Let P(x₁ , y₁) and Q(x₂ , y₂) are two points which divide AB in three equal parts.
By section formula
\(P\left( { x }_{ 1 }{ y }_{ 1 } \right) =\left( \frac { 1\times \left( -4 \right) +2\times \left( 2 \right) }{ 1+2 } ,\frac { 1\times \left( -6 \right) +2\times \left( -3 \right) }{ 1+2 } \right) \)
\(=\left( \frac { -4+4 }{ 3 } ,\frac { -6+(-6) }{ 3 } \right) \)
= (0, -4)
\(Q\left( { x }_{ 2 }{ y }_{ 2 } \right) =\left( \frac { 2\times \left( -4 \right) +1\times \left( 2 \right) }{ 2+1 } ,\frac { 2\times \left( -6 \right) +1\times \left( -3 \right) }{ 2+1 } \right) \)
\(=\left( \frac { -8+2 }{ 3 } ,\frac { -12+(-3) }{ 3 } \right) \)
= (- 2, - 5)

LHS = \(\frac { 1-\cos { A } +\sin { A } }{ \sin { A } +\cos { A } -1 } =\frac { \sec { A } +\tan { A } +1 }{ 1-(\sec { A } -\tan { A } ) } \)
\(=\frac { (\sec { A } +\tan { A } )-1 }{ 1-(\sec { A } -\tan { A } ) } \times \frac { 1+(\sec { A } -\tan { A } ) }{ 1+(\sec { A } -\tan { A } ) } \)
\(=\frac { (\sec { A } +\tan { A } )+(\sec ^{ 2 }{ A } -\tan ^{ 2 }{ A } )-1-(\sec { A } -\tan { A } ) }{ 1-(\sec ^{ 2 }{ A } -\tan ^{ 2 }{ A } -2\tan { A } \sec { A } ) } \)
\(=\frac { \tan { A } +1-1+\tan { A } }{ 1-(\tan ^{ 2 }{ A } -\tan ^{ 2 }{ A } -2\tan { A } \sec { A } ) } [\because \sec ^{ 2 }{ A } -\tan ^{ 2 }{ A } =1]\)
\(=\frac { 2\tan { A } }{ 2\tan { A } (\sec { A } -\tan { A } ) } =\frac { 1 }{ \sec { A } -\tan { A } } \)
\(=\frac { \sec { A } +\tan { A } }{ (\sec ^{ 2 }{ A } -\tan ^{ 2 }{ A } ) } \)
\(=\frac { \sec { A } +\tan { A } }{ 1 } =\frac { 1+\sin { A } }{ \cos { A } } =RHS\)
Hence proved.

LHS  =\(\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } \times \frac { 1+\cos { A } }{ 1+\cos { A } } } \)
\(=\frac { 1+\cos { A } }{ 1-\cos ^{ 2 }{ A } } =\frac { 1+\cos { A } }{ \sin { A } } \)   \([\therefore 1-\cos ^{ 2 }{ A } =\sin ^{ 2 }{ A } ]\)
\(=cosecA+\cot { A } \) 

0

Here, |AB|= \(\sqrt { (-2-2)^{ 2 }+(2-3)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(-1)^{ 2 } } \)
=\(\sqrt { 16+1 } =\sqrt { 17 } \)  Units
|BC|=  \(\sqrt { (-1+2)^{ 2 }+(-2-2)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(-4)^{ 2 } } \)
=\(\sqrt { 1+16 } =\sqrt { 17 } \)  units
|CD|= \(\sqrt { (3+1)^{ 2 }+(-1+2)^{ 2 } } \)
=\(\sqrt { 4^{ 2 }+1^{ 2 } } \)
=   \(\sqrt { 17 } \)=units
|DA|=\(\sqrt { (2-3)^{ 2 }+(3+1)^{ 2 } } \)
=  \(\sqrt { (-1)^{ 2 }+4^{ 2 } } \)=\(\sqrt { 1+16 } \)
= \(\sqrt { 17 } \) units
\(\Rightarrow \) AB=BC=CD=DA= units
Now,Diagonal |AC| =\(\sqrt { (-1-2)^{ 2 }+(-2-3)^{ 2 } } \)
= \(\sqrt { (-3)^{ 2 }+(-5)^{ 2 } } \)
=\(\sqrt { 9+25 } =\sqrt { 34 } \)  units
Diagonal |BD|= \(\sqrt { (3+2)^{ 2 }+(-1-2)^{ 2 } } \)
= \(\sqrt { { 5 }^{ 2 }+(-3)^{ 2 } } \)
= \(\sqrt { 25+9 } =\sqrt { 34 } \) units
\(\Rightarrow \) Diagonal AC=Diagonal BD= units 
Hence ABCD is a square

