Question \[ \text{If } \cosec x+\cot x=\frac{11}{2}, \] \[ \text{then the value of } \tan x \text{ is} \] Solution Using identity \[ (\cosec x+\cot x)(\cosec x-\cot x)=1 \] \[ \cosec x-\cot x=\frac{2}{11} \] Subtracting, \[ 2\cot x = \frac{11}{2}-\frac{2}{11} \] \[ = \frac{121-4}{22} = \frac{117}{22} \] \[ \cot x=\frac{117}{44} \] Therefore, \[ \tan x=\frac{44}{117}

Question \[ \text{If } \cosec x+\cot x=\alpha, \] \[ \text{then } \sin x=\ ? \] Solution Using identity \[ (\cosec x+\cot x)(\cosec x-\cot x)=1 \] \[ \cosec x-\cot x=\frac1\alpha \] Adding, \[ 2\cosec x = \alpha+\frac1\alpha \] \[ \cosec x = \frac{\alpha^2+1}{2\alpha} \] Therefore, \[ \sin x = \frac{2\alpha}{\alpha^2+1} \] Answer \[ \boxed{\frac{2\alpha}{\alpha^2+1}} \] Next

Question \[ \text{If } \sec x-\tan x=\frac23, \] \[ \text{then } \tan x=\ ? \] Solution Using identity \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] \[ (\sec x+\tan x)\times\frac23=1 \] \[ \sec x+\tan x=\frac32 \] Now, \[ 2\tan x = (\sec x+\tan x)-(\sec x-\tan x) \] \[ =\frac32-\frac23 \] \[ =\frac{9-4}{6} =\frac56 \] Therefore, \[

Question \[ \text{If } \sec x+\tan x=\sqrt3, \] \[ \text{then } \sec x-\tan x=\ ? \] Solution Using identity \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] Substituting \[ \sqrt3(\sec x-\tan x)=1 \] Therefore, \[ \sec x-\tan x=\frac1{\sqrt3} \] Answer \[ \boxed{\frac1{\sqrt3}} \] Next Question / Full Exercise

Question \[ \text{If } -\frac{\pi}{2}<x<\frac{\pi}{2}, \] \[ \sqrt{(1-\sin x)(1+\sin x)} \] \[ \text{is equal to} \] 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 \[ -\frac{\pi}{2}<x<\frac{\pi}{2} \] \(\cos x\) is positive in this interval. Hence, \[ |\cos x|=\cos x \]

Question \[ \text{If } \sin x+\cos x=a, \] \[ \text{then } \sin x-\cos x=\ ? \] Solution Given, \[ \sin x+\cos x=a \] Squaring both sides, \[ (\sin x+\cos x)^2=a^2 \] \[ \sin^2x+\cos^2x+2\sin x\cos x=a^2 \] \[ 1+2\sin x\cos x=a^2 \] \[ 2\sin x\cos x=a^2-1 \] Now, \[ (\sin x-\cos x)^2 = \sin^2x+\cos^2x-2\sin x\cos x

Question \[ f(x)=2\sin\sqrt{x^2+x+1} \] \[ \text{The values of } f(x) \text{ lie in the interval} \] Solution Since \[ -1\le\sin\theta\le1 \] therefore, \[ -2\le2\sin\theta\le2 \] Now, \[ x^2+x+1 = \left(x+\frac12\right)^2+\frac34 \] \[ x^2+x+1>0 \] So \[ \sqrt{x^2+x+1} \] is always real. Hence, \[ -2\le f(x)\le2 \] Answer \[ \boxed{[-2,\,2]} \] Next Question / Full Exercise

Question \[ \frac{\cos50^\circ}{\cos130^\circ} \] Solution Using identity \[ \cos(180^\circ-\theta)=-\cos\theta \] \[ \cos130^\circ = \cos(180^\circ-50^\circ) \] \[ =-\cos50^\circ \] Therefore, \[ \frac{\cos50^\circ}{\cos130^\circ} = \frac{\cos50^\circ}{-\cos50^\circ} =-1 \] Answer \[ \boxed{-1} \] Next Question / Full Exercise

Question \[ \frac{\sin70^\circ}{\sin110^\circ} \] Solution Using identity \[ \sin(180^\circ-\theta)=\sin\theta \] \[ \sin110^\circ = \sin(180^\circ-70^\circ) \] \[ =\sin70^\circ \] Therefore, \[ \frac{\sin70^\circ}{\sin110^\circ} = \frac{\sin70^\circ}{\sin70^\circ} =1 \] Answer \[ \boxed{1} \] Next Question / Full Exercise

Question \[ \text{If for real values of } x, \] \[ \cos\theta=x+\frac1x, \] \[ \text{then} \] (a) \(\theta\) is an acute angle (b) \(\theta\) is a right angle (c) \(\theta\) is an obtuse angle (d) No value of \(\theta\) is possible Solution For real \(x\), \[ x+\frac1x\ge2 \quad \text{or} \quad x+\frac1x\le-2 \] But \[ -1\le\cos\theta\le1