Chat with us, powered by LiveChat Trigonometry Part 2(Solution): Advance math 2000 important questions.

+91-9817390373 , +91-9729327755( 9am-5pm | Mon to Sat) | [email protected]

## 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.

## Subscribe For Latest Updates

Signup for our newsletter and get notified when we publish new articles for free!

error: Content is protected !!