k=-1

Given: AB the lighthouse 100 m above sea level and C is a ship sailing towards AB. 
⇒ ㄥEAC = 30\(\unicode{xb0} \)
After travelling from C to C' angle of depression changes from ㄥEAC = 30\(\unicode{xb0} \) to ㄥEAC' = 60\(\unicode{xb0} \)
To Find: CC'
Solution: AE||BC
[Line of sight and line of horizontal]
⇒ ㄥACC' = ㄥEAC = 30\(\unicode{xb0} \) [Alternate angles]
ㄥAC'B = ㄥEAC' = 60\(\unicode{xb0} \) [Alternate angles]
In right ΔABC', \(\frac { AB }{ BC } \) = tan30^o
\(\frac { 100 }{ BC } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow BC=100\sqrt { 3 } \)
CC' = BC -  BC'
=\(\left( 100\sqrt { 3 } -\frac { 100 }{ \sqrt { 3 } } \right) \)m
=\(\frac { 100\times 3-100 }{ \sqrt { 3 } } \)
=\(\frac { 200 }{ \sqrt { 3 } } m=\frac { 200\sqrt { 3 } }{ 3 } m\) = 115.466 m

Given: A boy is standing at a distance of 100 m from the bird flying at an elevation of 30^o. A girl is standing on the roof of 20 m high building finds the angle of elevation of the bird to be 45^o Boy and girl are on opposite side of the bird.
To find: Distance between the bird and the girl i.e., BE.
Solution: In ΔACB,
⇒ \(\frac { h }{ 100 } \) = sin 300 ⇒ h = \(\frac { 1 }{ 2 } \) x 100 = 50 m
⇒ BF = h - 20 = (50 - 20) m = 30 m
In ΔBFE,\(\frac { 30 }{ BE } =sin{ 45 }^{ 0 }\Rightarrow \frac { 30 }{ BE } =\frac { 1 }{ \sqrt { 2 } } \Rightarrow 30\sqrt { 2 } \) = BE
BE = 30 x 1.414 = 42.420 = 42.42 m

Let the height of the tower AB = h m,
We have PB = 4 m, QB = 9 m
Let ㄥAQB = θ   [Both are complementary angles]
Then ㄥAPB = 90\(\unicode{xb0} \) ^- θ
In ΔABP, \(\frac { AB }{ PB } \) = tan (90\(\unicode{xb0} \)- θ)
⇒ h/4 = cotθ  ....(i)
In ΔABQ, \(\frac { AB }{ QB } \) = tanፀ
h/9 = tanθ
h = 9tanθ  ....(ii)
From equation (i) and (ii) we get
h x h = 4 cotθ x 9 tanθ
⇒ h² = 36 cotθ x tanθ = 36 1/tanθ x tanθ
⇒ h² = 36 ⇒ h = 6m
Hence, the height of the tower is 6 m.

(i) (b): The person who makes small angle of elevation is more closer to the balloon.
\(\therefore\) Radlra is more closer to the balloon.
(ii) (b): \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \)
\(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\)
(iii) (a): In \(\Delta\)GCE,
\(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\)
(iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m
(v) (b)