\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

141.96 m

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

Here, we have
\(a_{2}-a_{1}=-\frac{1}{2}-\left ( -\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0,\)
\(a_{3}-a_{2}=-\frac{1}{2}-\left (-\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0,\)
\(a_{4}-a_{3}=-\frac{1}{2}-\left ( -\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0\)
and so on.
Since, the difference of any two consecutive terms is same. Therefore, the given list of numbers forms an AP and its common difference (d) is 0.
Now, next three terms of this AP are
a₅ = a₄ + d = -\(\frac{1}{2}\) + 0 = - \(\frac{1}{2}\)
a₆ = a₅ + d = -\(\frac{1}{2}\) + 0 = - \(\frac{1}{2}\)
and a₇ = a₆ + d = -\(\frac{1}{2}\) + 0 = -\(\frac{1}{2}\)

Given equation is, x²+x+1=0
When x=0, LHS=0²+0+1=1\(\ne\)0
x=0 is not the solution
When x=1, LHS=1²+1+1=3\(\ne\)0
x=1 is not the solution of given equation.

\(\frac{10}{x+y}+\frac{2}{x-y}=4\) 
\(\frac{15}{x+y}-\frac{5}{x-y}=-2\) 
Putting \(\frac{1}{x}+y=p\) ,in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
⇒ 15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
\(\frac{p}{4-20}=\frac{q}{-60-(-20)}=\frac{1}{-50-30}\) 

\(\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}\) 
\(\frac{p}{-16}=\frac{1}{-80}\) 
p = 1/5 and q = 1
\(p=\frac{1}{x+y}=\frac{1}{5}\) and \(q=\frac{1}{x-y}=1\) 
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

tan² 30° sin 30° + cos 60° sin² 90° tan² 60° - 2 tan 45° cos² 0° sin 90°
Expression = \(\frac{1}{3}\times\frac{1}{2}+\frac{1}{2}\times1\times3\times-2\times 1\times 1\times 1\)
\(=\frac{1}{6}+\frac{3}{2}-2=\frac{1+9-12}{6}\)
\(=-\frac{2}{6}=-\frac{1}{3}\)

Given: ABC is right angled at Band D is the midpoint of BC.
\(\therefore BD=DC=\frac { 1 }{ 2 } BC\)
In \(\triangle\)ABD, AD² = AB² + BD²   (Pythagoras theorem) ...(i) 
In \(\triangle\)ABC, AC² = AB² + BC² (Pythagoras theorem) ...(ii)
From eqn. (i), AD² = AB² +\({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)   (D is the mid-point of BC) 
\(\Rightarrow\) 4AD² = 4AB² + BC²
\(\Rightarrow\) BC2 = 4A02 - 4AB2
Using this in equation (ii),
\(\Rightarrow\)AC² = AB² + 4AD²-4AB²
\(\Rightarrow\)AC² = 4AD² - 3AB²
Hence proved.

Area of quadrilateral ABCD
=Area of traingle ABC+Area of traingle ACD    ...(i)
Now, ar (\(\triangle \)ABC)
=\(\frac { 1 }{ 2 } \) [-3(-7+8)-2(-8-5)+1(5+7)]
=\(\frac { 1 }{ 2 } \) [-3+26+12] = \(\frac { 35 }{ 2 } \)sq.units ...(ii)
Also, ar ( \(\triangle \)ACD)
=\(\frac { 1 }{ 2 } \) [-3(-8-3)-1(3-5)+6(5+8)]
= \(\frac { 1 }{ 2 } \)[33-2+78]=\(\frac { 109 }{ 2 } \) sq.units    ...(iii)
From (i),(ii),(iii) we get 
ar (ABCD)=\(\frac { 35 }{ 2 } +\frac { 109 }{ 2 } =\frac { 144 }{ 2 } \) =72 sq.units

Applying Pythagoras theorem in ΔABM, we obtain
AB² = AM² + MB²
= (AD² − DM²) + MB²
= (AD² − DM²) + (BD − MD)²
= AD² − DM² + BD² + MD² − 2BD x MD
= AD² + BD² − 2BD x MD
\( =A D^{2}+\left(\frac{B C}{2}\right)^{2}-2\left(\frac{B C}{2}\right) \times M D \\ \)

Similarly, on ray AY, mark points C₁, C₂, C₃, C₄ and C such that AC₁ = C₁C₂ = C₂C₃ = C₃C₄ = C₄C. Then join B₁C₁ and BC.

Note that \(\frac{\mathrm{AB}_{1}}{\mathrm{~B}_{1} \mathrm{~B}}=\frac{\mathrm{AC}_{1}}{\mathrm{C}_{1} \mathrm{C}}\) (Each equal to \(\frac{1}{4}\)) 
You can also see that lines B₁C₁ and BC are parallel to each other, i.e., 
B₁C₁ || BC                                                     (1)
Similarly, by joining B₂C₂, B₃C₃ and B₄C₄, you can see that:
\(\frac{\mathrm{AB}_{2}}{\mathrm{~B}_{2} \mathrm{~B}}=\frac{\mathrm{AC}_{2}}{\mathrm{C}_{2} \mathrm{C}}\left(=\frac{2}{3}\right) \text { and } \mathrm{B}_{2} \mathrm{C}_{2} \| \mathrm{BC}\)      (2)
\(\frac{\mathrm{AB}_{3}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{AC}_{3}}{\mathrm{C}_{3} \mathrm{C}}\left(=\frac{3}{2}\right) \text { and } \mathrm{B}_{3} \mathrm{C}_{3} \| \mathrm{BC}\)      (3)
\(\frac{\mathrm{AB}_{4}}{\mathrm{~B}_{4} \mathrm{~B}}=\frac{\mathrm{AC}_{4}}{\mathrm{C}_{4} \mathrm{C}}\left(=\frac{4}{1}\right) \text { and } \mathrm{B}_{4} \mathrm{C}_{4} \| \mathrm{BC}\)        (4)
From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY . Each time, you will arrive at the same result.

