Length of arc = Circumference of wire of radius = 3 cm
\(\therefore l=2\pi r=2\pi\times3\ cm=6\pi cm\)
The radius of the sector (r) = 48cm
\(\theta=\frac{l}{r}=\frac{6\pi}{48}=\frac{\pi}{8}\ radians\)
\(\therefore\) Angle in degrees which is subtended at the centre of the sector
\(=(\frac{\pi}{8}\times\frac{180^°}{\pi})^°=(\frac{45}{2})^°=22°30'\)

Let log₃x = y
Then x = 3^y and so, \({ 3 }^{ { y }^{ 2 } }\)= 9.
Thus, y² = 2, which implies y =\(\sqrt{2},-\sqrt{2}\). Hence, x = \({ 3 }^{ \sqrt { 2 } },{ 3 }^{ -\sqrt { 2 } }\)

¹⁰C₂ + 2 x ¹⁰C₃ + ¹⁰C₄ = ¹⁰C₂ + (¹⁰C₃ + ¹⁰C₃) + ¹⁰C₄
= (¹⁰C₂ + ¹⁰C₃) + (¹⁰C₃ + ¹⁰C₄)
= ¹¹C₃ + ¹¹C₄
= ¹²C₄.

Let \(\frac{x}{(x+3)(x-4)}=\frac{A}{x+3}+\frac{B}{x-4}\) where A and B are constants.
Then, \(\frac{x}{(x+3)(x-4)}=\frac{A(x-4)+B(x+3)}{(x+3)(x-4)}\) which gives x = A(x - 4) + B(x + 3)
When x = 4, we have B = \(\frac{4}{7}\)
When x = -3 we have A = \(\frac{3}{7}\)
Hence, \(\frac{x}{(x+3)(x-4)}=\frac{3}{7(x+3)}+\frac{4}{7(x-4)}\)

LHS = tan (360° - 45°) [-cot (360° + 45°)] + cot (360° + 135°) [-tan (360° + 225°)]
= [-tan 45°] [-cot 45°] + [- tan 45°] [-tan 45°]
= (-1) (-1) + (-1) (-1) = 2

sin 765° = sin (2 x 360° + 45°)
= sin 45⁰ = \(\frac{1}{\sqrt 2}\)

Observe that \(\sqrt{x+20}\) is defined only if x + 20 ≥ 0
By definition, \(\sqrt{x+20}\ge0\). So, x is positive.
Now squaring we get x²= x + 20. x²- x - 20 = 0
(x - 5)(x + 4) = 0, which gives x = 5, x = -4
Since, x is positive, the required solution is x = 5

Note that 3x - 9 ≥ 0 implies 3x ≥ 9, by multiplying both sides by \(\frac{1}{3}\) we get x ≥ 3.
Similarly, 4x - 10 ≤ 6 implies 4x ≤ 16 and hence x ≤ 4
So the solution set of 3x - 9 ≥ 0, 4x -10 ≤ 6 is the intersection of [3, ∞) and (-∞, 4]. Clearly, the intersection of these intervals give [3, 4].

Each of the 10 bulbs are operated independently means that each bulb can be operated in two ways. That is in off mode or on mode. The total number of doing this are 2¹⁰ which includes the case in which 10 bulbs are off. Keeping all 10 bulbs in "off" mode, the room cannot be illuminated. Hence, the total number of ways are 2¹⁰ - 1 = 1024 - 1 = 1023.

The following figure illustrates the electricity distribution network.

There are 110 smaller capacity transformer attached to a larger capacity transformer. As each smaller capacity transformer can be linked with only 100 consumers, we have for the 109 transformers, there will be 109 \(\times\)100 = 10900 links. For the 110^th transformer, there are only 29 consumers linked. Hence, the total number of consumer linked to the 238^th larger capacity transformer is 10900 + 29 = 10929.

(i) a particular teacher is included?
There are 5 teachers and 20 students 2 teachers out of 5 teachers can be selected in 5C₂ ways.
3 students out of 20 students can be selected in 20C₃ ways
Hence, total number of committees = 20C₃ \(\times \) 5C₂
= \(\frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= \(10\times 19\times 6\times 5\times 2\)
= 11400
Since a particular teacher is included, the committee will have 1 teacher and 3 students.
∴ 1 teacher can be selected from 4'teachers in 4C₁ = 4 ways.
3 students out of 20 students can be selected in 20C₃ ways.
Hence, required number of committees
= 4C₁ \(\times \) 20C₃
= \(4\times \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \)
= 4560
(ii) 2 teachers can be selected from 5 teachers in 5C₂ ways
Since a particular student is excluded, 3 students can be selected from 19 students in 19C₃ ways
Hence required number of committees = 19C₃ \(\times \) 5C₂
=\(\frac { 19\times 18\times 17 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= 19 \(\times \)6\(\times \)17\(\times \)5
= 9690

