Given that the negative y intercept is 5 i.e., b = -5 and a = 150⁰,

Slope m = tan 150⁰ = tan(180° - 30⁰) = -tan 30⁰ = -\(\frac{1}{\sqrt{3}}\)
Slope and intercept form of the equation is y = mx + b
That is y = -\(\frac{1}{\sqrt{3}}\)x - 5
\(x+\sqrt{3}y+5\sqrt{3}=0\)

cos \(\theta ={1\over 2}\)
Principal value of cos θ lies in the I and II quadrant.
Since cos \(\theta ={1\over 2}\) > 0, the principal value of cos \(\theta \) lies in the interval [ 0, \({\pi \over 2}\)]
cos \(\theta ={1\over 2}\)= cos \(={\pi \over 2}\)
 Thus, \(\theta\) = \(\pi \over 3\)is the principal solution.

LHS = \(\sin { 4\alpha } =sin2(2\alpha)\)
= \(2\left( \frac { 2\tan { \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \left( \frac { 1-\tan ^{ 2 }{ \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \)
= \(4\tan { \alpha } .\frac { 1-\tan ^{ 2 }{ \alpha } }{ { \left( 1+\tan ^{ 2 }{ \alpha } \right) }^{ 2 } } \) = RHS
Hence proved.

Given sin\(\theta\) = \(-\frac { 1 }{ \sqrt { 2 } } \)
sin\(\theta\) = \(-\frac { 1 }{ \sqrt { 2 } } \) < 0, so principal value lies in the IV quadrant.
sin \(\theta\) = -sin\(\frac { \pi }{ 4 } \)- sin(-\(\frac { \pi }{ 4 } \))
\(\Rightarrow \) \(\theta\) = - \(\frac { \pi }{ 4 } \) is the principal solution

Let \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \) = y, where \(-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
⇒ sin y = \(\frac { 1 }{ \sqrt { 2 } } \)
⇒ sin y = sin \(\frac { \pi }{ 4 } \)
⇒ y = \(\frac { \pi }{ 4 } \)
Thus the principal value of \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac { \pi }{ 4 } \) .

\({ LHS=tan }^{ -1 }\left( \frac { m }{ n } \right) -{ tan }^{ -1 }\left( \frac { m-n }{ m+n } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { m }{ n } -\frac { m-n }{ m+n } }{ 1+\frac { m }{ n } \times \frac { m-n }{ m+n } } \right) ={ tan }^{ -1 }\left( \frac { { m }^{ 2 }+mn-mn+{ n }^{ 2 } }{ mn+{ n }^{ 2 }+{ m }^{ 2 }-mn } \right) \)
\(={ tan }^{ -1 }\left( \frac { { m }^{ 2 }+{ n }^{ 2 } }{ { m }^{ 2 }+{ n }^{ 2 } } \right) ={ tan }^{ -1 }(1)=\frac { \pi }{ 4 } \) 

By sine formula \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =2R\)
\(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { c }^{ 2 } } =\frac { { (2RsinA) }^{ 2 }-{ (2RsinB) }^{ 2 } }{ { (2RsinC) }^{ 2 } } \)
\(=\frac { 4{ R }^{ 2 }{ sin }^{ 2 }A-4{ R }^{ 2 }{ sin }^{ 2 }B }{ 4{ R }^{ 2 }{ sin }^{ 2 }C } =\frac { { sin }^{ 2 }A-{ sin }^{ 2 }B }{ { sin }^{ 2 }C } \)
\(=\frac { sin(A+B)sin(A-B) }{ { sin }^{ 2 }C } [\because sinC=sin(A+B)]\)
\(=\frac { sin(A+B)sin(A-B) }{ { sin }^{ 2 }(A+B) } =\frac { sin(A-B) }{ sin(A+B) } \) 

3x²+10xy+8y²+14x+22y+15=0
a= 3, h = 5, b = 8, g = 7, f= 11, c = 15
The condition is af²+bg²+ch² abc-2fgh = 0
3(11)² +8(7)² +15(5)² -(3)(8)(15)-2(11)(7)(5)=363+392+375-360-770=0
Hence the equation represents a pair of straight lines
tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\pm \frac { 2 }{ 11 } \)
tanθ=\(\frac { 2 }{ 11 } \)
⇒ θ=tan-1\(\left( \frac { 2 }{ 11 } \right) \).

Equation of line through the intersection of straight lines 5x - 6y = 1 and 3x + 2y + 5
5x - 6y - 1 + k (3x + 2y + 5) = 0
x (5 + 3k) + y (-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x - 5y + 11 = 0
That is, the product of their slopes is -1
-\(\left( \frac { 5+3k }{ -6+2k } \right) \left( -\frac { 3 }{ -5 } \right) \)=-1
⇒ \(\frac { 15+9k }{ -30+10k } \)=1
45=k
Required equation is 5x - 6y -1 + 45 (3x + 2y + 5) = 0
140x + 84y + 224 = 0
20x + 12y + 32 = 0
5x+ 3y+ 8 = 0.

