In ABC, tan A = \(\frac{BC}{AB}\) = 1
BC = AB
Let AB = BC = k, where k is a positive number

Now, AC = \(\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
= \(\sqrt { { (k) }^{ 2 }+{ (k) }^{ 2 } } =k\sqrt { 2 } \)
Therefore,
\(sinA=\frac { BC }{ AC } =\frac { 1 }{ \sqrt { 2 } } \) and
\(cosA=\frac { AB }{ Ac } =\frac { 1 }{ \sqrt { 2 } } \)
So, \(2sinAcosA=2\left[ \frac { 1 }{ \sqrt { 2 } } \right] \left[ \frac { 1 }{ \sqrt { 2 } } \right] =1\), which is the required value

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

Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.
LHS = \(\frac { sin\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } =\frac { tan\theta -1+sec\theta }{ tan\theta +1-sec\theta } \)
= \(\frac { (tan\theta +sec\theta )-1 }{ (tan\theta -sec\theta )+1 } \)
= \(\frac { \{ (tan\theta +sec\theta )-1\} (tan\theta -sec\theta ) }{ \{ tan\theta -sec\theta )+1\} (tan\theta -sec\theta ) } \)
= \(\frac { ({ tan }^{ 2 }\theta -{ sec }^{ 2 }\theta )-(tan\theta -sec\theta ) }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1-tan\theta +sec\theta }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1 }{ tan\theta -sec\theta } \)
= \(\frac { 1 }{ sec\theta -tan\theta } \)

Let us first draw a right ΔABC.
Now, we know that tan A = \(\frac { BC }{ AB } =\frac { 4 }{ 3 } \)
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.

Now, by using the Pythagoras theorem, we have
AC² = AB² + BC²
= (4k)² + (3k)² = 25k²
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
\(sinA=\frac { BC }{ AC } =\frac { 4k }{ 5k } =\frac { 4 }{ 5 } \)
\(cosA=\frac { AB }{ AC } =\frac { 3k }{ 5k } =\frac { 3 }{ 5 } \)
Therefore, \(cotA=\frac { 1 }{ tanA } =\frac { 3 }{ 4 } \)
\(cosecA=\frac { 1 }{ sinA } =\frac { 5 }{ 4 } \) and,
\(secA=\frac { 1 }{ cosA } =\frac { 5 }{ 3 } \)

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),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

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.

σ =\(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
=\(\\ \sqrt { \frac { 20300 }{ 7 } -\left( \frac { 350 }{ 7 } \right) ^{ 2 } } \)
=\(\sqrt { 400 } \) = 20

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

An arrow diagram
 

A table

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

\(\frac { (k-12) }{ a } { x }^{ 2 }+\frac { 2(k-12) }{ b } x+\frac { 2 }{ c } =0\)
D² = b²- 4ac = (2(k - 12))² - 4(k - 12)(2)
= 4(k - 12)[(k - 12) - 2]
= 4(k-12)(k- 14)
The given equation will have equal roots, if D = 0
⇒ 4(k-12)(k-14) 0
k - 12 = 0 or k - 14 0
k 12, 14

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

x + y + z = 6            ....(1)
2x + 3y + 4z = 20     ...(2)
3x + 2y + 5z = 22   ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

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.

We have 6 = 2¹ x 3¹ and
20 = 2 x 2 x 5 = 2² x 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 x 2 x 3 x 5 = 60.
As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² x 3¹ x 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0  ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

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.

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

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

A = {1,2,3},B =  {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) =  2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(ii) An arrow diagram

(iv) A Graph f = \(\{(x, f(x) / x \in A\}\)
= {(0, 1), (1, 3), (2, 5), (3, 7)}

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

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.