Question
Find the value of the given trigonometric
expression: (sin 22°cos 68° + cos²22°) × sin 30° + (cos 60°tan 45°) × sec 60°Solution
(sin 22°cos 68° + cos²22°) × sin 30° + (cos 60°tan 45°) × sec 60° = {sin 22°cos (90° – 22°) + cos²22°} × sin 30° + cos 60°tan 45°sec 60° Using, cos (90° – A) = sin A = (sin²22° + cos²22°) sin 30° + cos 60°tan45°sec 60° Using, sin²A + cos²A = 1 = {1 × (1/2)} + {(1/2) × 1 × 2} = (1/2) + 1 = (3/2)
I. 6y2 – 23y + 20 = 0
II. 4x2 – 24 x + 35 = 0
I. 117x² + 250x + 117 = 0
II. 54y² -123y + 65 = 0
The equation x2 – px – 60 = 0, has two roots ‘a’ and ‘b’ such that (a – b) = 17 and p > 0. If a series starts with ‘p’ such...
I. 3p² + 13p + 14 = 0
II. 8q² + 26q + 21 = 0
I. 6 y² + 11 y – 7= 0
II. 21 x² + 5 x – 6 = 0
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I.8(x+3) +Â 8(-x) =72Â
II. 5(y + 5) + 5(-y) = 150Â
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 2x² - 8x + 6 = 0
Equation 2: y² - 7y + 10 = 0
I. 10x2 + 33x + 9 = 0
II. 2y2 + 13y + 21 = 0
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