Ravi Kant Kumar

Question Find the value of the following trigonometric ratio : \[ \cos\left(\frac{19\pi}{4}\right) \] Solution \[ \cos\left(\frac{19\pi}{4}\right) = \cos\left(\frac{19\pi}{4}-4\pi\right) \] \[ = \cos\frac{3\pi}{4} \] \[ = \cos\left(\pi-\frac{\pi}{4}\right) \] Using, \[ \cos(\pi-\theta)=-\cos\theta \] \[ = -\cos\frac{\pi}{4} = -\frac{1}{\sqrt2} = -\frac{\sqrt2}{2} \] Answer : \[ \boxed{ \cos\left(\frac{19\pi}{4}\right) = -\frac{\sqrt2}{2} } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \tan\left(-\frac{13\pi}{4}\right) \] Solution \[ \tan\left(-\frac{13\pi}{4}\right) = \tan\left(-\frac{13\pi}{4}+4\pi\right) \] \[ = \tan\frac{3\pi}{4} \] \[ = \tan\left(\pi-\frac{\pi}{4}\right) \] Using, \[ \tan(\pi-\theta)=-\tan\theta \] \[ = -\tan\frac{\pi}{4} = -1 \] Answer : \[ \boxed{ \tan\left(-\frac{13\pi}{4}\right)=-1 } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \cosec\left(-\frac{20\pi}{3}\right) \] Solution \[ \cosec\left(-\frac{20\pi}{3}\right) = \frac{1}{\sin\left(-\frac{20\pi}{3}\right)} \] \[ = \frac{1}{\sin\left(-\frac{20\pi}{3}+6\pi\right)} \] \[ = \frac{1}{\sin\left(-\frac{2\pi}{3}\right)} \] Using, \[ \sin(-\theta)=-\sin\theta \] \[ = \frac{1}{-\sin\frac{2\pi}{3}} \] \[ = \frac{1}{-\frac{\sqrt3}{2}} = -\frac{2}{\sqrt3} = -\frac{2\sqrt3}{3} \] Answer : \[ \boxed{ \cosec\left(-\frac{20\pi}{3}\right) = -\frac{2\sqrt3}{3} } \] Next Question /

Question Find the value of the following trigonometric ratio : \[ \sin\left(-\frac{11\pi}{6}\right) \] Solution \[ \sin\left(-\frac{11\pi}{6}\right) = -\sin\frac{11\pi}{6} \] Using, \[ \sin(-\theta)=-\sin\theta \] \[ = -\sin\left(2\pi-\frac{\pi}{6}\right) \] \[ = -\left(-\sin\frac{\pi}{6}\right) \] \[ = \frac{1}{2} \] Answer : \[ \boxed{ \sin\left(-\frac{11\pi}{6}\right)=\frac{1}{2} } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \cos \frac{19\pi}{6} \] Solution \[ \cos \frac{19\pi}{6} = \cos\left(2\pi+\frac{7\pi}{6}\right) \] Using, \[ \cos(2\pi+\theta)=\cos\theta \] \[ = \cos\frac{7\pi}{6} \] \[ = \cos\left(\pi+\frac{\pi}{6}\right) \] Using, \[ \cos(\pi+\theta)=-\cos\theta \] \[ = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2} \] Answer : \[ \boxed{ \cos\frac{19\pi}{6} = -\frac{\sqrt3}{2} } \] Next Question /

Question Find the value of the following trigonometric ratio : \[ \sin \frac{17\pi}{6} \] Solution \[ \sin \frac{17\pi}{6} = \sin\left(2\pi+\frac{5\pi}{6}\right) \] Using, \[ \sin(2\pi+\theta)=\sin\theta \] \[ = \sin\frac{5\pi}{6} \] \[ = \sin\left(\pi-\frac{\pi}{6}\right) \] \[ = \sin\frac{\pi}{6} = \frac{1}{2} \] Answer : \[ \boxed{ \sin\frac{17\pi}{6}=\frac{1}{2} } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \tan \frac{7\pi}{4} \] Solution \[ \tan \frac{7\pi}{4} = \tan\left(2\pi-\frac{\pi}{4}\right) \] Using, \[ \tan(2\pi-\theta)=-\tan\theta \] \[ = -\tan\frac{\pi}{4} \] \[ = -1 \] Answer : \[ \boxed{ \tan\frac{7\pi}{4}=-1 } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \cos\left(-\frac{25\pi}{4}\right) \] Solution \[ \cos\left(-\frac{25\pi}{4}\right) = \cos\left(-\frac{25\pi}{4}+6\pi\right) \] \[ = \cos\left(-\frac{25\pi}{4}+\frac{24\pi}{4}\right) \] \[ = \cos\left(-\frac{\pi}{4}\right) \] Using, \[ \cos(-\theta)=\cos\theta \] \[ = \cos\frac{\pi}{4} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \] Answer : \[ \boxed{ \cos\left(-\frac{25\pi}{4}\right) = \frac{\sqrt2}{2} } \] Next Question / Full Chapter

Question Find the values of the following trigonometric ratios : \[ \tan \frac{11\pi}{6} \] Solution \[ \tan \frac{11\pi}{6} = \tan \left(2\pi-\frac{\pi}{6}\right) \] Using, \[ \tan(2\pi-\theta)=-\tan\theta \] \[ = -\tan\frac{\pi}{6} \] \[ = -\frac{1}{\sqrt3} = -\frac{\sqrt3}{3} \] Answer : \[ \boxed{ \tan\frac{11\pi}{6} = -\frac{\sqrt3}{3} } \] Next Question / Full Chapter

Question Find the values of the following trigonometric ratios : \[ \sin 17\pi \] Solution \[ \sin 17\pi = \sin(n\pi) \] Using, \[ \sin(n\pi)=0 \] Therefore, \[ \sin 17\pi = 0 \] Answer : \[ \boxed{ \sin 17\pi = 0 } \] Next Question / Full Chapter