Ravi Kant Kumar

Question Prove that : \[ 3\sin\frac{\pi}{6}\sec\frac{\pi}{3} – 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1 \] Solution Using standard values, \[ \sin\frac{\pi}{6}=\frac12, \qquad \sec\frac{\pi}{3}=2 \] \[ \sin\frac{5\pi}{6}=\frac12, \qquad \cot\frac{\pi}{4}=1 \] Substituting these values, \[ \begin{aligned} &3\sin\frac{\pi}{6}\sec\frac{\pi}{3} – 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} \\[4pt] =& 3\left(\frac12\right)(2) – 4\left(\frac12\right)(1) \\[4pt] =& 3-2 \\[4pt] =& 1 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \tan\frac{11\pi}{3} – 2\sin\frac{4\pi}{6} – \frac34\cosec^2\frac{\pi}{4} + 4\cos^2\frac{17\pi}{6} = \frac{3-4\sqrt3}{2} \] Solution \[ \tan\frac{11\pi}{3} = \tan\left(2\pi-\frac{\pi}{3}\right) = -\tan\frac{\pi}{3} = -\sqrt3 \] \[ 2\sin\frac{4\pi}{6} = 2\sin\frac{2\pi}{3} = 2\times\frac{\sqrt3}{2} = \sqrt3 \] \[ \frac34\cosec^2\frac{\pi}{4} = \frac34\left(\sqrt2\right)^2 = \frac34\times2 = \frac32 \] \[ 4\cos^2\frac{17\pi}{6} = 4\cos^2\left(2\pi+\frac{5\pi}{6}\right) \] \[ = 4\cos^2\frac{5\pi}{6} = 4\left(-\frac{\sqrt3}{2}\right)^2 \] \[

Question Prove that : \[ \cos570^\circ\sin510^\circ+\sin(-330^\circ)\cos(-390^\circ)=0 \] Solution First reduce the angles. \[ \cos570^\circ = \cos(360^\circ+210^\circ) = \cos210^\circ = -\frac{\sqrt3}{2} \] \[ \sin510^\circ = \sin(360^\circ+150^\circ) = \sin150^\circ = \frac12 \] Also, \[ \sin(-330^\circ) = -\sin330^\circ = -\left(-\frac12\right) = \frac12 \] \[ \cos(-390^\circ) = \cos390^\circ = \cos(360^\circ+30^\circ) = \cos30^\circ = \frac{\sqrt3}{2} \] Substituting these values, \[

Question Prove that : \[ \tan(-225^\circ)\cot(-405^\circ)-\tan(-765^\circ)\cot(675^\circ)=0 \] Solution We know that tangent and cotangent have period \(180^\circ\). First, \[ \tan(-225^\circ) = \tan(-225^\circ+180^\circ) \] \[ = \tan(-45^\circ) = -1 \] Next, \[ \cot(-405^\circ) = \cot(-405^\circ+360^\circ) \] \[ = \cot(-45^\circ) = -1 \] Also, \[ \tan(-765^\circ) = \tan(-765^\circ+720^\circ) \] \[ = \tan(-45^\circ) = -1 \] And, \[

Question Prove that : \[ \cos24^\circ+\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ=\frac12 \] Solution \[ \begin{aligned} &\cos24^\circ+\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ \\[4pt] =& \cos24^\circ+\cos55^\circ-\cos55^\circ-\cos24^\circ+\cos60^\circ \\[4pt] =& \frac12 \end{aligned} \] Using, \[ \cos(180^\circ-\theta)=-\cos\theta, \qquad \cos(180^\circ+\theta)=-\cos\theta \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} = \frac12 \] Solution \[ \begin{aligned} &\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} \\[4pt] =& \sin\frac{2\pi}{3}\cos\frac{\pi}{6} + \cos\frac{\pi}{3}\sin\left(-\frac{\pi}{6}\right) \\[4pt] =& \left(\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}\right) + \left(\frac12\times-\frac12\right) \\[4pt] =& \frac34-\frac14 \\[4pt] =& \frac12 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \tan225^\circ \cot405^\circ + \tan765^\circ \cot675^\circ = 0 \] Solution \[ \tan225^\circ = \tan(180^\circ+45^\circ) = \tan45^\circ = 1 \] \[ \cot405^\circ = \cot(360^\circ+45^\circ) = \cot45^\circ = 1 \] \[ \tan765^\circ = \tan(720^\circ+45^\circ) = \tan45^\circ = 1 \] \[ \cot675^\circ = \cot(720^\circ-45^\circ) = \cot315^\circ = -1 \] Therefore, \[ =(1)(1)+(1)(-1) \] \[

Question Find the value of the following trigonometric ratio : \[ \sin\left(\frac{151\pi}{6}\right) \] Solution \[ \sin\left(\frac{151\pi}{6}\right) = \sin\left(\frac{151\pi}{6}-24\pi\right) \] \[ = \sin\frac{7\pi}{6} \] \[ = \sin\left(\pi+\frac{\pi}{6}\right) \] Using, \[ \sin(\pi+\theta)=-\sin\theta \] \[ = -\sin\frac{\pi}{6} = -\frac{1}{2} \] Answer : \[ \boxed{ \sin\left(\frac{151\pi}{6}\right) = -\frac{1}{2} } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \cos\left(\frac{39\pi}{4}\right) \] Solution \[ \cos\left(\frac{39\pi}{4}\right) = \cos\left(\frac{39\pi}{4}-8\pi\right) \] \[ = \cos\frac{7\pi}{4} \] \[ = \cos\left(2\pi-\frac{\pi}{4}\right) \] Using, \[ \cos(2\pi-\theta)=\cos\theta \] \[ = \cos\frac{\pi}{4} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \] Answer : \[ \boxed{ \cos\left(\frac{39\pi}{4}\right) = \frac{\sqrt2}{2} } \] Next Question / Full Chapter

Question Find the value of the following trigonometric ratio : \[ \sin\left(\frac{41\pi}{4}\right) \] Solution \[ \sin\left(\frac{41\pi}{4}\right) = \sin\left(\frac{41\pi}{4}-10\pi\right) \] \[ = \sin\frac{\pi}{4} \] \[ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \] Answer : \[ \boxed{ \sin\left(\frac{41\pi}{4}\right) = \frac{\sqrt2}{2} } \] Next Question / Full Chapter