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Trignometric Ratios, Functions and Equations Test - 1

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Trignometric Ratios, Functions and Equations Test - 1
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  • Question 1
    2 / -0.83

    cos 68 °cos 8 °+ sin 68 °sin 8 °= ?

    Solution

    We know, 
    cosA cosB + sinA sinB = cos(A-B)

    cos 68 °cos 8 °+ sin 68 °sin 8 °= Cos (68-8) = Cos60 °
    =1/2

  • Question 2
    2 / -0.83

    In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2  is equal to:

    Solution

    In triangle ABC,
    ►  Sum of all three angles = 1800
    ►  A + B + C = 180 = A + B = 180 - C  

    devide by 2 both side  

    A/2 + B/2 = 180/2 - C/2



  • Question 3
    2 / -0.83

    In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals:

    Solution

    cosec A (sin B cos C + cos B sin C) = cosec A ×sin(B+C)
    = cosec A ×sin(180 –A)
    = cosec A ×sin A
    = cosec A ×1/cosec A
    = 1

  • Question 4
    2 / -0.83

    The value of Cos75 °is equal to

    Solution

       

  • Question 5
    2 / -0.83

    Sin A = 1/√10 , Sin B = 1/√5  If A and B are both acute angles, then, A+B =?

    Solution

    We know that:
    Sin θ = Opposite / Hypotenuse

    ∴SinA = 1/√10
    CosA= 3/√10
    similarly, SinB = 1/√5
    CosB= 2/√5

    Multiply:
    Cos(A+B)= CosA x CosB - SinA x SinB

    Substituting the value in above equation we get:

    = 3/√10 x 2/√5 - 1/√10 x 1/√5
    = 6/√50 - 1/√50
    = 6-1/5 √2. ........(√50=5 √2)
    = 1/ √2

    we know that, sin 45 = 1/ √ 2 therefore
    sinθ / cosθ = 45

  • Question 6
    2 / -0.83

    If 3 ×tan(x –15) = tan(x + 15), then the value of x is:

    Solution

    3 ×tan (x –15) = tan (x + 15)
    ⇒tan(x + 15) / tan(x –15) = 3/1

    ►{tan (x + 15) + tan (x –15)} / {tan (x + 15) –tan (x –15)} = (3 + 1) / (3 –1)

    ►{tan (x + 15) + tan (x –15)} / {tan (x + 15) –tan (x –15)} = 2

    sin(x + 15 + x –15) / sin(x + 15 –x + 15) = 2
    sin 2x / sin 30 = 2
    sin 2x / (1/2) = 2
    2 ×sin 2x = 2
    Sin 2x = 1

    ∴Sin 2x = Sin 90
    2x = 90
    x = 45

  • Question 7
    2 / -0.83

    The value of tan 20 °×tan 40 °×tan 80 °is

    Solution

  • Question 8
    2 / -0.83

    Chose which of the following expressions equals sinA + cosA.

    Solution

    sinA + cosA = √2(sinA/√2+cosA/√2)
    =√2(sinAcos(π/4)+cosAsin(π/4))
    =√2sin(A+π/4)
    by using sin(A+B) formula

  • Question 9
    2 / -0.83

    If acos x + bsin x = c, then the value of (asin x –bcos x)² is:

    Solution

    (acos x + bsin x)²+ (asin x –bcos x)²= a ²+ b ²
    ⇒c ²+ (asin x –bcos x)²= a ²+ b ²
    ⇒(asin x –bcos x)²= a ²+ b ²–c ²

  • Question 10
    2 / -0.83

    sin(60 °+ A) cos(30 °–B) + cos(60 °+ A) sin(30 °–B) is equal to:

    Solution

    L.H.S. = sin(60+A)cos(30 −B)+cos(60+A)sin(30 −B)     
    = sin[(60+A)+(30 −B)]       (Using, sin(A+B)sinAcosB+cosAsinB)       
    = sin(90+A −B)       
    = sin(90+(A −B))       
    = cos(A −B)       (Using, sin(90+θ)=cos θ)    = R.H.S.Hence Proved.

  • Question 11
    2 / -0.83

    If cos a + 2cos b + cos c = 2 then a, b, c are in

    Solution

    cos A + 2 cos B + cos C = 2
    ⇒cos A + cos C = 2(1 –cos B)

    2 cos((A + C)/2) ×cos((A-C)/2 = 4 sin ²(B/2)

    2 sin(B/2) cos((A-C)/2) = 4 sin ²(B/2) 
    cos((A-C)/2) = 2 sin (B/2)
    cos((A-C)/2) = 2 cos((A+C)/2)
    cos((A-C)/2) –cos((A+C)/2) = cos((A+C)/2)

    2 sin(A/2) sin(C/2) = sin(B/2)
    ⇒2 {√(s-b)(s-c)√bc} ×{√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac

    2(s –b) = b
    a + b + c –2b = b
    a + c –b = b
    a + c = 2b


  • Question 12
    2 / -0.83

    sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A =

    Solution

    sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

  • Question 13
    2 / -0.83

    cos(π/4 - x) cos (π/4 - y) -sin(π/4 - x) sin(π/4 - y) =

    Solution

    Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

    = CosA*Cos B - Sin A*Sin B
    = Cos (A+B)
    = cos(π/4-x+π/4-y)
    = cos(π/2-x-y)
    = cos{π/2 - (x+y)}
    = sin(x+y)

  • Question 14
    2 / -0.83

    The value of tan 3A –tan 2A –tan A is

    Solution

    3A= A+ 2A
    ⇒ tan 3A = tan (A + 2A)
    ⇒ tan 3A = (tan A + tan 2A) / (1 –tan A . tan 2A)
    ⇒ tan A + tan 2A = tan 3A –tan 3A x  tan 2A . tan A
    ⇒ tan 3 A –tan 2A –tan A = tan 3A . tan 2A . tan A

  • Question 15
    2 / -0.83

    Which of the following is incorrect:

    Solution

    D is not a trigonometric identity.

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