2x − 2y − 2 = 0
4x − 4y − 5 = 0
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{-2}{-4}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{2}{5}\) 
Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) 
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent

\(\tan \theta=\frac{P}{B} \text { and } B C=2 C D\)
= 2 

111222333

Let A be the centre of the balloon (spherical) whose radius is r.
Let AB be the height of the ballon i.e., h units, such that
ㄥDPC=ፀ, ㄥAPB-ф
Since ΔPDA and ΔPCA are congruent, therefore
ㄥAPC=ㄥAPD=ፀ/2
In rt. ㄥed ΔPCA,
\(\frac { AC }{ AP } =sin\frac { \theta }{ 2 } \)
⇒ AP=\(AC\frac { 1 }{ sin\frac { \theta }{ 2 } } =r.cosec\frac { \theta }{ 2 } \) or
Consider rt . ㄥed ΔPBA, we have
\(\frac { AB }{ AP } \)=sin ф
AB=AP.sinф
h=r.cosec\( \frac { \theta }{ 2 } \).sinф
h=r sinф.cosec\( \frac { \theta }{ 2 } \).

We have, AB = BC = 6\(\sqrt{2}\) m and AC=12m .
(i) (d):\(\because\)  Dis mid point of AC.
\(\therefore\) AD=DC=6m
Now, AB² = BD² + AD² (\(\therefore\) \(\Delta\)ABD is a right triangle)
\(\Rightarrow B D^{2}=(6 \sqrt{2})^{2}-6^{2}=72-36=36 \)
\(\Rightarrow B D=6 \mathrm{~m}\)
(ii) (c) : \(\operatorname{In} \Delta A B D, \sin A=\frac{B D}{A B}=\frac{6}{6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin A=\sin 45^{\circ} \Rightarrow \angle A=45^{\circ}\)
(iii) (c) : \(\operatorname{In} \Delta B D C, \tan C=\frac{B D}{D C}=\frac{6}{6}\)
\(\Rightarrow \tan C=1=\tan 45^{\circ} \Rightarrow \angle C=45^{\circ}\)
(iv) (d) : \(\sin A=\frac{1}{\sqrt{2}}, \cos C=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\therefore \quad \sin A+\cos C=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
(v) (c): \((v) \quad(c): \tan C=1, \tan A=\tan 45^{\circ}=1\)
\(\Rightarrow \tan ^{2} C+\tan ^{2} A=1+1=2\)
 

(i) (a): Since the diagonals of a parallelogram bisect each other.

\(\therefore\) By mid-point formula, we have
\(\left(\frac{x+3}{2}, \frac{3+7}{2}\right)=\left(\frac{5+6}{2}, \frac{6+y}{2}\right) \)
\(\Rightarrow x+3=11 \text { and } y+6=10 \Rightarrow x=8 \text { and } y=4\)
(ii) (d): Distance between (3, 3) and (6, 4)
\(=\sqrt{(6-3)^{2}+(4-3)^{2}}=\sqrt{9+1}=\sqrt{10} \text { units }\)
And distance between (6, 4) and (8, 7)
\(=\sqrt{(8-6)^{2}+(7-4)^{2}}=\sqrt{4+9}=\sqrt{13} \text { units }\)
Now, required perimeter \(=2(\sqrt{10}+\sqrt{13}) \text { units }\)
(iii) (a): Let A(l, 3), B(2, 6), C(S, 7) and D(4, 4) be the given points. Then length of diagonal
\(A C =\sqrt{(5-1)^{2}+(7-3)^{2}=\sqrt{16+16}} \)
\(=\sqrt{32}=4 \sqrt{2} \text { units } \)
\(\text { and } B D =\sqrt{(4-2)^{2}+(4-6)^{2}}=\sqrt{4+4} \)
\(=\sqrt{8}=2 \sqrt{2} \text { units }\)
(iv) (d): Length of one of the sides
\(=\sqrt{(2-1)^{2}+(6-3)^{2}}=\sqrt{1+9}=\sqrt{10} \text { units }\)
\(\therefore \quad \text { Perimeter }=4 \sqrt{10} \text { units }\)
(v) (c)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2).
(i) (d): Required distance \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\)
\(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
(ii) (d): Required distance \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\)
\(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\)
(iii) (b): Distance between Band C
\(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\)
Thus, distance travelled by first bus to reach to B
\(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\)
and distance travelled by second bus to reach to B
\(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
\(\therefore\) Distance of first bus is greater than distance of the
second bus, therefore second bus should be chosen.
(iv) (d): Distance travelled by first bus = 8.48 km
\(\therefore\) Total fare = 8.48 x 10 = Rs 84.80
(v) (b): Distance travelled by second bus = 7. 2 km
\(\therefore\) Total fare = 7.2 x 15 = Rs 108