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
\(=(\frac{\pi}{8}\times\frac{180^°}{\pi})^°=(\frac{45}{2})^°=22°30'\)

There are two straight lines making 30° with the y-axis.
From the figure, it is clear that the two lines make the angles 60° and 120° with the x-axis
Let m₁ be tan 60° = √3 and 
m₂ be tan 120° = tan( 180° - 60°)

= -tan60°= -√3
m₁ = √3, m₂ = -√3 and b = 7
Equations of lines are y = m₁x + b and y = m₂x + b
y = √3x + 7 and y = -√3x + 7

Let the equation of the line in intercept form be \(\frac{x}{a}+\frac{y}{b}=1\)
Its intercepts on x and y axes are a and b.
Given that \(\frac{1}{a}+\frac{1}{b}\) = constant = K
\(\therefore \frac{1}{K_a}+\frac{1}{K_b}=1\)
\(\Rightarrow \frac{\frac{1}{K}}{a}+\frac{\frac{1}{K}}{b}=1\)
\(\Rightarrow (\frac{1}{k},\frac{1}{k})\) satisfies the equation \(\frac{x}{a}+\frac{y}{b}=1\)
Hence the equation (1), passes through the fixed point \( (\frac{1}{k},\frac{1}{k})\).

One completed rotation = 360⁰
∴ Number of degree taken in 1 = 1000 \(\times\) 360⁰
ie Number of degree taken in 60'' = 1000 \(\times\) 360⁰ [∵ 1' = 60'']
Number of degree taken in 1 = \(\frac { 1000\times 360 }{ 60 } \)
= 1000 \(\times\) 60^o
= 6000^o

Let A'and B be the availability of first and second fire engine respectively, then A and B are independent.
Then \(P(A)=P(B)=0.96 \)
\(P(\bar{A})=P(\bar{B})=1-0.96=0.04\)
(i) P(a fire engine is available when needed)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(A \cap B)\)
\(=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B)\)
\(=0.96 \times 0.04+0.04 \times 0.96+0.96 \times 0.96\)
\(=0.96(0.04+0.04+0.96) \)
\(=0.96 \times 1.04 \)
\(=0.9984\)
(ii) P (Neither is available when needed)
\(=P(\bar{A} \cap \bar{B}) \)
\(=P(\bar{A}) \cdot P(\bar{B})\)
\(=0.04 \times 0.04\)
\(=0.0016\)

Let A, B, C be the events that the problems solved by 3 students. Then,
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}\)
(i) P (Problem is solved) \(=P(A \cup B \cup C)\)
\(=1-P(\overline{A \cup B \cup C})\)
\(=1-P(\bar{A} \cap \bar{B} \cap \bar{C})\)
\(=1-P(\bar{A}) P(\bar{B}) P(\bar{C})\)
\(=1-\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \)
\(=1-\frac{2}{5}=\frac{3}{5}\)
(ii) P (exactly one of them will solve)
\(=P(A \bar{B} \bar{C} \cup \bar{A} B \bar{C} \cup \bar{A} \bar{B} C)\)
\(=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C}) +P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)\)

\(=\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{1}{5}\left(1+\frac{2}{3}+\frac{1}{2}\right) \)
\(=\frac{6+4+3}{30}=\frac{13}{30}\)

Let x-axis be the time in hours and y-axis be the distance in kilometer.
From the given information, we have
y = \(6\left( x-\frac { 10 }{ 60 } \right) \) ⇒ y = 6x - 1
y = \(4\left( x+\frac { 5 }{ 60 } \right) \) ⇒ y = 4x + \(\frac { 1 }{ 3 } \)
Solving the above two equations, we get (x, y) = \(\left( \frac { 2 }{ 3 } ,3 \right) \)
x = \(\frac { 2 }{ 3 } \) hour = 40 minutes, y = 3 km
(i) the distance between house and the school is 3km
(ii) the minimum average speed to reach the school on time is \(\frac { 60 }{ 40 } \) \(\times\) 3 = 4.5 kmph and time taken is hours \(\frac { 2 }{ 3 } \) or 40 miniutes
(iii) the school gate closes at 8.40 AM
(iv) the pair of straight lines of his path of walk to school is
(6x - y - 1)(4x - y + \(\frac { 1 }{ 3 } \)) = 0
72x²- 30xy + 3y²- 6x + 2y - 1 = 0

(i) Let x-axis be the time in seconds and y-axis be the distance in meters.
Let the train be at the origin
ஃ Length of train = 150m is the negative y-intercept
Slope of the motion of the train m = 12.5m/sec
Since we are given slope and y-intercept, the equation of the line is y = mx - c
Equation of the motion of the train ஃ y = 12.5x-150
(ii) To find the Time taken to cross the pole, put y = 0
⇒ ஃ 12.5x = 150
⇒ x = \(\frac{150}{12.5}\) = 12 sec
(iii) Time taken to cross the bridge of length 850m is,
850 = 12.5x - 150
⇒ 850 + 150 = 12.5x
= \(\frac{1000}{12.5}=x\)
⇒ x = 80 sec

