Educational

Question \[ x\sin45^\circ\cos^260^\circ = \frac{\tan^260^\circ\cosec30^\circ} {\sec45^\circ\cot^230^\circ} \] then \(x=\) (a) \(2\) (b) \(4\) (c) \(8\) (d) \(16\) Solution Using standard values, \[ \sin45^\circ=\frac1{\sqrt2}, \quad \cos60^\circ=\frac12 \] \[ \tan60^\circ=\sqrt3, \quad \cosec30^\circ=2 \] \[ \sec45^\circ=\sqrt2, \quad \cot30^\circ=\sqrt3 \] Substituting, \[ x\left(\frac1{\sqrt2}\right)\left(\frac12\right)^2 = \frac{(\sqrt3)^2\times2} {\sqrt2\times(\sqrt3)^2} \] \[ x\left(\frac1{4\sqrt2}\right) = \frac{3\times2}{\sqrt2\times3} \] \[ x\left(\frac1{4\sqrt2}\right) = \frac2{\sqrt2} \] \[ x […]

Question \[ \text{If } \tan A+\cot A=4, \] \[ \text{then } \tan^4A+\cot^4A \text{ is equal to} \] (a) \(110\) (b) \(191\) (c) \(80\) (d) \(194\) Solution Since \[ \tan A\cdot \cot A=1 \] \[ (\tan A+\cot A)^2 = \tan^2A+\cot^2A+2 \] \[ 4^2 = \tan^2A+\cot^2A+2 \] \[ \tan^2A+\cot^2A=14 \] Now, \[ (\tan^2A+\cot^2A)^2 = \tan^4A+\cot^4A+2\tan^2A\cot^2A \] \[

Question \[ \sin^2\frac{\pi}{18} + \sin^2\frac{2\pi}{9} + \sin^2\frac{7\pi}{18} + \sin^2\frac{4\pi}{9} \] is equal to (a) \(1\) (b) \(4\) (c) \(2\) (d) \(0\) Solution Using identity \[ \sin^2\theta+\sin^2\left(\frac{\pi}{2}-\theta\right)=1 \] \[ \sin^2\frac{\pi}{18} + \sin^2\frac{4\pi}{9} =1 \] because \[ \frac{\pi}{18}+\frac{4\pi}{9} = \frac{\pi}{2} \] \[ \sin^2\frac{2\pi}{9} + \sin^2\frac{7\pi}{18} =1 \] because \[ \frac{2\pi}{9}+\frac{7\pi}{18} = \frac{\pi}{2} \] Therefore, \[ 1+1=2 \]

Question \[ \sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\cdots+\sin^285^\circ+\sin^290^\circ \] is equal to (a) \(7\) (b) \(8\) (c) \(9.5\) (d) \(10\) Solution Using identity \[ \sin^2\theta+\sin^2(90^\circ-\theta)=1 \] Pair the terms: \[ \sin^25^\circ+\sin^285^\circ=1 \] \[ \sin^210^\circ+\sin^280^\circ=1 \] \[ \sin^215^\circ+\sin^275^\circ=1 \] Similarly, \[ 8 \text{ such pairs } =8 \] Middle term: \[ \sin^245^\circ=\frac12 \] Last term: \[ \sin^290^\circ=1 \] Therefore, \[ 8+\frac12+1

Question \[ \text{If } x \text{ is an acute angle and } \tan x=\frac{1}{\sqrt7}, \] \[ \text{then the value of } \frac{\cosec^2x-\sec^2x}{\cosec^2x+\sec^2x} \] is (a) \(\frac34\) (b) \(\frac12\) (c) \(2\) (d) \(\frac54\) Solution \[ \tan x=\frac{1}{\sqrt7} \] Take \[ \text{Perpendicular}=1,\quad \text{Base}=\sqrt7 \] \[ \text{Hypotenuse} = \sqrt{1+7} = 2\sqrt2 \] \[ \sin x=\frac{1}{2\sqrt2} \Rightarrow \cosec^2x=8 \]

Question \[ \sec^2x=\frac{4xy}{(x+y)^2} \] \[ \text{is true if and only if} \] (a) \(x+y\neq0\) (b) \(x=y,\ x\neq0\) (c) \(x=y\) (d) \(x\neq0,\ y\neq0\) Solution Since \[ \sec^2x\ge1 \] So, \[ \frac{4xy}{(x+y)^2}\ge1 \] \[ 4xy\ge(x+y)^2 \] \[ 4xy\ge x^2+2xy+y^2 \] \[ 0\ge x^2-2xy+y^2 \] \[ 0\ge(x-y)^2 \] But \[ (x-y)^2\ge0 \] Hence, \[ (x-y)^2=0 \] \[ x=y

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) \(\frac{117}{44}\) 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 \[ \text{If } \cosec x-\cot x=\frac12,\ 0<x<\frac{\pi}{2}, \] \[ \text{then } \cos x \text{ is equal to} \] (a) \(\frac53\) (b) \(\frac35\) (c) \(-\frac35\) (d) \(-\frac53\) Solution Using identity \[ (\cosec x-\cot x)(\cosec x+\cot x)=1 \] \[ \frac12(\cosec x+\cot x)=1 \] \[ \cosec x+\cot x=2 \] Now, \[ 2\cosec x = (\cosec x+\cot x)+(\cosec

Question \[ \sin^6A+\cos^6A+3\sin^2A\cos^2A= \] (a) \(0\) (b) \(1\) (c) \(2\) (d) \(3\) Solution Using identity \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Take \[ a=\sin^2A,\quad b=\cos^2A \] \[ \sin^6A+\cos^6A = (\sin^2A+\cos^2A)^3 -3\sin^2A\cos^2A(\sin^2A+\cos^2A) \] \[ = 1^3-3\sin^2A\cos^2A(1) \] \[ = 1-3\sin^2A\cos^2A \] Therefore, \[ \sin^6A+\cos^6A+3\sin^2A\cos^2A \] \[ =1 \] Answer \[ \boxed{1} \] Correct Option: (b) Next Question / Full

Question \[ \text{If } \frac{3\pi}{4}