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

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 = \(\frac { sin\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } =\frac { tan\theta -1+sec\theta }{ tan\theta +1-sec\theta } \)
= \(\frac { (tan\theta +sec\theta )-1 }{ (tan\theta -sec\theta )+1 } \)
= \(\frac { \{ (tan\theta +sec\theta )-1\} (tan\theta -sec\theta ) }{ \{ tan\theta -sec\theta )+1\} (tan\theta -sec\theta ) } \)
= \(\frac { ({ tan }^{ 2 }\theta -{ sec }^{ 2 }\theta )-(tan\theta -sec\theta ) }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1-tan\theta +sec\theta }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1 }{ tan\theta -sec\theta } \)
= \(\frac { 1 }{ sec\theta -tan\theta } \)

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)

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

An arrow diagram
 

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.

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
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

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

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.

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.

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

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.