Answer :

$\cos x=-\dfrac{1}{2}$

$\therefore \sec x=\dfrac{1}{\cos x}=\dfrac{1}{(-\dfrac{1}{2})}=-2$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \sin ^{2} x=1-(-\dfrac{1}{2})^{2}$

$\Rightarrow \sin ^{2} x=1-\dfrac{1}{4}=\dfrac{3}{4}$

$\Rightarrow \sin x= \pm \dfrac{\sqrt{3}}{2}$

Since $x$ lies in the $3^{\text{rd }}$ quadrant, the value of $\sin x$ will be negative.

$\therefore \sin x=-\dfrac{\sqrt{3}}{2}$

$cosec x=\dfrac{1}{\sin x}=\dfrac{1}{(-\dfrac{\sqrt{3}}{2})}=-\dfrac{2}{\sqrt{3}}$

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{(-\dfrac{\sqrt{3}}{2})}{(-\dfrac{1}{2})}=\sqrt{3}$

$\cot x=\dfrac{1}{\tan x}=\dfrac{1}{\sqrt{3}}$

Answer :

$\sin x=\dfrac{3}{5}$

$cosec x=\dfrac{1}{\sin x}=\dfrac{1}{(\dfrac{3}{5})}=\dfrac{5}{3}$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \cos ^{2} x=1-\sin ^{2} x$

$\Rightarrow \cos ^{2} x=1-(\dfrac{3}{5})^{2}$

$\Rightarrow \cos ^{2} x=1-\dfrac{9}{25}$

$\Rightarrow \cos ^{2} x=\dfrac{16}{25}$

$\Rightarrow \cos x= \pm \dfrac{4}{5}$

Since $x$ lies in the $2^{\text{nd }}$ quadrant, the value of $\cos x$ will be negative

$\therefore \cos x=-\dfrac{4}{5}$

$\sec x=\dfrac{1}{\cos x}=\dfrac{1}{(-\dfrac{4}{5})}=-\dfrac{5}{4}$

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{(\dfrac{3}{5})}{(-\dfrac{4}{5})}=-\dfrac{3}{4}$

$\cot x=\dfrac{1}{\tan x}=-\dfrac{4}{3}$

Answer :

$\cot x=\dfrac{3}{4}$

$\tan x=\dfrac{1}{\cot x}=\dfrac{1}{(\dfrac{3}{4})}=\dfrac{4}{3}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+(\dfrac{4}{3})^{2}=\sec ^{2} x$

$\Rightarrow 1+\dfrac{16}{9}=\sec ^{2} x$

$\Rightarrow \dfrac{25}{9}=\sec ^{2} x$

$\Rightarrow \sec x= \pm \dfrac{5}{3}$

Since $x$ lies in the $3^{\text{rd }}$ quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\dfrac{5}{3}$

$\cos x=\dfrac{1}{\sec x}=\dfrac{1}{(-\dfrac{5}{3})}=-\dfrac{3}{5}$

$\tan x=\dfrac{\sin x}{\cos x}$

$\Rightarrow \dfrac{4}{3}=\dfrac{\sin x}{(\dfrac{-3}{5})}$

$\Rightarrow \sin x=(\dfrac{4}{3}) \times(\dfrac{-3}{5})=-\dfrac{4}{5}$

$cosec x=\dfrac{1}{\sin x}=-\dfrac{5}{4}$

Answer :

$\sec x=\dfrac{13}{5}$

$\cos x=\dfrac{1}{\sec x}=\dfrac{1}{(\dfrac{13}{5})}=\dfrac{5}{13}$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \sin ^{2} x=1-(\dfrac{5}{13})^{2}$

$\Rightarrow \sin ^{2} x=1-\dfrac{25}{169}=\dfrac{144}{169}$

$\Rightarrow \sin x= \pm \dfrac{12}{13}$

Since $x$ lies in the $4^{\text{th }}$ quadrant, the value of $\sin x$ will be negative.

$\therefore \sin x=-\dfrac{12}{13}$

$cosec x=\dfrac{1}{\sin x}=\dfrac{1}{(-\dfrac{12}{13})}=-\dfrac{13}{12}$

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{(\dfrac{-12}{13})}{(\dfrac{5}{13})}=-\dfrac{12}{5}$

$\cot x=\dfrac{1}{\tan x}=\dfrac{1}{(-\dfrac{12}{5})}=-\dfrac{5}{12}$

Answer :

$\tan x=-\dfrac{5}{12}$ $\cot x=\dfrac{1}{\tan x}=\dfrac{1}{(-\dfrac{5}{12})}=-\dfrac{12}{5}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+(-\dfrac{5}{12})^{2}=\sec ^{2} x$

$\Rightarrow 1+\dfrac{25}{144}=\sec ^{2} x$

$\Rightarrow \dfrac{169}{144}=\sec ^{2} x$

$\Rightarrow \sec x= \pm \dfrac{13}{12}$

Since $x$ lies in the $2^{\text{nd }}$ quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\dfrac{13}{12}$

$\cos x=\dfrac{1}{\sec x}=\dfrac{1}{(-\dfrac{13}{12})}=-\dfrac{12}{13}$

$\tan x=\dfrac{\sin x}{\cos x}$

$\Rightarrow-\dfrac{5}{12}=\dfrac{\sin x}{(-\dfrac{12}{13})}$

$\Rightarrow \sin x=(-\dfrac{5}{12}) \times(-\dfrac{12}{13})=\dfrac{5}{13}$

$cosec x=\dfrac{1}{\sin x}=\dfrac{1}{(\dfrac{5}{13})}=\dfrac{13}{5}$

Answer :

It is known that the values of $\sin x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore \sin 765^{\circ}=\sin (2 \times 360^{\circ}+45^{\circ})=\sin 45^{\circ}=\dfrac{1}{\sqrt{2}}$

Answer :

It is known that the values of $cosec x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore cosec(-1410^{\circ})=cosec(-1410^{\circ}+4 \times 360^{\circ})$

$=cosec(-1410^{\circ}+1440^{\circ})$

$=cosec 30^{\circ}=2$

Answer :

It isknown that the values of $\tan x$ repeat after an interval of $\pi$ or $180^{\circ}$.

$\therefore \tan \dfrac{19 \pi}{3}=\tan 6 \dfrac{1}{3} \pi=\tan (6 \pi+\dfrac{\pi}{3})=\tan \dfrac{\pi}{3}=\tan 60^{\circ}=\sqrt{3}$

Answer :

It is known that the values of $\sin x$ repeat after an interval of $2 \pi$ or $360^{\circ}$.

$\therefore \sin (-\dfrac{11 \pi}{3})=\sin (-\dfrac{11 \pi}{3}+2 \times 2 \pi)=\sin (\dfrac{\pi}{3})=\dfrac{\sqrt{3}}{2}$

Answer :

It is known that the values of $\cot x$ repeat after an interval of $\pi$ or $180^{\circ}$.

$\therefore \cot (-\dfrac{15 \pi}{4})=\cot (-\dfrac{15 \pi}{4}+4 \pi)=\cot \dfrac{\pi}{4}=1$