Ravi Kant Kumar

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

Question Prove that : \[ \sec\left(\frac{3\pi}{2}-x\right)\sec\left(x-\frac{5\pi}{2}\right) + \tan\left(\frac{5\pi}{2}+x\right)\tan\left(x-\frac{3\pi}{2}\right) =-1 \] Solution \[ \begin{aligned} &\sec\left(\frac{3\pi}{2}-x\right)\sec\left(x-\frac{5\pi}{2}\right) + \tan\left(\frac{5\pi}{2}+x\right)\tan\left(x-\frac{3\pi}{2}\right) \\[8pt] =& (-\cosec x)(\cosec x) + (-\cot x)(\cot x) \\[8pt] =& -\cosec^2x-\cot^2x \\[8pt] =& -(1+\cot^2x)-\cot^2x \\[8pt] =& -1 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \sin^2\frac{\pi}{18} + \sin^2\frac{\pi}{9} + \sin^2\frac{7\pi}{18} + \sin^2\frac{4\pi}{9} = 2 \] Solution \[ \frac{7\pi}{18} = \frac{\pi}{2}-\frac{\pi}{9} \] \[ \sin^2\frac{7\pi}{18} = \cos^2\frac{\pi}{9} \] Also, \[ \frac{4\pi}{9} = \frac{\pi}{2}-\frac{\pi}{18} \] \[ \sin^2\frac{4\pi}{9} = \cos^2\frac{\pi}{18} \] Therefore, \[ \begin{aligned} &\sin^2\frac{\pi}{18} +\sin^2\frac{\pi}{9} +\sin^2\frac{7\pi}{18} +\sin^2\frac{4\pi}{9} \\[8pt] =& \sin^2\frac{\pi}{18} +\cos^2\frac{\pi}{18} +\sin^2\frac{\pi}{9} +\cos^2\frac{\pi}{9} \\[8pt] =& 1+1 \\[8pt] =&

Question Prove that : \[ \frac{\tan\left(\frac{\pi}{2}-x\right)\sec(\pi-x)\sin(-x)} {\sin(\pi+x)\cot(2\pi-x)\cosec\left(\frac{\pi}{2}-x\right)} =1 \] Solution \[ \begin{aligned} &\frac{\tan\left(\frac{\pi}{2}-x\right)\sec(\pi-x)\sin(-x)} {\sin(\pi+x)\cot(2\pi-x)\cosec\left(\frac{\pi}{2}-x\right)} \\[8pt] =& \frac{\cot x(-\sec x)(-\sin x)} {(-\sin x)(\cot x)(\sec x)} \\[8pt] =& \frac{\cot x\sec x\sin x} {\sin x\cot x\sec x} \\[8pt] =& 1 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \left(1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right) \left(1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right) = 2\cot x \] Solution \[ \begin{aligned} &\left(1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right) \left(1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right) \\[8pt] =& (1+\cot x)^2-\sec^2\left(\frac{\pi}{2}+x\right) \\[8pt] =& (1+\cot x)^2-\cosec^2x \\[8pt] =& 1+\cot^2x+2\cot x-\cosec^2x \\[8pt] =& 1+\cot^2x+2\cot x-(1+\cot^2x) \\[8pt] =& 2\cot x \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)} {\sin(2\pi-x)\cos(2\pi+x)\cosec(-x)\sin\left(\frac{3\pi}{2}-x\right)} =1 \] Solution \[ \begin{aligned} &\frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)} {\sin(2\pi-x)\cos(2\pi+x)\cosec(-x)\sin\left(\frac{3\pi}{2}-x\right)} \\[8pt] =& \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)} {(-\sin x)(\cos x)(-\cosec x)(-\cos x)} \\[8pt] =& \frac{-\sin^2x\cot^2x} {-\sin x\cos x\cosec x\cos x} \\[8pt] =& \frac{\sin^2x\cdot\frac{\cos^2x}{\sin^2x}} {\sin x\cos^2x\cdot\frac1{\sin x}} \\[8pt] =& \frac{\cos^2x}{\cos^2x} \\[8pt] =& 1 \end{aligned} \] Hence Proved. Next Question / Full Chapter

Question Prove that : \[ \frac{\cosec(90^\circ+x)+\cot(450^\circ+x)} {\cosec(90^\circ-x)+\tan(180^\circ-x)} + \frac{\tan(180^\circ+x)\sec(180^\circ-x)} {\tan(360^\circ+x)-\sec(-x)} =2 \] Solution \[ \begin{aligned} &\frac{\cosec(90^\circ+x)+\cot(450^\circ+x)} {\cosec(90^\circ-x)+\tan(180^\circ-x)} + \frac{\tan(180^\circ+x)\sec(180^\circ-x)} {\tan(360^\circ+x)-\sec(-x)} \\[8pt] =& \frac{\sec x-\tan x} {\sec x-\tan x} + \frac{\tan x(-\sec x)} {\tan x-\sec x} \\[8pt] =& 1+ \frac{\tan x\sec x} {\sec x-\tan x} \\[8pt] =& 1+1 \\[8pt] =& 2 \end{aligned} \] Hence Proved. Next

Question Prove that : \[ \frac{\cos(2\pi+x)\cosec(2\pi+x)\tan\left(\frac{\pi}{2}+x\right)} {\sec\left(\frac{\pi}{2}+x\right)\cos x\cot(\pi+x)} =1 \] Solution Using standard identities, \[ \cos(2\pi+x)=\cos x \] \[ \cosec(2\pi+x)=\cosec x \] \[ \tan\left(\frac{\pi}{2}+x\right)=-\cot x \] \[ \sec\left(\frac{\pi}{2}+x\right)=-\cosec x \] \[ \cot(\pi+x)=\cot x \] Substituting these values, \[ \begin{aligned} &\frac{\cos x\cdot\cosec x\cdot(-\cot x)} {(-\cosec x)\cos x\cot x} \\[4pt] =& \frac{-\cos x\cosec x\cot x} {-\cos x\cosec