Use \(\sec A=\frac{1}{\cos A}, \quad \tan A=\frac{\sin A}{\cos A}\) and \(\operatorname{cosec} A=\frac{1}{\sin A}\) in LHS of given equation and simplify.

Use Pythagoras theorem to find QR and then find sec P, tan P and put in given expression
\(\frac{9}{2}\) 

To prove \(\sec ^{ 4 }{ \theta } -\sec ^{ 2 }{ \theta } =\tan ^{ 4 }{ \theta } +\tan ^{ 2 }{ \theta } \)
LHS = \(\sec ^{ 4 }{ \theta } -\sec ^{ 2 }{ \theta } \)
\(=\sec ^{ 2 }{ \theta } \left( \sec ^{ 2 }{ \theta } -1 \right) \)
\(\left[ \because \quad 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=\sec ^{ 2 }{ \theta } \left( \tan ^{ 2 }{ \theta } \right) \)
\(=\left( 1+\tan ^{ 2 }{ \theta } \right) \tan ^{ 2 }{ \theta } \)
\(=\tan ^{ 2 }{ \theta } \tan ^{ 4 }{ \theta } \)
= RHS

Let P(x,0) divide the line segment AB in the ratio k:1

y-coordinate  \(y=\frac { k\times \left( -4 \right) +1\times \left( 1 \right) }{ k+1 } ,\quad y=\frac { -4k+1 }{ k+1 } \)
on x-axis y=0 ,  \(\therefore \quad 0=\frac { k(5)+1(-5) }{ k-1 } \)
\(0=5k-5\quad \Rightarrow \quad 5k=5\quad \Rightarrow \quad k=1\)
Hence the required ration is 1:1, at point  \(\left( \frac { -3 }{ 2 } ,0 \right) \) .

OA = 10 units
\(\Rightarrow \ OA=\sqrt { { (2\alpha -1+3 })^{ 2 }+(7+1)^{ 2 } } \)
\(\Rightarrow \ 10=\sqrt { 4{ \alpha }^{ 2 }+4+8\alpha +64 } \)
Squaring \(100\ =\ { 4\alpha }^{ 2 }+8\alpha +68\)
\(\Rightarrow \ { 4\alpha }^{ 2 }+8\alpha -32=0\ \Rightarrow \ { \alpha }^{ 2 }+2\alpha -8=0\)
\(\Rightarrow \ { \alpha }^{ 2 }+4\alpha -2\alpha -8=0\ \Rightarrow \alpha (\alpha +4)-2(\alpha +4)=0\ \Rightarrow \ (\alpha +4)(\alpha -2)=0\)
\(\\ \alpha =-4\ ,\ \alpha =2\)

\(\frac { \sec ^{ 2 }{ \left( { 90 }^{ ° }-\theta \right) -\cot ^{ 2 }{ \theta } } }{ 2\left( \sin ^{ 2 }{ { 25 }^{ ° } } +\sin ^{ 2 }{ { 65 }^{ ° } } \right) } -\frac { 2\cos ^{ 2 }{ { 60 }^{ ° }\tan ^{ 2 }{ { 28 }^{ ° }\tan ^{ 2 }{ { 62 }^{ ° } } } } }{ 3\left( \sec ^{ 2 }{ { 43 }^{ ° } } -\cot ^{ 2 }{ { 47 }^{ ° } } \right) } \)
\(=\frac { \left( { cosec }^{ 2 }\theta -\cot ^{ 2 }{ \theta } \right) }{ 2\left( \sin ^{ 2 }{ { 25 }^{ ° } } +\cos ^{ 2 }{ { 25 }^{ ° } } \right) } -\frac { 2\times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \tan ^{ 2 }{ { 28 }^{ ° } } \times \cot ^{ 2 }{ { 28 }^{ ° } } }{ 3\left[ \sec ^{ 2 }{ { 43 }^{ ° } } -\tan ^{ 2 }{ { 43 }^{ ° } } \right] } \)
\(=\frac { 1 }{ 2\times \left( 1 \right) } -\frac { \frac { 1 }{ 2 } \times \tan ^{ 2 }{ { 28 }^{ ° }\times \frac { 1 }{ \tan ^{ 2 }{ { 28 }^{ ° } } } } }{ 3 } \)
\(=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\)