(i) Let the x - axis represent the weight and the y-axis represent the length

(ii) Consider the point \(\begin{pmatrix} { x }_{ 1 } & { y }_{ 1 } \\ 2 & 3 \end{pmatrix}\begin{pmatrix} { x }_{ 2 } & { y }_{ 2 } \\ 4 & 4 \end{pmatrix}\)
Equation of the straight line using two point form is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ 4-3 } =\frac { x-2 }{ 4-2 } \)
\(\Rightarrow \quad \frac { y-3 }{ 1 } =\frac { x-2 }{ 2 } \)
\(\Rightarrow\) 2y - 6 = x-2
 \(\Rightarrow\) x - 2y + 4 = 0
(iii) To find the actual length of the spring,
Put x = 0
\(\Rightarrow\) 0 - 2y + 4 = 0
\(\Rightarrow\) -2y = -4
\(\Rightarrow\) y = 2 cm
(iv) Put y = 9 in (1) we get
\(\Rightarrow\) x - 18 + 4 = 0
\(\Rightarrow\) x - 14 = 0
\(\Rightarrow\) x = 14 kg
\(\therefore\) 14 Kg must be added
(v) Put x = 6 in (1) we get,
\(\Rightarrow\)  6 - 2y + 4 = 0
\(\Rightarrow\) 10 - 2y = 0
\(\Rightarrow\) 10 = 2y 
\(\Rightarrow\) y = 5
\(\therefore\) Strength of the spring = 5 cm

\(2cos\left( \frac { 5x+x }{ 2 } \right) .sin\left( \frac { 5x-x }{ 2 } \right) =cos3x\)
2cos 3x.sin 2x = cos 3x
2cos 3x sin 2x - cos 3x = 0
cos 3x = 0 or 2sin 2x = 1
sin 2x =\(\frac{1}{2}\)
case (i)
when cos 3x = 0
3x = (2n +1)\(\frac{\pi}{2}\), n∈z
x = (2n + 1)\(\frac{\pi}{6}\), n∈z
case (ii)
sin 2x = \(\frac{1}{2}\)
sin 2x = sin\(\frac{\pi}{6}\)
2x = nπ + (-1)ⁿ, \(\frac{\pi}{6}\), n∈z
x = \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\), n∈z
Hence the solutions are x = (2n + 1)\(\frac{\pi}{6}\)or \(\frac{n\pi}{2}\)+(-1)ⁿ\(\frac{\pi}{12}\), n∈z

LHS = \(cot{ \left( 7\frac { 1 }{ 2 } \right) }^{ ° }\)
= \(\frac { cos{ 7\frac { 1 }{ 2 } }^{ ° } }{ sin{ 7\frac { 1 }{ 2 } }^{ ° } } \)
Multiplying the numerator and denominator by 2sin(\({ 7\frac { 1 }{ 2 } }^{ ° }\))
\(\frac { 2sin{ 7\frac { 1 }{ 2 } }^{ ° }cos{ 7\frac { 1 }{ 2 } }^{ ° } }{ 2{ sin }^{ 2 }{ 7\frac { 1 }{ 2 } }^{ ° } } =\frac { sin15° }{ 1-cos15° } \)
\(\frac { sin\left( 45-30° \right) }{ 1-cos\left( 45-30° \right) } =\frac { sin45cos30-cos45°sin30° }{ 1-\left( cos45°cos30°+sin45sin30 \right) } \)
= \(\frac { \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } }{ 1-\left( \frac { 1 }{ 2 } .\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } /1-\left( \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) \)
= \(\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } \times \frac { 2\sqrt { 2 } }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } \)
= \(\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } \times \frac { 2\sqrt { 2 } +\sqrt { 3 } +1 }{ 2\sqrt { 2 } +\sqrt { 3 } +1 } \)
\(\frac { \left( \sqrt { 3 } -1 \right) \left( 2\sqrt { 2 } +\sqrt { 3 } +1 \right) }{ { \left( 2\sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 3 } +1 \right) }^{ 2 } } =\frac { 2\sqrt { 6 } +3-2\sqrt { 3 } -\sqrt { 3 } -1 }{ 8-\left( 1+3+2\sqrt { 3 } \right) } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } }{ 4-2\sqrt { 3 } } =\frac { \sqrt { 6 } +1-\sqrt { 2 } }{ 2-\sqrt { 3 } } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } +\sqrt { 18 } +\sqrt { 3 } -\sqrt { 6 } }{ 4-2\sqrt { 3 } } =\frac { \sqrt { 6 } +1-\sqrt { 2 } }{ 2-\sqrt { 3 } } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } +\sqrt { 18 } +\sqrt { 3 } -\sqrt { 6 } }{ 4-3 } =2\sqrt { 6 } +2-2\sqrt { 2 } +3\sqrt { 2 } +\sqrt { 3 } -\sqrt { 6 } \)
= \(\sqrt { 6 } +\sqrt { 3 } +2-2\sqrt { 2 } +3\sqrt { 2 } =\sqrt { 6 } +\sqrt { 3 } +2+2\sqrt { 2 } \)
= \(\sqrt { 2 } +\sqrt { 3 } +\sqrt { 4 } +\sqrt { 6 } \) = RHS
Hence proved.