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#### The value of    is : Option: 1   Option: 2   Option: 3   Option: 4

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

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Allied Angles (Part 1) -

Allied Angles (Part 1)

Two angles or numbers are called allied iff their sum or difference is a multiple of π/2

• sin (900 - θ) = cos (θ)

• cos (900 - θ) = sin (θ)

• tan (900 - θ) = cot (θ)

• csc (900 - θ) = sec (θ)

• sec (900 - θ) = csc (θ)

• cot (900 - θ) = tan (θ)

• sin (900 + θ) = cos (θ)

• cos (900 + θ) = - sin (θ)

• tan (900 + θ) = - cot (θ)

• csc (900 + θ) = sec (θ)

• sec (900 + θ) = - csc (θ)

• cot (900 + θ) = - tan (θ)

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#### If  and   then  is equal to ________. Option: 1   Option: 2  Option: 3  Option: 4

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

-

Double Angle Formula and Reduction Formula -

Double Angle Formula and Reduction Formula

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Correct Option (2)

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#### Two poles, of length a metres and  of length metres are erected at the same horizontal level with bases at . If and , then : Option: 1 Option: 2 Option: 3 Option: 4

$\angle ACB= \angle ACD-\angle BCD\\$

$\Rightarrow \tan\angle ACB=\tan\left ( \angle ACD-\angle BCD \right )\\$

$\tan\angle ACD= \frac{x}{b}, \tan\angle BCD= \frac{x}{a+b}\\$

$\Rightarrow \frac{1}{2}= \frac{\frac{x}{b}-\frac{x}{a+b}}{1+\frac{x}{b}\cdot \frac{x}{a+b}}\\$

$\Rightarrow \frac{1}{2}= \frac{\left ( a+b \right )x-bx}{b\left ( a+b \right )+x^{2}}\\$

$\Rightarrow x^{2}-2ax+b\left ( a+b \right )= 0$

#### If is the number of solutions of the equation and  is the sum of all these solutions, then the ordered pair isOption: 1Option: 2Option: 3Option: 4

$2\cos x\left [ 4\left ( \sin ^{2} \frac{\pi }{4}-\sin ^{2}x\right ) -1\right ]= 1$
$\Rightarrow 2\cos x\left [ 4\left ( \frac{1}{2} -\sin ^{2}x\right )-1\right ]= 1$
$\Rightarrow 2\cos x\left [2-4\sin ^{2}x-1\right ]= 1$
$\Rightarrow 2\cos x\left [1-4\sin ^{2}x\right ]= 1$
$\Rightarrow 2\cos x\left [1-4+4\cos ^{2}x\right ]= 1$
$\Rightarrow 2\cos x\left [4\cos ^{2}x-3\right ]= 1$
$\Rightarrow 2\left ( 4\cos ^{3}x-3\cos x \right )= 1$
$\Rightarrow \cos 3x= \frac{1}{2}$
$\Rightarrow 3x= \frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3}$
$\Rightarrow x= \frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$
$\Rightarrow n= 3,S= \frac{13\pi }{9}$

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#### The range of the function isOption: 1Option: 2Option: 3Option: 4

$f(x)=\log _{\sqrt{5}}\left ( 3+\cos \left ( \frac{3 \pi}{4} \right )\cos x-\sin \left ( \frac{3 \pi}{4} \right )\sin x+\cos \left ( \frac{ \pi}{4} \right )\cos x-\sin \left ( \frac{ \pi}{4} \right ) \cdot \sin x+\cos \left ( \frac{ \pi}{4} \right )\cos x+\sin \left ( \frac{ \pi}{4} \right )\cdot \sin x-\cos \left ( \frac{3 \pi}{4} \right )\cos x-\sin \left ( \frac{3 \pi}{4} \right )\sin x\right )$
$=\log _{\sqrt{5}}\left ( 3+2\cdot \cos \left ( \frac{\pi}{4} \right )\cos x-2\sin \left ( \frac{3\pi}{4} \right )\sin x \right )$
$=\log _{\sqrt{5}}\left ( 3+\sqrt{2}\left ( \cos x-\sin x \right ) \right )$
$Now\, -\sqrt{2}\leq \cos x-\sin x\leq \sqrt{2}$
$-2\leq \sqrt{2} \left ( \cos x-\sin x \right )\leq 2$
$1\leq3+ \sqrt{2} \left ( \cos x-\sin x \right )\leq 5$
$\log _{\sqrt{5}}\left ( 1 \right )\leq \log _{\sqrt{5}}\left ( 3+\sqrt{2} \left ( \cos x-\sin x \right )\right )\leq \log _{\sqrt{5}}\left ( 5 \right )$
$0\leq f\left ( x \right )\leq 2$

