Educational

Question Prove that : \[ \sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = -1 \] Solution Reducing the angles, \[ \sin\frac{10\pi}{3} = \sin\left(2\pi+\frac{4\pi}{3}\right) = \sin\frac{4\pi}{3} = -\frac{\sqrt3}{2} \] \[ \cos\frac{13\pi}{6} = \cos\left(2\pi+\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2} \] \[ \cos\frac{8\pi}{3} = \cos\left(2\pi+\frac{2\pi}{3}\right) = \cos\frac{2\pi}{3} = -\frac12 \] \[ \sin\frac{5\pi}{6} = \frac12 \] Substituting these values, \[ \begin{aligned} &\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} […]

Question Prove that : \[ \sin\frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac12 \] Solution Step 1: Reduce angles using periodicity: \[ \sin\frac{13\pi}{3} = \sin\left(4\pi + \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \] \[ \sin\frac{2\pi}{3} \text{ is already in the first cycle.} \] \[ \cos\frac{4\pi}{3} = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\frac{\pi}{3} = -\frac12 \] \[ \sin\frac{13\pi}{6} = \sin\left(2\pi + \frac{\pi}{6}\right) =

Question Prove that : \[ \sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac12 \] Solution \[ \begin{aligned} &\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} \\[8pt] =& \sin\frac{\pi}{3}\sin\frac{2\pi}{3} + \left(-\frac12\right)\left(\frac12\right) \\[8pt] =& \left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right) -\frac14 \\[8pt] =& \frac34-\frac14 \\[8pt] =& \frac12 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \tan4\pi-\cos\frac{3\pi}{2}-\sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac14 \] Solution \[ \begin{aligned} &\tan4\pi-\cos\frac{3\pi}{2}-\sin\frac{5\pi}{6}\cos\frac{2\pi}{3} \\[8pt] =& 0-0-\left(\frac12\right)\left(-\frac12\right) \\[8pt] =& \frac14 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Find \(x\) from the following equation : \[ x\cot\left(\frac{\pi}{2}+\theta\right) + \tan\left(\frac{\pi}{2}+\theta\right)\sin\theta + \cosec\left(\frac{\pi}{2}+\theta\right) = 0 \] Solution \[ \begin{aligned} &x\cot\left(\frac{\pi}{2}+\theta\right) + \tan\left(\frac{\pi}{2}+\theta\right)\sin\theta + \cosec\left(\frac{\pi}{2}+\theta\right) = 0 \\[8pt] \Rightarrow\;& -x\tan\theta-\cot\theta\sin\theta+\sec\theta=0 \\[8pt] \Rightarrow\;& -x\tan\theta-\cos\theta+\sec\theta=0 \\[8pt] \Rightarrow\;& x\tan\theta=\sec\theta-\cos\theta \\[8pt] \Rightarrow\;& x= \frac{\sec\theta-\cos\theta}{\tan\theta} \\[8pt] \Rightarrow\;& x= \frac{\frac1{\cos\theta}-\cos\theta}{\frac{\sin\theta}{\cos\theta}} \\[8pt] \Rightarrow\;& x= \frac{1-\cos^2\theta}{\sin\theta} \\[8pt] \Rightarrow\;& x= \frac{\sin^2\theta}{\sin\theta} \\[8pt] \Rightarrow\;& x=\sin\theta

Question Find \( \theta \) from the following equation : \[ \cosec\left(\frac{\pi}{2}+\theta\right) + \cos\theta\cot\left(\frac{\pi}{2}+\theta\right) = \sin\left(\frac{\pi}{2}+\theta\right) \] Solution \[ \begin{aligned} &\cosec\left(\frac{\pi}{2}+\theta\right) + \cos\theta\cot\left(\frac{\pi}{2}+\theta\right) = \sin\left(\frac{\pi}{2}+\theta\right) \\[8pt] \Rightarrow\;& \sec\theta-\cos\theta\tan\theta = \cos\theta \\[8pt] \Rightarrow\;& \sec\theta-\sin\theta = \cos\theta \\[8pt] \Rightarrow\;& \frac1{\cos\theta}-\sin\theta = \cos\theta \\[8pt] \Rightarrow\;& 1-\sin\theta\cos\theta = \cos^2\theta \\[8pt] \Rightarrow\;& \sin^2\theta = \sin\theta\cos\theta \\[8pt] \Rightarrow\;& \sin\theta(\sin\theta-\cos\theta)=0 \end{aligned} \]

Question If \(A, B, C, D\) be the angles of a cyclic quadrilateral, taken in order, prove that : \[ \cos(180^\circ-A)+\cos(180^\circ+B)+\cos(180^\circ+C)-\sin(90^\circ+D)=0 \] Solution Using identities, \[ \cos(180^\circ-A)=-\cos A \] \[ \cos(180^\circ+B)=-\cos B \] \[ \cos(180^\circ+C)=-\cos C \] \[ \sin(90^\circ+D)=\cos D \] Therefore, \[ \begin{aligned} &\cos(180^\circ-A)+\cos(180^\circ+B) \\ &+\cos(180^\circ+C)-\sin(90^\circ+D) \\[8pt] =& -\cos A-\cos B-\cos C-\cos D \end{aligned}

Question In a \( \triangle ABC \), prove that : \[ \tan\frac{A+B}{2} = \cot\frac{C}{2} \] Solution In a triangle, \[ A+B+C=\pi \] \[ A+B=\pi-C \] Dividing by \(2\), \[ \frac{A+B}{2} = \frac{\pi}{2}-\frac{C}{2} \] Therefore, \[ \begin{aligned} \tan\frac{A+B}{2} &= \tan\left(\frac{\pi}{2}-\frac{C}{2}\right) \\[8pt] &= \cot\frac{C}{2} \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question In a \( \triangle ABC \), prove that : \[ \cos\frac{A+B}{2} = \sin\frac{C}{2} \] Solution In a triangle, \[ A+B+C=\pi \] \[ A+B=\pi-C \] Dividing by \(2\), \[ \frac{A+B}{2} = \frac{\pi-C}{2} = \frac{\pi}{2}-\frac{C}{2} \] Therefore, \[ \begin{aligned} \cos\frac{A+B}{2} &= \cos\left(\frac{\pi}{2}-\frac{C}{2}\right) \\[8pt] &= \sin\frac{C}{2} \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question In a \( \triangle ABC \), prove that : \[ \cos(A+B)+\cos C=0 \] Solution In a triangle, \[ A+B+C=\pi \] \[ A+B=\pi-C \] Therefore, \[ \begin{aligned} \cos(A+B)+\cos C &= \cos(\pi-C)+\cos C \\[8pt] &= -\cos C+\cos C \\[8pt] &= 0 \end{aligned} \] Hence Proved. Next Question / Full Chapter