Educational

Question \[ \text{If } \tan x+\cot x=4, \] \[ \text{then } \tan^4x+\cot^4x= \] Solution Given, \[ \tan x+\cot x=4 \] Squaring, \[ \tan^2x+\cot^2x+2\tan x\cot x=16 \] \[ \tan x\cot x=1 \] \[ \tan^2x+\cot^2x+2=16 \] \[ \tan^2x+\cot^2x=14 \] Again squaring, \[ (\tan^2x+\cot^2x)^2 = \tan^4x+\cot^4x+2\tan^2x\cot^2x \] \[ 14^2 = \tan^4x+\cot^4x+2 \] \[ 196 = \tan^4x+\cot^4x+2 \] Therefore, […]

Question \[ \text{If } \sec x=t+\frac1{4t}, \] \[ \text{then the value of } \sec x+\tan x \text{ is} \] Solution Let \[ \sec x+\tan x=y \] Using identity \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] \[ \sec x-\tan x=\frac1y \] Adding, \[ 2\sec x = y+\frac1y \] Given, \[ \sec x=t+\frac1{4t} \] \[ 2\left(t+\frac1{4t}\right) =

Question \[ \text{If } \sin x=\frac{2t}{1+t^2} \] \[ \text{and } x \text{ lies in the second quadrant,} \] \[ \text{then } \cos x= \] Solution Using identity \[ \sin^2x+\cos^2x=1 \] \[ \cos^2x = 1-\left(\frac{2t}{1+t^2}\right)^2 \] \[ = \frac{(1+t^2)^2-4t^2}{(1+t^2)^2} \] \[ = \frac{1-2t^2+t^4}{(1+t^2)^2} \] \[ = \frac{(1-t^2)^2}{(1+t^2)^2} \] \[ \cos x = \pm\frac{1-t^2}{1+t^2} \] Since \(x\)

Question \[ \text{If } \cos^2x+\sin x+1=0, \] \[ 0<x<2\pi, \] \[ \text{then } x= \] Solution Using identity \[ \cos^2x=1-\sin^2x \] Substituting, \[ 1-\sin^2x+\sin x+1=0 \] \[ -\sin^2x+\sin x+2=0 \] \[ \sin^2x-\sin x-2=0 \] \[ (\sin x-2)(\sin x+1)=0 \] Since \[ \sin x\neq2 \] therefore, \[ \sin x=-1 \] Now, \[ x=\frac{3\pi}{2} \] Answer \[

Question \[ \text{If } \sec x=m \text{ and } \tan x=n, \] \[ \frac1m\left\{(m+n)+\frac1{m+n}\right\} \] \[ \text{is equal to} \] Solution Given, \[ m=\sec x,\quad n=\tan x \] So, \[ m+n=\sec x+\tan x \] Using identity \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] \[ \frac1{m+n} = \sec x-\tan x \] Therefore, \[ (m+n)+\frac1{m+n} \] \[

Question \[ 9\tan^2\theta+4\cot^2\theta \] \[ \text{Find the minimum value} \] Solution Let \[ \tan^2\theta=x \] Then \[ \cot^2\theta=\frac1x \] So, \[ 9\tan^2\theta+4\cot^2\theta = 9x+\frac4x \] Using AM ≥ GM, \[ 9x+\frac4x \ge 2\sqrt{9x\cdot\frac4x} \] \[ = 2\sqrt{36} =12 \] Hence minimum value is \[ 12 \] Answer \[ \boxed{12} \] Next Question / Full Exercise

Question \[ \text{If } \pi<x<2\pi, \] \[ \sqrt{(1+\cos x)(1-\cos x)} + \sqrt{(1-\cos x)(1+\cos x)} = k\cosec x \] \[ \text{then } k= \] Solution Using identity \[ (1+\cos x)(1-\cos x)=1-\cos^2x \] \[ =\sin^2x \] Therefore, \[ \sqrt{\sin^2x}+\sqrt{\sin^2x} = |\sin x|+|\sin x| \] \[ =2|\sin x| \] Since \[ \pi<x<2\pi \] \(x\) lies in III or

Question \[ \cos1^\circ \cos2^\circ \cos3^\circ \cdots \cos179^\circ \] \[ \text{is equal to} \] Solution Observe that the product contains \[ \cos90^\circ \] But \[ \cos90^\circ=0 \] Therefore, \[ \cos1^\circ \cos2^\circ \cos3^\circ \cdots \cos179^\circ=0 \] Answer \[ \boxed{0} \] Next Question / Full Exercise

Question \[ \text{If } \frac{\pi}{2}<x<\pi, \] \[ \sqrt{(1+\sin x)(1-\sin x)} = k\sec x \] \[ \text{then } k= \] Solution Using identity \[ (1+\sin x)(1-\sin x)=1-\sin^2x \] \[ =\cos^2x \] Therefore, \[ \sqrt{(1+\sin x)(1-\sin x)} = \sqrt{\cos^2x} = |\cos x| \] Since \[ \frac{\pi}{2}<x<\pi \] \(x\) lies in second quadrant, so \[ \cos x<0 \]

Question \[ \text{If } \frac{\pi}{2}<x<\pi, \] \[ \sqrt{(1+\sin x)(1-\sin x)} + \sqrt{(1-\sin x)(1+\sin x)} = k\sec x \] \[ \text{then } k= \] Solution Using identity \[ (1+\sin x)(1-\sin x)=1-\sin^2x \] \[ =\cos^2x \] Therefore, \[ \sqrt{\cos^2x}+\sqrt{\cos^2x} = |\cos x|+|\cos x| \] \[ =2|\cos x| \] Since \[ \frac{\pi}{2}<x<\pi \] \(x\) lies in second quadrant,