RHS = \(\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4cotA } \)
\(=\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4\frac { cosA }{ SinA } } =\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4cosA } \times sinA=\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4\left( \frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 2bc } \right) } sina\)
= \(\frac { sinA }{ 4 } \times 2bc=\frac { 1 }{ 2 } bcsinA=area\quad of\quad ABC\)
∴ △ = \(\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4cotA } \)

by Napier's formula we have
\(\left( \frac { A-B }{ 2 } \right) = \frac { a-b }{ a+b } cot\frac { C }{ 2 } =\frac { (\sqrt { 3 } +1)-(\sqrt { 3 } -1) }{ \sqrt { 3 } +1)+(\sqrt { 3 } -1) } cot\frac { 60 }{ 2 } \)
\(\frac { 2 }{ 2\sqrt { 3 } } cot30=\frac { 1 }{ \sqrt { 3 } } x\sqrt { 3 } =1\)
\(\frac { A-B }{ 2 } =45\quad A-B=90\)
Also A+ B = 180⁰ - C = 180 - 60⁰ = 120⁰
Adding (1) and (2), 2A = 210⁰  Substituting A = 105⁰ in (2) we get
105⁰ + B = 120⁰ B = 120 - 105 = 15⁰
Now by sine Formula
\(\frac { c }{ sinC } =\frac { a }{ sinA } C=a\frac { sinC }{ SinA } \)
\(=\frac { (\sqrt { 3 } +1)sin60 }{ sin105 } =\frac { (\sqrt { 3 } +1)\frac { \sqrt { 3 } }{ 2 } }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } =\frac { \sqrt { 3 } }{ 2 } \times 2\sqrt { 2 } =6\)
0, 0 and C = \(\sqrt { 6 } \)
 

(i) First place can be given to anyone of the 4 children and second place can be given to anyone of the 3 remain children.
(ii) The race can be finished in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 4! = 24 ways.

Given \(cos\theta =\frac { 1 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
LHS = cos 3θ = 4cos³θ-3cosθ
= \(4\left[ \frac { 1 }{ 2 } { \left( a+\frac { 1 }{ a } \right) }^{ 3 } \right] -3.\frac { 1 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(4\left( \frac { 1 }{ 8 } \left( { a }^{ 3 }+{ 3a }^{ 2 }\left( \frac { 1 }{ a } \right) +3a\left( \frac { 1 }{ { a }^{ 2 } } \right) +\left( \frac { 1 }{ { a }^{ 3 } } \right) \right) \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(\frac { 1 }{ 2 } \left( { a }^{ 3 }+3a+\frac { 3 }{ a } +\frac { 1 }{ { a }^{ 3 } } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) =\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ a } \right) +\frac { 1 }{ 2 } \left( 3a+\frac { 3 }{ a } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ { a }^{ 3 } } \right) +\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
\(\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ { a }^{ 3 } } \right) \) = RHS
Hence proved.

sin (30^o + \(\theta\)) + cos (60^o + \(\theta\))
LHS = sin (30^o + \(\theta\)) + cos (60^o + \(\theta\))
= sin 30 cos \(\theta\) + cos 30 . sin \(\theta\) + cos 60 cos \(\theta\) - sin 60 sin \(\theta\)
= \(\frac { 1 }{ 2 } \cos { \theta } +\frac { \sqrt { 3 } }{ 2 } \sin { \theta } +\frac { 1 }{ 2 } \cos { \theta } -\frac { \sqrt { 3 } }{ 2 } \sin { \theta } =\frac { 1 }{ 2 } \cos { \theta } +\frac { 1 }{ 2 } \cos { \theta } =\frac { 2\cos { \theta } }{ 2 } =\cos { \theta } \) LHS
Hence proved.

Sum of the roots = sin 15^o + cos 15^o
= sin (45 - 30) + cos (45 - 30)
= (sin 45 cos 30 - cos 45 sin 30) + (cos 45 cos 30 + sin 45 sin 30)
= \(\left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) \)
= \(\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } -\frac { 1 }{ 2\sqrt { 2 } } +\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } =\frac { 2\sqrt { 3 } }{ 2\sqrt { 2 } } =\frac { \sqrt { 3 } }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 6 } }{ 2 } \)
Product of the roots = sin 15^o cos 15^o  = \(\frac { 2\sin { { 15 }^{ o } } \cos { { 15 }^{ o } } }{ 2 } \)
= \(\frac { \sin { { 30 }^{ o } } }{ 2 } =\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 4 } \left[ \begin{matrix} since & \sin { 2A } \\ =2\sin { A } & \cos { A } \end{matrix} \right] \)
Required equation is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { \sqrt { 6 } }{ 2 } \right) +\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get \({ 4x }^{ 2 }-2\sqrt { 6x } +1=0\)

Since \(0 lies in the I quadrant
ஃ All the trigonometric ratios are positive.
Also \(\pi  y lies in the III quadrant
Only tan y and cot y are positive
  
