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Trigonometry Part 2(Solution): Advance math series 2000 most important questions.

Trigonometry Part 2(Solution): Advance math series 2000 most important questions. Solution 41. (b) Given expression is  Solution 42. (a) sinθ+sin(θ+120°)+sin(θ+240°) is equal to 0. Solution 43.(d) Given 2sin2θ+cos245° and 0°≤θ≤90° Solution 44.(c) Given 2sin2x+cos245° and 0°≤x≤90° Solution 45.(d) From Pythagoras theorem ; AC2 = AB2 + BC2 ; (17)2 = AB2 + (8)2  ; AB2 = 225 ;

AB = 15 Solution 46.(b) In first quadrant, sinθ is positive; Solution 47. (b) Here  y = 2 sin t cos t  ; → y2 = 4 sin2 t cos2 t

→ y2 = 4x2 (1 – cos2t)  ;  → y2 = 4x2 – 4x4

∴ x= cot t ; → y2= 4x2– 4x2

Solution 48. (b) The value of is 1 Solution 49.(a) The value of θ is 30° Solution 50.(b) Solution 51.(a) Solution 52. (c)  Since, Sin2 θ + cos2 θ =1 ∴  Sin2 38° + cos2 38°=1

Solution 53. (a) 8 tan A = 15, → tan A =15/8 ; Solution 54.(c) Given that x = a cos2θsinθ , y=  a sin2θcosθ then Solution 55.(a) Solution 56.(b) Solution 57.(c) Solution 58.(b)

. Solution 59.(d) x2 + y2 + z2 = (3 cosA. CosB)2 + (3 cosA. sinB)2 + (3 sinA)2

= 9cos2 A cos2B + 9 cos2A sin2B + 9 sin2 A;= 9cos2 A (cos2B + sin2B) + 9 sin2 A

= 9 cos2 A1+9 sin2A (∴sin2 B + cos2B = 1 ); = 9 (cos2 A + sin2A) (∴cos2A + sin2 A = 1) ;= 91=9

Solution 60.(d) Solution 61.(c) Solution 62. (d) Solution 63.(d) Solution 64.(d) Solution 65. (b) Given lines are x cos a + y sin a = p and x cos β+ysinβ=p1

Slope of line x cos a + y sin a = p is Slope of line x cos β+ysinβ=p1 is Let the angle between the lines be θ Solution 66.(b) Solution 67.(d) Given   2 [(sin6 x + cos6x) + t (sin4x + cos4x) = – 1

= 2 (sin2x + cos2x)3 – 3 sin2x cos2x (sin2x + cos2x)]+ t [(sin2x + cos2x)2 – 2sin2x cos2x)] = – 1

A3 + b3  = (a + b)3 – 3 ab (a + b)  ;  Use a2 + b2 = (a + b)2 – 2 ab

Where a = sin2 x, b = cos2x  ;   → 2 [1 – 3 sin2x cos2x] + t [ 1 – 2 sin2x cos2x] = – 1

2 – 6 sin2x cos2x + t 2t sin2x cos2x = – 1

→ t (1 – 2 sin2x cos2x) = – 3 (1 – 2 sin2 x cos2x) → t = – 3

Solution 68.(b) Solution 69.(c) Solution 70.(b) Area fo parallelogram: Solution 71.(d) Solution 72.(a) The value of sin θ + cos θ lies between -√2 and √2

Hence equation has no solution.

Solution 73.(a) a3 + b3 + c3 – 3abc

= (a3 + b3 + c3– ab – bc – ca) (a + b + c) = [(a2 + b2 + c2 ) – (ab+ bc + ca)] (a + b + c)

But, a2 + b2 + c2 = ab + bc + ca (Given)

∴a3 + b3 + c3 – 3abc= (a + b + c) *0=0  ∴a3 + b3 + c3 = 3abc

Solution 74.(c) Solution 75. (b) Solution 76.(a) Solution 77.(d) Solution 78.(c) Solution 79.(a) Solution 80.(b) In first quadrant cosθ decreases when θ  increases.