Question

${\cot ^2}\dfrac{\pi }{6} + cosec\dfrac{{5\pi }}{6} + 3{\tan ^2}\dfrac{\pi }{6} = 6$

Answer 1

Hint: Note that, \[{\text{since }}\dfrac{{5\pi }}{6}{\text{ is in second quadrant, therefore }}\operatorname{cosec} \dfrac{{5\pi }}{6}{\text{ is positive}}{\text{.}}\]
Then in the left hand side, substitute the values of \[\cot \dfrac{\pi }{6}\], \[\operatorname{cosec} \dfrac{{5\pi }}{6}\] and \[\tan \dfrac{\pi }{6}\].
On solving them we will reach our required solution.

Complete step-by-step answer:
Given that, to prove ${\cot ^2}\dfrac{\pi }{6} + cosec\dfrac{{5\pi }}{6} + 3{\tan ^2}\dfrac{\pi }{6} = 6$
Left hand side:
$ = {\cot ^2}\dfrac{\pi }{6} + cosec\dfrac{{5\pi }}{6} + 3{\tan ^2}\dfrac{\pi }{6}$
The above expression can be written as,
$ = {\cot ^2}\dfrac{\pi }{6} + cosec\left( {\pi - \dfrac{\pi }{6}} \right) + 3{\tan ^2}\dfrac{\pi }{6}$
Since $\dfrac{{5\pi }}{6}$is in the second quadrant, therefore $\operatorname{cosec} \dfrac{{5\pi }}{6}$ is positive,
$ = {\cot ^2}\dfrac{\pi }{6} + cosec\dfrac{\pi }{6} + 3{\tan ^2}\dfrac{\pi }{6}$
Using, \[\cot \dfrac{\pi }{6} = \sqrt 3 ,\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }},\cos ec\dfrac{\pi }{6} = 2\], we get,
$ = {\left( {\sqrt 3 } \right)^2} + 2 + 3 \times {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2}$
On squaring we get,
$ = 3 + 2 + \left( {3 \times \dfrac{1}{3}} \right)$
On further simplification we get,
$ = 3 + 2 + 1$
$ = 6$
= Right hand side
Hence, ${\cot ^2}\dfrac{\pi }{6} + cosec\dfrac{{5\pi }}{6} + 3{\tan ^2}\dfrac{\pi }{6} = 6$ (proved).

Note:
Note the following important formulae:
$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
${\sin ^2}x + {\cos ^2}x = 1$
\[{\sec ^2}x - {\tan ^2}x = 1\]
\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
$\sin ( - x) = - \sin x$
$\cos ( - x) = \cos x$
$\tan ( - x) = - \tan x$
$\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
$\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
$\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }$
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