Question

[ Question ]

1.) Using quadratic formula , solve for 'x'

9x² – 9(a + b)x + (2a² + 5ab + 2b²) = 0

2.) Prove

$(tan ^{2} A - tan ^{2} B) = \frac{(sin ^{2}A - sin ^{2} B )}{cos ^{2} A \: cos ^{2}B } = \frac{(cos ^{2}B - cos ^{2} A }{cos ^{2}B \: cos ^{2} A } )$
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Answer 1

refer Attachment For Q.2

1)

9x² – 9(a + b)x + (2a² + 5ab + 2b²) = 0

Comparing with general form of quadratic equation ax² + bx + c = 0.

We get, a = 9, b = -9(a + b) and

c = (2a² + 5ab + 2b²)

now,

b² - 4ac = [-9(a + b)]² - 4 × 9 × (2a² + 5ab + 2b²)

= 81(a² + 2ab + b²) - 36(2a² + 5ab + 2b²)

= 81a² + 162ab + 81b² - 72a² - 180ab - 72b²

= 9a² - 18ab + 9b²

= (3a - 3b)²

•°• b² - 4ac = (3a - 3b)²

•°• √(b² - 4ac) = √((3a - 3b)² .....(1)

Now,

x = [-b +- √(b² - 4ac) ] / 2a

= [- (-9(a + b)) +- √((3a - 3b)²) ] / 2 × 9

= [ 9a + 9b +- (3a - 3b) ] / 18

•°• x = (9a + 9b + 3a - 3b) / 18

or x = [ 9a + 9b - (3a - 3b) ] / 18

•°• x = 3(3a + 3b + a - b) / 18

or x = 3(3a + 3b - a + b) / 18

•°• x = (4a + 2b)/6 or x = (2a + 4b)/6

•°• x = 2(2a + b)/6 or x = 2(a + 2b)/6

•°• x = (2a + b)/3 or x = (a + 2b)/3

Answer:-

x = (2a + b)/3 or x = (a + 2b)/3

Answer 2

ANSWER:

1) x = (2a + b)/3 , x = (a + 2b) / 3

FORMULA USED:

$x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

EXPLANATION:(REFER ATTACHMENT)

2) TO PROVE:

${(tan ^{2} A - tan ^{2} B) = \frac{(sin ^{2}A - sin ^{2} B )}{cos ^{2} A \: cos ^{2}B } = \frac{cos ^{2}B - cos ^{2} A }{cos ^{2}B \: cos ^{2} A } }$

FORMULAE:

★ Tan x = Sin x / cos x

★ Sin² x = 1 - cos² x

★ Cos² x = 1 - sin² x

PROOF:

${(tan ^{2} A - tan ^{2} B) = \dfrac{sin ^{2}A}{ {cos}^{2}A }- \dfrac {sin ^{2} B }{ cos ^{2}B } }$

$\dfrac{sin ^{2}A \: cos ^{2}B -sin ^{2} B \: {cos}^{2}A }{ {cos}^{2}A \: {cos}^{2}B}$

Sin² A cos² B = (1 - cos² A)cos² B

Sin² A cos² B = cos² B - cos² Acos² B

- Sin² B cos² A = - (1 - cos² B)cos² A

Sin² B cos² A = - cos² A + cos² Bcos² A

Sin² A cos² B - Sin² B cos² A = cos² B - cos² A

${(tan ^{2} A - tan ^{2} B) = \dfrac{cos ^{2}B - cos ^{2} A }{cos ^{2}B \: cos ^{2} A } }$

Cos² B - cos² A = 1 - sin² B - 1 + sin² A

Cos² B - cos² A = sin² A - sin² B

${(tan ^{2} A - tan ^{2} B) = \dfrac{sin ^{2}A - sin ^{2} B }{cos ^{2} A \: cos ^{2}B }}$

HENCE PROVED.