= \(\sqrt { { 25 }^{ 2 }-{ 24 }^{ 2 } } \)
= 7
\(sinx=\frac { 8 }{ 17 } \quad siny=-\frac { 7 }{ 25 } \)
\(cosx=\frac { 15 }{ 17 } \quad cosy=-\frac { 24 }{ 25 } \)
ஃ sin (x - y) = sin x cos y - cos x sin y
\(\left( \frac { 8 }{ 17 } \right) \left( -\frac { 24 }{ 25 } \right) -\left( \frac { 15 }{ 17 } \right) \left( -\frac { 7 }{ 25 } \right) \)
\(=-\frac { 192 }{ 425 } +\frac { 105 }{ 425 } =-\frac { 87 }{ 425 } \)

Since 0\(\frac { \pi }{ 2 } \) and 0\(\frac { \pi }{ 2 } \), A, B lies in the I quadrant
ஃ All the trigonometric ratios are positive

=\(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\)

=\(\sqrt { { 41 }^{ 2 }-{ 9 }^{ 2 } } \)
=40
\(sinA=\frac { 3 }{ 5 } \quad sinB=\frac { 40 }{ 41 } \)
\(cosA=\frac { 4 }{ 5 } \quad cosB=\frac { 9 }{ 41 } \)
sin (A + B) = sinA cosB + cosA sinB
=\(\frac { 3 }{ 5 } .\frac { 9 }{ 41 } +\frac { 4 }{ 5 } .\frac { 40 }{ 41 } \)
\(=\frac { 27 }{ 205 } +\frac { 160 }{ 205 } =\frac { 187 }{ 205 } \)

We have A + B + C = 180°
B + C = 180° - A = 180° - 60° = 120°
\(\Rightarrow \frac { B+C }{ 2 } =60°\)...(1)
Using sine formula, \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sin B, c = k sin C
RHS = 2a cos\(\left( \frac { B-C }{ 2 } \right) \)...(2)
= 2.K sin A cos\(\left( \frac { B-C }{ 2 } \right) \)
=2.K sin 60° cos \(\left( \frac { B-C }{ 2 } \right) \)
= 2K.sin\(\left( \frac { B+C }{ 2 } \right) \)cos\(\left( \frac { B-C }{ 2 } \right) \)
= K\(\left[ sin\left( \frac { B+C+B-C }{ 2 } \right) +sin\left( \frac { B+C-B-C }{ 2 } \right) \right] \)
= K[sin B + sin C] 
= K sin B + K sin C
= b + c [From(2)]
LHS Hence proved.

h(t) = -5t² + 100t , \(0\le t\le 20\)
Let the time be 't' see, when the rocket is 495 feet above the ground
\(\therefore\) 0 < h(t) < 495
\(\Rightarrow\) 0< -5t² + 100t < 495
\(\Rightarrow\) 0<- 5t² + 100t- 495 < 0 [subtracting 495]
\(\Rightarrow\)  -5t² + 100t- 495 = 0
\(\Rightarrow\)  t²- 20t + 99 =0 [Divided by -5]
\(\Rightarrow\)  (t-11) (t-9) = 0
\(\Rightarrow\)  t = 11 or 9.
\(\therefore\)  At 11 or 9 sec, the rocket is 495 feet above the ground.

Given sin θ + cos θ = m
LHS = cos⁶ θ + sin⁶θ
= (cos²θ)³+ (sin² θ)³
= (cos²θ + sin²θ)(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= 1(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= (cos²θ)²+ (sin²θ)² - cos²θsin²θ
= (cos²θ)²+ (sin²θ)²- cos²θ sin²θ
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ....(1)
RHS = \(\frac { 4-3{ \left( { m }^{ 2 }-1 \right) }^{ 2 } }{ 4 } \)
= \(\frac { 4-3{ \left[ { \left( sin\theta +cos\theta \right) }^{ 2 }-1 \right] }^{ 2 } }{ 4 } =\frac { 4-3\left[ { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta +2sin\theta cos\theta -1 \right] }{ 4 } \)
= \(\frac { 4-12{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ 4 } =\frac { 4 }{ 4 } -\frac { 12 }{ 4 } { sin }^{ 2 }\theta { cos }^{ 2 }\theta \)
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ...(2)
From (1) and (2), LHS = RHS

As we are finding the square root of x² - 5x + 6, we must have \(x^2-5x+6\ge0\) for all x in the domain. 
Solving x² - 5x + 6 = 0, we get x = 2 and 3. Now draw the number line as below:

Now we have three intervals. (\(-\infty,\)2), (2, 3) and (3, \(\infty\)).
(i) Take any point in (\(-\infty,\) 2), say x = 1. Clearly x² - 5x + 6 is positive.
(ii) Take any point in (2, 3), say x = 2.5. Clearly x² - 5x + 6 is negative.
(iii) Take anypoint in (3, \(\infty\)) say x = 4. Clearly x² - 5x + 6 is positive.
For all x, in the intervals \((-\infty,2)\) and \((3,\infty)\),x² - 5x + 6 is positive. At x = 2,3 the value of x² -5x + 6 is zero. Thus, \(\sqrt{x^2-5x+6}\) is defined for all x in \((-\infty,2]\cup[3,\infty).\) Hence the domain of \(\sqrt{x^2-5x+6}\) is \((-\infty,2]\cup[3,\infty).\)