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,

Question \[ \tan1^\circ \tan2^\circ \tan3^\circ \cdots \tan89^\circ \] \[ \text{is equal to} \] Solution Using identity \[ \tan\theta \tan(90^\circ-\theta)=1 \] \[ \tan1^\circ \tan89^\circ=1 \] \[ \tan2^\circ \tan88^\circ=1 \] \[ \tan3^\circ \tan87^\circ=1 \] Similarly all pairs are equal to \(1\). Also, \[ \tan45^\circ=1 \] Therefore, \[ \tan1^\circ \tan2^\circ \tan3^\circ \cdots \tan89^\circ=1 \] Answer \[ \boxed{1} \]

Question \[ \text{If } \sin x+\cos x=a, \] \[ \text{then } \sin^6x+\cos^6x= \] Solution Squaring, \[ (\sin x+\cos x)^2=a^2 \] \[ 1+2\sin x\cos x=a^2 \] \[ \sin x\cos x=\frac{a^2-1}{2} \] Now, \[ \sin^6x+\cos^6x \] \[ =(\sin^2x+\cos^2x)^3 -3\sin^2x\cos^2x(\sin^2x+\cos^2x) \] \[ =1-3\sin^2x\cos^2x \] \[ =1-3\left(\frac{a^2-1}{2}\right)^2 \] \[ =1-\frac34(a^2-1)^2 \] Answer \[ \boxed{1-\frac34(a^2-1)^2} \] Next Question / Full

Question \[ \text{Given } x>0, \] \[ f(x)=-3\cos\sqrt{3+x+x^2} \] \[ \text{lies in the interval} \] Solution We know that \[ -1\le\cos\theta\le1 \] Multiplying by \(-3\), \[ -3\le-3\cos\theta\le3 \] Now, \[ 3+x+x^2>0 \] for all \(x>0\). Hence \[ \sqrt{3+x+x^2} \] is always real. Therefore, \[ -3\le f(x)\le3 \] Answer \[ \boxed{[-3,\,3]} \] Next Question / Full

Question \[ 3(\sin x-\cos x)^4 +6(\sin x+\cos x)^2 +4(\sin^6x+\cos^6x) \] Solution Using identities \[ (\sin x-\cos x)^2 = 1-2\sin x\cos x \] \[ (\sin x+\cos x)^2 = 1+2\sin x\cos x \] Let \[ \sin x\cos x=t \] Then \[ 3(1-2t)^2+6(1+2t) +4(\sin^6x+\cos^6x) \] Now, \[ \sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 -3\sin^2x\cos^2x(\sin^2x+\cos^2x) \] \[ =1-3t^2 \] Substituting, \[ 3(1-4t+4t^2)+6+12t+4(1-3t^2)

Question \[ \tan x+\cot(\pi+x)+\cot\left(\frac{\pi}{2}+x\right)+\cot(2\pi-x) \] \[ \text{is equal to} \] Solution Using identities \[ \cot(\pi+x)=\cot x \] \[ \cot\left(\frac{\pi}{2}+x\right)=-\tan x \] \[ \cot(2\pi-x)=-\cot x \] Substituting, \[ \tan x+\cot x-\tan x-\cot x \] \[ =0 \] Answer \[ \boxed{0} \] Next Question / Full Exercise

Question \[ \text{If } \sin x+\cosec x=2, \] \[ \text{then } \sin^2x+\cosec^2x= \] Solution Let \[ a=\sin x \] Then \[ a+\frac1a=2 \] Squaring both sides, \[ \left(a+\frac1a\right)^2=4 \] \[ a^2+\frac1{a^2}+2=4 \] Therefore, \[ a^2+\frac1{a^2}=2 \] \[ \sin^2x+\cosec^2x=2 \] Answer \[ \boxed{2} \] Next Question / Full Exercise

Question \[ \text{If } \sin x=-\frac{24}{25}, \] \[ \text{then the value of } \tan x \text{ is} \] Solution Using identity \[ \sin^2x+\cos^2x=1 \] \[ \left(-\frac{24}{25}\right)^2+\cos^2x=1 \] \[ \frac{576}{625}+\cos^2x=1 \] \[ \cos^2x=\frac{49}{625} \] \[ \cos x=\pm\frac{7}{25} \] Now, \[ \tan x=\frac{\sin x}{\cos x} \] \[ =\frac{-24/25}{\pm7/25} \] \[ \tan x=\pm\frac{24}{7} \] Since quadrant is not

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}