Educational

Question \[ \text{If } \tan\theta=-\frac43, \] \[ \text{then } \sin\theta \text{ is equal to} \] (a) \(-\frac45\) but not \(\frac45\) (b) \(-\frac45\) or \(\frac45\) (c) \(\frac45\) but not \(-\frac45\) (d) none of these Solution \[ \tan\theta=\frac{\text{Perpendicular}}{\text{Base}} =-\frac43 \] Take \[ \text{Perpendicular}=4, \quad \text{Base}=-3 \] or \[ \text{Perpendicular}=-4, \quad \text{Base}=3 \] \[ \text{Hypotenuse} = \sqrt{4^2+3^2} =5 […]

Question \[ \text{Which of the following is correct?} \] (a) \(\sin1^\circ>\sin1\) (b) \(\sin1^\circ

Question \[ \tan1^\circ \tan2^\circ \tan3^\circ \cdots \tan89^\circ \] is equal to (a) \(0\) (b) \(1\) (c) \(\frac12\) (d) not defined 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

Question \[ \cos1^\circ \cos2^\circ \cos3^\circ \cdots \cos179^\circ \] is equal to (a) \(\frac1{\sqrt2}\) (b) \(0\) (c) \(1\) (d) \(-1\) Solution Observe that the product contains \[ \cos90^\circ \] But \[ \cos90^\circ=0 \] Therefore, \[ \cos1^\circ \cos2^\circ \cos3^\circ \cdots \cos179^\circ=0 \] Answer \[ \boxed{0} \] Correct Option: (b) Next Question / Full Exercise

Question \[ \text{Which of the following is incorrect?} \] (a) \(\sin x=-\frac15\) (b) \(\cos x=1\) (c) \(\sec x=\frac12\) (d) \(\tan x=20\) Solution We know that \[ -1\le\sin x\le1 \] \[ -1\le\cos x\le1 \] So options (a) and (b) are possible. Also, \[ \tan x \] can take any real value, so option (d) is possible.

Question \[ \text{If } f(x)=\cos^2x+\sec^2x, \] \[ \text{then} \] (a) \(f(x)<1\) (b) \(f(x)=1\) (c) \(1<f(x)<2\) (d) \(f(x)\ge2\) Solution Let \[ a=\cos^2x \] Then \[ \sec^2x=\frac1a \] So, \[ f(x)=a+\frac1a \] Using AM ≥ GM, \[ a+\frac1a\ge2 \] Therefore, \[ f(x)\ge2 \] Answer \[ \boxed{f(x)\ge2} \] Correct Option: (d) Next Question / Full Exercise

Question \[ \text{If } \sec x+\tan x=k, \] \[ \text{then } \cos x= \] (a) \(\dfrac{k^2+1}{2k}\) (b) \(\dfrac{2k}{k^2+1}\) (c) \(\dfrac{k}{k^2+1}\) (d) \(\dfrac{k}{k^2-1}\) Solution Using identity \[ (\sec x+\tan x)(\sec x-\tan x)=1 \] \[ \sec x-\tan x=\frac1k \] Adding, \[ 2\sec x = k+\frac1k \] \[ \sec x = \frac{k^2+1}{2k} \] \[ \cos x = \frac{2k}{k^2+1}

Question \[ \text{If } \tan\theta+\sec\theta=e^x, \] \[ \text{then } \cos\theta \text{ equals} \] (a) \(\dfrac{e^x+e^{-x}}{2}\) (b) \(\dfrac{2}{e^x+e^{-x}}\) (c) \(\dfrac{e^x-e^{-x}}{2}\) (d) \(\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\) Solution Using identity \[ (\sec\theta+\tan\theta) (\sec\theta-\tan\theta)=1 \] \[ \sec\theta-\tan\theta=e^{-x} \] Adding, \[ 2\sec\theta = e^x+e^{-x} \] \[ \sec\theta = \frac{e^x+e^{-x}}{2} \] \[ \cos\theta = \frac{2}{e^x+e^{-x}} \] Answer \[ \boxed{\frac{2}{e^x+e^{-x}}} \] Correct Option: (b) Next

Question \[ \text{If } \cosec x+\cot x=\frac{11}{2}, \] \[ \text{then } \tan x= \] (a) \(\frac{21}{22}\) (b) \(\frac{15}{16}\) (c) \(\frac{44}{117}\) (d) \(\frac{117}{43}\) Solution Using identity \[ (\cosec x+\cot x)(\cosec x-\cot x)=1 \] \[ \frac{11}{2}(\cosec x-\cot x)=1 \] \[ \cosec x-\cot x=\frac{2}{11} \] Now, \[ 2\cot x = (\cosec x+\cot x)-(\cosec x-\cot x) \] \[ =\frac{11}{2}-\frac{2}{11}

Question \[ \text{If } A \text{ lies in second quadrant and } 3\tan A+4=0, \] \[ \text{then the value of } 2\cot A-5\cos A+\sin A \] is equal to (a) \(-\frac{53}{10}\) (b) \(\frac{23}{10}\) (c) \(\frac{37}{10}\) (d) \(\frac{7}{10}\) Solution \[ 3\tan A+4=0 \] \[ \tan A=-\frac43 \] Since \(A\) lies in second quadrant, \[ \sin A>0,\quad