Question
If tanθ – cotθ = a and cosθ + sinθ = b, then
(b2 –1)(a2 + 4) = ?Solution
Put θ = 45°
then, a = tanθ – cotθ = 1 –1 = 0
and, b = cosθ + sinθ
= 1/√2 + 1/√2 =√2
Now, (b 2 –1) (a 2 + 4) = (2 – 1) (0 + 4) = 4
l. p2Â - 3p - 54 = 0
II. q2Â - 19q + 90 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 31x² - 170x + 216 = 0
Equation 2: 22y² - 132y + ...
I. 2x2 + 13x + 21 = 0
II. 3y2 + 34y + 63 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 32x + 207 = 0
Equation 2: y² - 51y + 648 = 0
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0Â
I. 3p² - 17p + 22 = 0
II. 5q² - 21q + 22 = 0
I. 8x – 3y = 85
II. 4x – 5y = 67
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 13x² - 60x + 47 = 0
Equation 2: 17y² - 80y + 63 = 0
I. 5x2 – 7x – 6 =0
II. 2y2 – 5y – 7 =0
I. Â x2 = 10 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
II. Â y2 - 9y + 20 = 0
...
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