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\left( {\pi + \dfrac{\pi }{6}} \right){\cos ^2}\dfrac{\pi }{3}$
Since, $\dfrac{{7\pi }}{6}$ is in the third quadrant, therefore $\operatorname{cosec} \dfrac{{7\pi }}{6}{\text{ }}$ is negative,
$= 2{\sin ^2}\dfrac{\pi }{6} + {\left( { - \operatorname{cosec} \dfrac{\pi }{6}} \right)^2}{\cos ^2}\dfrac{\pi }{3}{\text{ }}$
On simplification we get,
$= 2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\dfrac{\pi }{6}{\cos ^2}\dfrac{\pi }{3}$
Using $\sin \dfrac{\pi }{6} = \dfrac{1}{2},\cos \dfrac{\pi }{3} = \dfrac{1}{2},\cos ec\dfrac{\pi }{6} = 2$, we get,
$= 2 \times {\left( {\dfrac{1}{2}} \right)^2} + {\left( 2 \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2}$
On squaring we get,
$= \left( {2 \times \dfrac{1}{4}} \right) + \left( {4 \times \dfrac{1}{4}} \right)$
On simplification we get,
$= \dfrac{1}{2} + 1$
$= \dfrac{3}{2}$
= Right hand side
Hence, $2{\sin ^2}\dfrac{\pi }{6} + {\operatorname{cosec} ^2}\dfrac{{7\pi }}{6}{\cos ^2}\dfrac{\pi }{3} = \dfrac{3}{2}$ (proved).

Note: Note the following important formulae:
1.$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
2.${\sin ^2}x + {\cos ^2}x = 1$
3.${\sec ^2}x - {\tan ^2}x = 1$
4.${\operatorname{cosec} ^2}x - {\cot ^2}x = 1$
5.$\sin ( - x) = - \sin x$
6.$\cos ( - x) = \cos x$
7.$\tan ( - x) = - \tan x$
8.$\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
9.$\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
10.$\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }$
Sign convention:

$\sin 2x = 2\sin x\cos x$
$\cos 2x = {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1$
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} = \dfrac{2}{{\cot x - \tan x}}$
Also, the trigonometric ratios of the standard angles are given by

| $0^\circ$| $30^\circ$| $45^\circ$| $60^\circ$| $90^\circ$
---|---|---|---|---|---
$\operatorname{Sin} x$| 0| $\dfrac{1}{2}$ | $\dfrac{1}{{\sqrt 2 }}$ | $\dfrac{{\sqrt 3 }}{2}$ | 1
$\operatorname{Cos} x$| 1| $\dfrac{{\sqrt 3 }}{2}$| $\dfrac{1}{{\sqrt 2 }}$| $\dfrac{1}{2}$| 0
$\operatorname{Tan} x$| 0| $\dfrac{1}{{\sqrt 3 }}$ | 1| $\sqrt 3$| Undefined
$Cotx$| undefined| $\sqrt 3$| 1| $\dfrac{1}{{\sqrt 3 }}$| 0
$\cos ecx$| undefined| 2| $\sqrt 2$| $\dfrac{2}{{\sqrt 3 }}$| 1
$\operatorname{Sec} x$| 1| $\dfrac{2}{{\sqrt 3 }}$| $\sqrt 2$| 2| Undefined