Answer :

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

$\tan x=-\dfrac{5}{12}$

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

$\therefore \ \ \sin 765^{\circ}=\sin \left(2 \times 360^{\circ}+45^{\circ}\right)$

$\hspace{1.8cm}=\sin 45^{\circ}=\dfrac{1}{\sqrt{2}}$

Answer :

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

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

$\hspace{2.7cm}= \text{cosec }\left(-1410^{\circ}+1440^{\circ}\right)$

$\hspace{2.7cm}= \text{cosec } 30^{\circ}=2$

Answer :

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

$\therefore \ \ \tan \dfrac{19 \pi}{3}=\tan 6 \dfrac{1}{3} \pi=\tan \left(6 \pi+\dfrac{\pi}{3}\right)$

$\hspace{2cm}=\tan \dfrac{\pi}{3}=\tan 60^{\circ}=\sqrt{3}$

Answer :

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

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

$\hspace{2.5cm}=\sin \left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}$

Answer :

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

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

$\hspace{2.5cm}=\cot \dfrac{\pi}{4}=1$