In 3600 see distance travelled by plane = 648000 m
In 10 see distance travelled by plane -\(\frac { 648000 }{ 360 } \)
= 18000 m
In \(\Delta\)ABC,
\(\frac { h }{ x } =tan\quad 60°\)
\(\frac { h }{ x } =\sqrt { 3 } ,\)
\(\Rightarrow \quad h=x\sqrt { 3 } \quad ...(i)\)
In \(\Delta\)ABC,
\(\frac { h }{ x+1800 } =tan\quad 30°\)
\(\frac { h }{ x+1800 } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \quad h=\frac { x+1800 }{ \sqrt { 3 } } \) ....(ii)
From equations (i) and (ii), we get
\(x\sqrt { 3 } =\frac { x+1800 }{ \sqrt { 3 } } \)
\(\Rightarrow\) 3x = x + 1800
\(\Rightarrow\)     2x = 1800
\(\Rightarrow\)       x = 900 m
h = x\(\sqrt { 3 } \)
= 900 \(\times\) 1.732
= 1558.8 m
\(\therefore\) Height of jet = 1558.8 m.

LHS = \(\sqrt { \frac { 1-\cos { \theta } }{ 1+\cos { \theta } } } =\sqrt { \frac { 1-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } } } \left( \because \quad \cos { { 60 }^{ ° } } =\frac { 1 }{ 2 } \right) \)
\(=\sqrt { \frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
RHS \(=\frac { \sin { \theta } }{ 1+\cos { \theta } } =\frac { \sin { { 60 }^{ ° } } }{ 1+\cos { { 60 }^{ ° } } } \)
\(=\frac { \frac { \sqrt { 3 } }{ 2 } }{ 1+\frac { 1 }{ 2 } } =\frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 3 }{ 2 } } \)
\(=\frac { 1 }{ \sqrt { 3 } } =LHS\)
Hence relation is verified for \(\theta ={ 60 }^{ ° }\)

Let points are A(2,y) and B(-4,3). Here, (x₁,y₁)=(2,y) and (x₂,y₂) = (-4,3)
\(\therefore\) Distance between two points,
\(AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) }^{ 2 } } \)  [ by distance formula]
\(\Rightarrow AB\sqrt { ({ 2+4) }^{ 2 }+{ (y-3 })^{ 2 } } \)
\(\Rightarrow 10=\sqrt { { (6) }^{ 2 }+{ (y) }^{ 2 }+9-6y } \)
 [\(\because \)AB-10 and (a-b)²-a²+b² -2ab]
On squariing both sides, we get
100=36+y²+9-6y
\(\Rightarrow \)100 = 45 + y²  - 6y
\(\Rightarrow \)y²-6y-55 = 0    
\(\Rightarrow \)y²-11y+5y-55=0  [by factorisation]
\(\Rightarrow \)y(y-11) +5(y-11) = 0
\(\Rightarrow \) (y-11)(y+5) = 0
\(\Rightarrow \) y-11=0 or y+5=0
\(\Rightarrow \)y = 11 or y = -5
Hence, the required value of y are 11 and -5.

Let the line 2x+3-5=0 divides the line segement joining the points (8,-9)and (2,1) in the ration k:1

\(\therefore \)  \(P\left( \frac { 2k+8 }{ k+1 } ,\frac { k-9 }{ k+1 } \right) \)  =P(x,y)
Thus, x=  \(\frac { 2k+8 }{ k+1 } \) and y= \(\frac { k-9 }{ k+1 } \)
point P(x,y) lies on the given line
\(\therefore \)\(2\left( \frac { 2k+8 }{ k+1 } \right) +3\left( \frac { k-9 }{ k+1 } \right) -5=0\)      
4k+16+3k-27-5k-5=0
2k=16 \(\Rightarrow \) k=8
Thus,the required ration is 8:1
Point of division is given as
\(P\left( \frac { 2(8)+8 }{ 8+1 } ,\frac { 8-9 }{ 8+1 } \right) i.e\quad P\left( \frac { 8 }{ 3 } ,\frac { -1 }{ 9 } \right) \)

