Educational

Question \[ \text{If } \tan x=-\frac{1}{\sqrt5} \text{ and } x \text{ lies in IV quadrant,} \] \[ \text{then the value of } \cos x \text{ is} \] (a) \(\frac{\sqrt5}{\sqrt6}\) (b) \(\frac{2}{\sqrt6}\) (c) \(\frac12\) (d) \(\frac1{\sqrt6}\) Solution \[ \tan x=\frac{\text{Perpendicular}}{\text{Base}} = -\frac1{\sqrt5} \] Take \[ \text{Perpendicular}=-1,\quad \text{Base}=\sqrt5 \] \[ \text{Hypotenuse} = \sqrt{(-1)^2+(\sqrt5)^2} \] \[ = \sqrt{1+5} […]

Question \[ \text{If } \tan x+\sec x=\sqrt3,\ 0<x<\pi, \] \[ \text{then } x \text{ is equal to} \] (a) \(\frac{5\pi}{6}\) (b) \(\frac{2\pi}{3}\) (c) \(\frac{\pi}{6}\) (d) \(\frac{\pi}{3}\) Solution Using identity, \[ \sec x+\tan x = \tan\left(\frac{\pi}{4}+\frac{x}{2}\right) \] So, \[ \tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=\sqrt3 \] \[ \frac{\pi}{4}+\frac{x}{2}=\frac{\pi}{3} \] \[ \frac{x}{2}=\frac{\pi}{3}-\frac{\pi}{4} \] \[ \frac{x}{2}=\frac{\pi}{12} \] \[ x=\frac{\pi}{6} \] Answer \[ \boxed{\frac{\pi}{6}}

Question \[ x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta \] Then \[ x^2+y^2+z^2 \] is independent of (a) \(\theta,\phi\) (b) \(r,\theta\) (c) \(r,\phi\) (d) \(r\) Solution \[ x^2+y^2+z^2 \] \[ =r^2\sin^2\theta\cos^2\phi +r^2\sin^2\theta\sin^2\phi +r^2\cos^2\theta \] \[ =r^2\sin^2\theta(\cos^2\phi+\sin^2\phi) +r^2\cos^2\theta \] \[ =r^2\sin^2\theta+r^2\cos^2\theta \] \[ =r^2(\sin^2\theta+\cos^2\theta) \] \[ =r^2 \] Hence, it is independent of \[ \theta \text{ and } \phi \]

Question \[ \text{If } \frac{\pi}{2} < x < \pi,\ \text{then} \] \[ \sqrt{\frac{1-\sin x}{1+\sin x}} + \sqrt{\frac{1+\sin x}{1-\sin x}} \] (a) \(2\sec x\) (b) \(-2\sec x\) (c) \(\sec x\) (d) \(-\sec x\) Solution \[ = \frac{1-\sin x}{\sqrt{1-\sin^2 x}} + \frac{1+\sin x}{\sqrt{1-\sin^2 x}} \] \[ = \frac{1-\sin x}{|\cos x|} + \frac{1+\sin x}{|\cos x|} \] \[ =

Question \[ \text{If } 0

If \( \pi < x < 2\pi \), then \( \sqrt{\frac{1+\cos x}{1-\cos x}} \) is equal to (a) \( \cosec x+\cot x \) (b) \( \cosec x-\cot x \) (c) \( -\cosec x+\cot x \) (d) \( -\cosec x-\cot x \) Solution: \[ \sqrt{\frac{1+\cos x}{1-\cos x}} = \sqrt{\frac{(1+\cos x)^2}{1-\cos^2 x}} \] \[ = \sqrt{\frac{(1+\cos x)^2}{\sin^2

If \( \frac{\pi}{2} < x < \frac{3\pi}{2} \), then \( \sqrt{\frac{1-\sin x}{1+\sin x}} \) is equal to (a) \( \sec x-\tan x \) (b) \( \sec x+\tan x \) (c) \( \tan x-\sec x \) (d) none of these Solution: \[ \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}} \] \[ = \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}} \]

If \( \sec x = x + \frac{1}{4x} \), then \( \sec x + \tan x = \) (a) \( x,\ \frac{1}{x} \) (b) \( 2x,\ \frac{1}{2x} \) (c) \( -2x,\ \frac{1}{2x} \) (d) \( -\frac{1}{x},\ x \) Solution: \[ \sec x=x+\frac{1}{4x} =\frac{4x^2+1}{4x} \] Using identity, \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] Now, \[

If \( \tan x = x – \frac{1}{4x} \), then \( \sec x – \tan x \) is equal to (a) \( -2x,\ \frac{1}{2x} \) (b) \( -\frac{1}{2x},\ 2x \) (c) \( 2x \) (d) \( 2x,\ \frac{1}{2x} \) Solution: \[ \tan x=x-\frac{1}{4x} =\frac{4x^2-1}{4x} \] \[ =\frac{\left(2x-\frac{1}{2x}\right)}{2} \] Using identity, \[ \sec x+\tan x=2x \]

Question Prove that : \[ \tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0 \] Solution Reducing the angles, \[ \tan\frac{5\pi}{4} = \tan\frac{\pi}{4} = 1 \] \[ \cot\frac{9\pi}{4} = \cot\frac{\pi}{4} = 1 \] \[ \tan\frac{17\pi}{4} = \tan\frac{\pi}{4} = 1 \] \[ \cot\frac{15\pi}{4} = \cot\frac{7\pi}{4} = -1 \] Substituting these values, \[ \begin{aligned} &\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} \\[8pt] =& (1)(1)+(1)(-1) \\[8pt]