$\therefore Range\, is\, \left [ 0,2\right ]$

#### is equal to : (The inverse trigonometric functions take the principal values)Option: 1Option: 2Option: 3Option: 4

$\cos^{-1}\left ( \cos \left ( -5 \right ) \right )+\sin^{-1}\left ( \sin \left ( 6 \right ) \right )-\tan^{-1}\left ( \tan \left ( 12 \right ) \right )$$= \cos^{-1}\left ( \cos \left ( 5 \right ) \right )+\sin^{-1}\left ( \sin \left ( 6 \right ) \right )-\tan^{-1}\left ( \tan \left ( 12 \right ) \right )$

Using graphs of $y= T^{-1}\left ( T\left ( x \right ) \right )$

$= \left ( 2\pi -5 \right )+\left ( 6-2\pi \right )-\left ( 12-4\pi \right )$
$= 4\pi -11$

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#### The domain of the function is :Option: 1Option: 2Option: 3Option: 4

Domain

$-1\leq \frac{3x^{2}+x-1}{\left ( x-1 \right )^{2}}\leq 1;x\neq1\\$

$\Rightarrow -\left ( x-1 \right )^{2}\leq 3x^{2}+x-1\leq \left ( x-1 \right )^{2}\\$

$\Rightarrow 4x^{2}-x\geq 0\; and\; 2x^{2}+3x-2\leq 0\\$

$\Rightarrow 4x\left ( x-\frac{1}{4} \right )\geq 0\; and\; 2x^{2}+4x-x-2\leq 0\\$

$\Rightarrow x\in\left ( -\infty,0 \right ]\cup \left [\frac{1}{4},\infty \right )\; and\; x\in\left [ -2,\frac{1}{2} \right ]\\$

$So\; x\in\; \left [ -2,0 \right ]\cup \left [ \frac{1}{4},\frac{1}{2} \right ]\\$            ..........(1)

$Also\; -1\leq \frac{x-1}{x+1}\leq 1\Rightarrow \frac{x-1}{x+1}+1\geq 0\; and\; \frac{x-1}{x+1}-1\leq 0\\$

$\Rightarrow \frac{2x}{x+1}\geq 0\; and\; \frac{-2}{x+1}\leq 0\Rightarrow \frac{1}{x+1}\geq 0,x\neq -1\\$

$\Rightarrow x \in\left ( -\infty,-1 \right ]\cup \left [ 0,\infty \right )\; and \; x\in\left ( -1,\infty \right )\\$

$\Rightarrow x\in\left [ 0,\infty \right )$                          .......(2)

From (1) and (2)

$x\in \{0\} \cup \left [ \frac{1}{4},\frac{1}{2} \right ]$

#### The number of solutions of the equation is :Option: 1Option: 2Option: 3Option: 4

$32^{\tan^{2}x}+32^{1+\tan^{2}x}= 81\\$

$32^{\tan^{2}x}=t\\$

$\Rightarrow t+32t=81\\$

$\Rightarrow 33t=81\Rightarrow t=\frac{81}{33}\\$

$\Rightarrow 32^{\tan^{2}x}=\frac{81}{33}=\frac{27}{11}\\$

$\Rightarrow \tan^{2}x=\log_{32}\frac{27}{11}\in\left ( 0,1 \right )\\$

$\Rightarrow x=\tan^{-1}\sqrt{\log_{32}\frac{27}{11}}For \; 0\leq x\leq \pi/4\\$

$\Rightarrow only\; 1 \; solution$

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#### A vertical pole fixed to the horizontal ground is divided in the ratio by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground away from the base of the pole, then the height of the pole (in meters) is :Option: 1Option: 2Option: 3Option: 4

$In\: \triangle ABC ;\tan \theta = \frac{3x}{18}= \frac{x}{6}---\left ( 1 \right )$
$In\: \triangle ABD ;\tan 2\theta = \frac{10x}{18}= \frac{5x}{9}$
$\Rightarrow \frac{2\tan \theta }{1-\tan ^{2}\theta }= \frac{5x}{9}$
$\Rightarrow \frac{2\times \frac{x}{6}}{1-\frac{x^{2}}{36}}= \frac{5x}{9}$
$\Rightarrow 108= 180-5x^{2}\Rightarrow 5x^{2}= 72$
$\Rightarrow x = 6\sqrt{\frac{2}{5}}\Rightarrow 10x = 6\sqrt{\frac{2}{5}}\times 10= 12\sqrt{10}$
option (3)

#### is a root of the equation :Option: 1Option: 2Option: 3Option: 4

$\operatorname{cosec} 18^{\circ}= \frac{1}{\sin 18^{\circ}}= \frac{4}{\sqrt{5}-1}= \sqrt{5}+1= \alpha$
$Let\, \beta = 1-\sqrt{5}$
$\alpha +\beta = 2 ;\alpha \, \beta = -4$
$\Rightarrow x^{2}-2x-4= 0$
option (3)