P(3,-1)

P is the mid point of side AB
Therefore the coordinates of P are ((-1-1)/2,(-1+4)/2) = (-1, 3/2)
Similary the coordinates of Q , R and S are  (2,4),(5, 3/2), and (2, -1) respectilvely
Length of \(\mathrm{PQ}=\sqrt{(-1-2)^{2}+\left(\frac{3}{2}-4\right)^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\) 
Length of \(\mathrm{QR}=\sqrt{(2-5)^{2}+\left(4-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\) 
Length of \(R S=\sqrt{(5-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\) 
Length of \(S P=\sqrt{(2+1)^{2}+\left(-1-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\) 
 Length of  \(Q S=\sqrt{(2-2)^{2}+(4+1)^{2}}=5\) 
It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.

The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C (7, – 4) is given by
\(\frac{1}{2}[5(7+4)+4(-4-2)+7(2-7)]\) 
\(=\frac{1}{2}(55-24-35)=\frac{-4}{2}=-2\) 
Since area is a measure, which cannot be negative, we will take the numerical value of – 2, i.e., 2. Therefore, the area of the triangle = 2 square units.

In \(\Delta ABD\),
tan 60°=\(\frac{AB}{BD}\)
\(\sqrt{3}\)=\(\frac{60}{x}\)
x=\(\frac { 60 }{ \sqrt { 3 } } \)
=20\(\sqrt{3}\)
Now, in \(\Delta ABCD\),
tan 30° =\(\frac{CD}{BD}\)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 20\sqrt { 3 } } \)
\(h=\frac { 20\sqrt { 3 } }{ \sqrt { 3 } } \)
h = 20m
Height of the building is 20 m.

\(A+B+C={ 180 }^{ 0 }\Rightarrow A+B={ 180 }^{ 0 }-C\Rightarrow A+B={ 90 }^{ 0 }\) 

Given, \(\tan A=\frac{3}{4}\)
\(\tan A=\frac{P}{B}\)
\(\Rightarrow \quad P=3 k \text { and } B=4 k\)
\(\text { As, } \quad H=\sqrt{(3 k)^2+(4 k)^2}=5 k\)
\(\sin A=\frac{P}{H}=\frac{3 k}{5 k}=\frac{3}{5}\)
\(\cos A=\frac{B}{H}=\frac{4 k}{5 k}=\frac{4}{5}\)
\(\frac{1}{\sin A}+\frac{1}{\cos A}=\frac{5}{3}+\frac{5}{4}=\frac{20+15}{12}=\frac{35}{12}\)

(a) (ii) Distance covered by Niharika 
= Niharika runs \(\frac{1}{4}\) distance AD on the 2nd line
= \(\frac{1}{4} \times 100=25 \mathrm{~m}\) 
(b) (i)  Distance covered by Preet
= Preet runs \(\frac{1}{5}\) th distance AD on the 8  th line 
= \(\frac{1}{5} \times 100=20 \mathrm{~m}\) 
(c) (i) Distance between Green and Red flags
= \(\)\(\sqrt{(8-2)^{2}+(20-25)^{2}}\) 
= \(\sqrt{36+25}=\sqrt{61} \mathrm{~m}\) 
(d) (iii)Position of blue flag
= Mid-point of green flag and red flag
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)=(5,22.5)\) 
(e) (iv) (1,12,5)

We have, KL = 4 cm, ML = 4\(\sqrt{3}\)m, KM = 8 cm
(i) (a): \(\tan M=\frac{K L}{L M}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan M=\tan 30^{\circ} \Rightarrow \angle M=30^{\circ}\)
(ii) (c) : \(\tan K=\frac{M L}{K L}=\frac{4 \sqrt{3}}{4}=\sqrt{3}=\tan 60^{\circ}\)
\(\Rightarrow \angle K=60^{\circ}\)
(iii) (b)
(iv) (c)
(v) (a) : \(\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}=\frac{(1)^{2}-1}{1^{2}+1}=\frac{0}{2}=0\)