EVALUATE THE GIVEN TRIGONOMETRIC FUNCTION BY TRIGONOMETRIC RATIOS

Question 1 :

If cos A  =  3/5, then find the value of

(sin A - cos A)/2 tan A

Solution :

cos A  =  3/5  =  Adjacent side / Hypotenuse side

(Opposite side)²  =  (Hypotenuse side)² - (Adjacent side)²

  =  5² - 3²

(Opposite side)²  =  25 - 9  =  16

Opposite side  =  √16  =  4

sin A  =  opposite side/ hypotenuse side  =  4/5

tan A  =  opposite side / Adjacent side

tan A  =  4/3

(sin A - cos A)/2 tan A  =  (4/5) - (3/5) / 2(4/3)

  =  (1/5) / (8/3)

  =  3/40

Question 2 :

If cos A  =  2x/(1 + x²), then find the values of sin A and tan A in terms of x.

Solution :

cos A  =  2x/(1 + x²)  =  Adjacent side / Hypotenuse side

Opposite side²  =  (1 + x²)² - (2x)²

  =  1 + x⁴ + 2x² - 4x²

  =  1 + x⁴ - 2x²

  =  (1 - x²)²

Opposite side =  1 - x²

sin A  =  Opposite side / Hypotenuse side

sin A  =   (1 - x²)/ (1 + x²)

tan A  =  Opposite side / Adjacent side

tan A  =   (1 - x²)/2x

Question 3 :

If sin  θ  =  a/√a² + b², then show that

b sin θ  =  a cos θ

Solution :

Given that, sin  θ  =  a/√a² + b² 

sin θ  =  Opposite side / Hypotenuse side

(Adjacent side)²  =  (Hypotenuse)² - (Opposite side)²

(Adjacent side)²  =  (√a² + b² )² - (a)²

Adjacent side  =  b

b sin θ  =  ab/√a² + b²  ----(1)

cos θ  =  b/√a² + b²

a cos θ  =  ab/√a² + b² ----(2)

(1)  =  (2)

Hence proved.

Question 4 :

If 3cot A = 2 , then find the value of

(4sin A - 3cos A)/(2sin A + 3cos A)

Solution :

cot A  =  2/3  =  Adjacent side / Opposite side

(Hypotenuse side)²  =  (Opposite side)² + (Adjacent side)²

(Hypotenuse side)²  = 3² + 2²

Hypotenuse side  =  √13

sin A  =  Opposite side / Hypotenuse side

sin A  =  3/√13

cos A  =  Adjacent side / Hypotenuse side

cos A  =  2/√13

(4 sin A - 3 cos A)/(2 sin A + 3 cos A) :

  =  [4(3/√13) - 3(2/√13)]/[2(3/√13) + 3(2/√13)]

  =  [(12-6)/√13]/[(6+6)/√13]

  =  6/12

  =  1/2

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