LHS = sec A (1 - sin A) (sec A + tan A)
= \(\left[ \frac { 1 }{ cosA } \right] (1-sinA)\left[ \frac { 1 }{ cosA } +\frac { sinA }{ cosA } \right] \)
= \(\frac { (1-sinA)(1+cosA) }{ { cos }^{ 2 }A } \)
= \(\frac { 1-{ sin }^{ 2 }A }{ { cos }^{ 2 }A } \)
= \(\frac { { cos }^{ 2 }A }{ { cos }^{ 2 }A } \) = 1 = RHS

Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

PQ II RS


Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

Let assumed mean A = 30
C = 5

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, C = 5
σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

A table

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

p²x² + (P² -Comparing this with ax' + bx + c = 0, we have
a=p²
b=p²-q²
c =-q²
D = b²-4ac
= (P²-q²)-4xp²x-q²
= (P²-q²)²+ 4p² q²
= (P²+q²)²>0
So, the given equation has real roots given by
\(\alpha =\frac { -b-\sqrt { D } }{ 2a } =\frac { -({ p }^{ 2 }-{ q }^{ 2 })+({ p }^{ 2 }+{ q }^{ 2 }) }{ { 2p }^{ 2 } } \)
\(=\frac { { q }^{ 2 } }{ { p }^{ 2 } } \)
\(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -({ p }^{ 2 }-{ q }^{ 2 })+({ p }^{ 2 }+{ q }^{ 2 }) }{ { 2p }^{ 2 } } \)
=-1

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get
12576 = 4052 x 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 x 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 x 1 + 148, 148 ≠ 0
∴ Again by division lemma
272 = 148 x 1 + 124, here 124 ≠ 0
∴ Again by division lemma
148 = 124 x 1 + 24, Here 24 ≠ 0
∴ Again by division lemma
124 = 24 x 5 + 4, Here 4 ≠ 0
∴ Again by division lemma
24 = 4 x 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

It is not a function as the vertical line PQ cuts the graph at two points

Let P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.

in right triangle ABC [see fig.(b)]
tan\(\theta \) =\(\frac { 8 }{ 3 } \)
= tan^-1(2.66)
\(\theta \) = \(69.4°\)(since tan \(69.4°\)=2.6604)
\(\angle \)BAC = \(69.4°\)

1 + 2 + 3 +...+ n = \(\frac{n(n+1)}{2}\)
51 + 52 + 53 + ... + 92: (1 + 2 + 3 + ...+ 92) - (1 + 2 +...+ 50)
\(=\frac{92 \times(92+1)}{2}-\frac{50 \times(50+1)}{2} \)
\(=\frac{92 \times 93}{2}-\frac{50 \times 51}{2} \)
= 4278 - 1275 = 3003
51 + 52 + 53 +...+ 92 = 3003

\(\frac { { x }^{ 3 } }{ x-y } +\frac { { y }^{ 3 } }{ y-x } =\frac { { x }^{ 3 } }{ x-y } -\frac { { y }^{ 3 } }{ y-x } =\frac { { x }^{ 3 }-{ y }^{ 3 } }{ (x-y) } \)
\(=\frac { (x-y)\left( { x }^{ 2 }+xy+{ y }^{ 2 } \right) }{ x-y } \)
x² + xy + y²

 
Here each term is divided by 10. Hence , the next three terms are
\({ a }_{ 4 }=\frac { 0.01 }{ 10 } =0.001\)
\({ a }_{ 5 }=\frac { 0.001 }{ 10 } =0.0001\)
\({ a }_{ 6 }=\frac { 0.0001 }{ 10 } =0.00001\)

The standard deviation of a given data is 4.5. If we subtract some fixed constant from all the data, the standard deviation will not change.
Each value of the data decreased by 5, the new standard deviation will not change.
New standard deviation = 4.5

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

\(y=\frac { 3 }{ 5 } \)
\(y=0x+\frac { 3 }{ 5 } \therefore Slope\quad m=0\)
(or) Comparing with ax + by + c = 0
\(\text { Slope } =-\frac{a}{b} \)
\(=-\frac{0}{5}=0 \)

