# Find the value of

Question:

Find the value of

(i) $\cos 840^{\circ}$

(ii) $\sin 870^{\circ}$

(iii) $\tan \left(-120^{\circ}\right)$

(iv) $\sec \left(-420^{\circ}\right)$

(v) $\operatorname{cosec}\left(-690^{\circ}\right)$

(vi) $\tan \left(225^{\circ}\right)$

(vii) $\cot \left(-315^{\circ}\right)$

(viii) $\sin \left(-1230^{\circ}\right)$

(ix) $\cos \left(495^{\circ}\right)$

Solution:

(i) $\operatorname{Cos} 840^{\circ}=\operatorname{Cos}\left(2.360^{\circ}+120^{\circ}\right) \ldots \ldots \ldots \ldots($ using $\operatorname{Cos}(2 \varpi+x)=\operatorname{Cos} x)$

$=\operatorname{Cos}\left(120^{\circ}\right)$

$=\operatorname{Cos}\left(180^{\circ}-60^{\circ}\right)$

$=-\operatorname{Cos} 60^{\circ} \ldots \ldots \ldots \ldots \ldots($ using $\operatorname{Cos}(\varpi-x)=-\operatorname{Cos} x)$

$=-\frac{1}{2}$

(ii) $\sin 870^{\circ}=\sin \left(2.360^{\circ}+150^{\circ}\right)$ .(using $\sin (2 \varpi+x)=\sin x$ )

$=\sin 150^{\circ}$

$=\sin \left(180^{\circ}-30^{\circ}\right) \ldots \ldots \ldots($ using $\sin (\varpi-x)=\sin x)$

$=\sin 30^{\circ}$

$=\frac{1}{2}$

(iii) $\tan \left(-120^{\circ}\right)=-\tan 12 \ldots \ldots .(\tan (-x)=\tan x)$

$=-\tan \left(180^{\circ}-60^{\circ}\right)$ ……. (in II quadrant tanx is negative)

$=-\left(-\tan 60^{\circ}\right)$

$=\tan 60^{\circ}$

$=\sqrt{3}$

(iv) $\sec \left(-420^{\circ}\right)=\frac{1}{\cos \left(-420^{\circ}\right)}$

$=\frac{1}{-\cos 420^{\circ}}$ .............(using $\cos (-x)=-\cos x$ )

$=\frac{-1}{-\cos \left(360^{\circ}+60\right)}$ ………... (using $\cos (2 \omega+x)=\cos x$ )

$=\frac{-1}{\cos 60^{\circ}} \Rightarrow \frac{-1}{1 / 2}=-2$

(v) $\operatorname{cosec}\left(690^{\circ}\right)=\frac{1}{\sin \left(-690^{\circ}\right)} \Rightarrow \frac{1}{-\sin \left(690^{\circ}\right)}=\frac{1}{-\sin \left(2.360-30^{\circ}\right)}$

$=\frac{1}{-\left(-\sin 30^{\circ}\right)} \Rightarrow \frac{1}{1 / 2}=2$

(vi) $\tan 225^{\circ}=\tan \left(180^{\circ}+45^{\circ}\right)$ …………(in III quadrant tanx is positive)

$\Rightarrow \tan 45^{\circ}=1$

(vii)  $\cot \left(-315^{\circ}\right)=\frac{1}{\tan (-315)^{\circ}} \Rightarrow \frac{1}{-\tan \left(315^{\circ}\right)}=\frac{1}{-\tan \left(360^{\circ}-45^{\circ}\right)}$

$\ldots .(\tan (-x)=-\tan x)$

$=\frac{1}{-\left(-\tan 45^{\circ}\right)} \Rightarrow 1$  …..(in IV quadrant tanx is negative)

(viii) $\sin \left(-1230^{\circ}\right)=\sin 1230^{\circ}$ ………….(using sin( - x) = sinx)

$=\sin \left(3.360^{\circ}+150^{\circ}\right)$

$=\sin 150^{\circ}$

$=\sin \left(180^{\circ}-30^{\circ}\right)$ ………… (using $\sin \left(180^{\circ}-x\right)=\sin x$ )

= sin30

$=\frac{1}{2}$

(ix) $\cos 495^{\circ}=\cos \left(360^{\circ}+135^{\circ}\right)$............(using $\left.\cos \left(360^{\circ}+x\right)=\cos x\right)$

$=\cos 135^{\circ}$

$=\cos \left(180^{\circ}-45^{\circ}\right) \ldots \ldots \ldots \ldots\left(\right.$ using $\left.\cos \left(180^{\circ}-x\right)=-\cos x\right)$

$=-\cos 45^{\circ}$

$=-\frac{1}{\sqrt{2}}$