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Matrix Test 150
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Matrix Test 150

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  • Question 1/10
    1 / -0.25
    Find the value of  
    Solutions


  • Question 2/10
    1 / -0.25
    If cosecθ−sinθ = a3, secθ−cosθ = b3; then a2b2(a2+b2) is equal to
    Solutions


    …(i)

    …(ii)
    Put in Eq. (i),

    …(iii)
    Similarly, ….(iv)
    Squaring and adding Eqs. (iii) and (iv), we get
    We have,
  • Question 3/10
    1 / -0.25
    If 5tanθ = 4; then find the value of
    Solutions
  • Question 4/10
    1 / -0.25
    The minimum value of cos2θ + sec2θ is
    Solutions
    We know that the minimum value of acos2x+bsec2x =
    Hence minimum value of cos2 θ + sec2θ = = 2
  • Question 5/10
    1 / -0.25
    If sin8θ + cos8θ – 1 = 0, then what is the value of cos2θ.sin2θ (If θ ≠ 0 or π/2)?
    Solutions

    sin8 θ + cos8 θ – 1 = 0
    ⇒ sin2 θ + cos2 θ - sin8 θ – cos8 θ = 0
    ⇒ sin2 θ(1-sin6θ) + cos2θ(1-cos6θ) = 0
    ⇒ sin2 θ(1-sin2θ) (1+sin2θ+sin4θ)+ cos2θ(1-cos2θ) (1+cos2θ+cos4θ) = 0
    ⇒ sin2θ cos2θ (1+sin2θ+sin4θ)+ sin2θ cos2θ (1+cos2θ+cos4θ) = 0
    ⇒ sin2θ cos2θ(1+sin2θ+sin4θ + 1+cos2θ+cos4θ) = 0
    ⇒ sin2θ cos2θ(3+ sin4θ + cos4θ) = 0 
    ⇒ sin2θ cos2θ(3+ 1 - 2 sin2θ cos2θ) = 0 
    ⇒ sin2θ cos2θ(4 - 2 sin2θ cos2θ) = 0 

    since, θ ≠ 0 or π/2
    sin2θ.cos2θ ≠ 0
    So, 4 - 2 sin2θ cos2θ = 0
    ⇒ sin2θ cos2θ = 2 

  • Question 6/10
    1 / -0.25
    The value of (sin21° + sin23° + sin25° +.....+ sin287° + sin289°) is
    Solutions
    We know that sin(90-x) =cosx & sin2x + cos2x= 1
    Hence sin21 ° + sin23 ° + sin25 ° + ….+ sin287 ° + sin289 °
    = sin2(90 – 89) ° + sin2(90 – 87) ° + sin2(90 – 85) ° + …. sin2(90 – 47) ° + sin245 ° + sin2 47 °+ sin287 ° + sin289 °
    = cos289 ° + cos287 ° + cos2 85 ° + ….+ cos2 47 °+ sin2 45 °+sin2 47 °+….+sin287 ° + sin289 °
    = (cos289 °+ sin289 °)+(cos287 °+ sin287 °) +…+ (cos247 °+ sin247 °)+ sin2 45 °
    = 22 + 1/2 =
  • Question 7/10
    1 / -0.25
    If P lies in the first quadrant and 4tan P = 3, then the value of  will be:
    Solutions

    Let us divide numerator and denominator of the given expression by cosP.

    ∴

    Put tanP = 3/4

    =

    Hence, option A is the correct answer.

  • Question 8/10
    1 / -0.25
    Simplify:
    Solutions

    =

    =  = 1

    Hence, option B is the correct answer.

  • Question 9/10
    1 / -0.25
    if θ is the positive acute angle and sin θ – cos θ = 0, then find value of sec θ + cosec θ
    Solutions

    Given, sinθ – cosθ = 0 ⇒ sin θ = cos θ

    ⇒     tanθ = 1

    ⇒ θ = 45o

    So, sec 45o + cosec 45o

    = √2 + √2 

    = 2√2.
  • Question 10/10
    1 / -0.25
    Simplify the following expression: 
    Solutions

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