Slope of a line passing through (- 2,3) and (8, 5) is 
\(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{3-5}{-2-8}=\frac{-2}{-10}=\frac{1}{5}=m_{1}\)
Slope of the line y = ax + 2 is 'a' = m₂
Given that the lines are perpendicular
\(m_{1} \times m_{2} =-1 \)
\(\frac{1}{5} \times a =-1 \)
a = -5

Range R = 13.67
Largest value L = 70.08
Range R = L - S
13.67 = 70.08-S
S = 70.08 - 13.67 = 56.41
Therefore, the smallest value is 56.41

A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f:A \(\longrightarrow \)N f(n) = the highest prime factor of n \(\in \) A
f(9) = 3 , f(10) = 5, f(11) = 11
f(12) = 3, f(13) = 3, f(14) = 7
f(15) = 5, f(16) = 2, f(17) = 7
Set of ordered pairs
f = {(9, 3), (10, 5), (11,11), (12, 3),(13, 13), (14, 7), (15, 5), (16, 2),(17, 17)}
Range = {2, 3, 5, 7, 11, 13, 17}

Given C(t) = 3t
To prove that C(t) is linear
Now, C(at₁) = 3at₁ and C(bt₂) = 3bt₂
C(at₁) +C(bt₂) = 3at₁ + 3bt₂= a(3t₁) +b(3t₂)
 = aC(t₁) + bC(t₂)  
Superposition principle is satisfied
Hence C(t) = 3t is linear.

External radius of hollow cylinder R = 16 cm
length h = 13 cm
Thickness R - r = 4
16 - r = 4
r = 12 cm
Total surface area of hollow cylinder \(=2 \pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r}+\mathrm{h}) \text { sq. units }\)
\(=2 \times \frac{22}{7} \times(16+12)(4+13) \)
\(=2 \times \frac{22}{7} \times 28 \times 17 \)
= 2992 sq. cm

Let f(x) = 21x²y
g(x) = 35 xy²
GCD = 7 xy
LCM of 21, 35 = 105
LCM of x²y, xy² = x²y²
LCM = 105 x²y²
Now, f(x) x g(x) = (21 x²y) (35 xy²)
= 735 x³ y³
LCM x GCD = (105 x²y²) (7 xy)
= 735 x³ y³
f(x) x g(x) = LCM x GCD
Hence verified

We have AB = 56.cm, AD = 14. cm, AC = 72. cm and AE = 18.cm.
BD = AB - AD = 5.6 –1.4 = 4.2 cm
and EC = AC – AE = 7.2–1.8 = 5.4 cm
\(\frac { AD }{ DB } =\frac { 1.4 }{ 4.2 } =\frac { 1 }{ 3 } \) and \(\frac { AE }{ EC } =\frac { 1.8 }{ 5.4 } =\frac { 1 }{ 3 } \)
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC. Hence proved.

\(\Delta AOQ\) and \(\Delta BOP,\angle OAQ=\angle OBP=90^{ 0 } \)
\(\angle AOQ=\angle BOP\) (Vertically opposite angles)
Therefore, by AA Criterion of similarity,
\(\Delta AOQ\sim \Delta BOP\)
\(\frac { AO }{ BO } =\frac { OQ }{ OP } =\frac { AQ }{ BP } \)
\(\frac { 10 }{ 6 } =\frac { AQ }{ 9 } \) gives \(AQ=\frac { 10\times 9 }{ 6 } =15cm\)

Let h₁ and h₂ be the heights of a cone and cylinder respectively.
Also, let r be the radius of the cone.
Given that, height of the cone h₁ = 24 cm; radius of the cone and cylinder r = 6 cm
Since, Volume of cylinder = Volume of cone
\({ \pi r }^{ 2 }=\frac { 1 }{ 3 } { \pi r }^{ 2 }{ h }_{ 1 }\)
\({ h }_{ 2 }=\frac { 1 }{ 3 } \times { h }_{ 1 }\quad gives\quad { h }_{ 2 }=\frac { 1 }{ 3 } \times 24=8\)
Therefore, height of cylinder is 8 cm