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 } )(tan
2
A−tan
2
B)=
cos
2
Acos
2
B
(sin
2
A−sin
2
B)
​
=
cos
2
Bcos
2
A
(cos
2
B−cos
2
A
​
)
​

Answer 1

Answer:

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

9(1²) – 9(a + b)1 + (2a² + 5ab + 2b²)

9×1-9(tan-cos)1+(2tan²+5tan.cos+2cos²)

9×8(sin/c/os-co/s)1+(2s/in/cos2+5si/n/c/os.c/os+2cos²)

/-it define wrong

Step-by-step explanation:

9×8(sin)1+(2cos²+5)

72(sin)1+(cos+5)

72(sin/cos)6

sin/cos=72/6

tan=12

Answer 2

Answer:

hope it helps you dear xd