Ravi Kant Kumar

Question \[ \text{Which of the following is incorrect?} \] (a) \(\sin\theta=-\frac15\) (b) \(\cos\theta=1\) (c) \(\sec\theta=\frac12\) (d) \(\tan\theta=20\) Solution We know that \[ -1\le\sin\theta\le1 \] \[ -1\le\cos\theta\le1 \] So options (a) and (b) are possible. Also, \[ \tan\theta \] can take any real value, so option (d) is possible. But \[ \sec\theta=\frac1{\cos\theta} \] Since \[ -1\le\cos\theta\le1

Question \[ \text{If } \sin\theta+\cosec\theta=2, \] \[ \text{then } \sin^2\theta+\cosec^2\theta \] is equal to (a) \(1\) (b) \(4\) (c) \(2\) (d) none of these Solution Let \[ x=\sin\theta \] Then \[ x+\frac1x=2 \] Squaring both sides, \[ \left(x+\frac1x\right)^2=4 \] \[ x^2+\frac1{x^2}+2=4 \] \[ x^2+\frac1{x^2}=2 \] Therefore, \[ \sin^2\theta+\cosec^2\theta=2 \] Answer \[ \boxed{2} \] Correct Option:

Question \[ \text{If } \sin\theta \text{ and } \cos\theta \] \[ \text{are the roots of the equation } ax^2-bx+c=0, \] \[ \text{then } a,b \text{ and } c \text{ satisfy the relation} \] (a) \(a^2+b^2+2ac=0\) (b) \(a^2-b^2+2ac=0\) (c) \(a^2+c^2+2ab=0\) (d) \(a^2-b^2-2ac=0\) Solution Since roots are \(\sin\theta\) and \(\cos\theta\), \[ \sin\theta+\cos\theta=\frac{b}{a} \] \[ \sin\theta\cos\theta=\frac{c}{a} \] Using

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}