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HomeMathConsider the Given Trigonometric Operate by Trigonometric Ratios

# Consider the Given Trigonometric Operate by Trigonometric Ratios

Query 1 :

If cos A  =  3/5, then discover the worth of

(sin A – cos A)/2 tan A

Resolution :

cos A  =  3/5  =  Adjoining facet / Hypotenuse facet

(Reverse facet)2  =  (Hypotenuse facet)2 – (Adjoining facet)2

=  52 – 32

(Reverse facet)2  =  25 – 9  =  16

Reverse facet  =  √16  =  4

sin A  =  reverse facet/ hypotenuse facet  =  4/5

tan A  =  reverse facet / Adjoining facet

tan A  =  4/3

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

=  (1/5) / (8/3)

=  3/40

Query 2 :

If cos A  =  2x/(1 + x2), then discover the values of sin A and tan A by way of x.

Resolution :

cos A  =  2x/(1 + x2)  =  Adjoining facet / Hypotenuse facet

Reverse facet2  =  (1 + x2)2 – (2x)2

=  1 + x4 + 2x2 – 4x2

=  1 + x4 – 2x2

=  (1 – x2)2

Reverse facet =  1 – x2

sin A  =  Reverse facet / Hypotenuse facet

sin A  =   (1 – x2)/ (1 + x2)

tan A  =  Reverse facet / Adjoining facet

tan A  =   (1 – x2)/2x

Query 3 :

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

b sin θ  =  a cos θ

Resolution :

Provided that, sin  θ  =  a/√a2 + b

sin θ  =  Reverse facet / Hypotenuse facet

(Adjoining facet)2  =  (Hypotenuse)2 – (Reverse facet)2

(Adjoining facet)2  =  (√a2 + b)2 – (a)2

Adjoining facet  =  b

b sin θ  =  ab/√a2 + b —-(1)

cos θ  =  b/√a2 + b2

a cos θ  =  ab/√a2 + b2 —-(2)

(1)  =  (2)

Therefore proved.

Query 4 :

If 3cot A = 2 , then discover the worth of

(4sin A – 3cos A)/(2sin A + 3cos A)

Resolution :

cot A  =  2/3  =  Adjoining facet / Reverse facet

(Hypotenuse facet) =  (Reverse facet)2 + (Adjoining facet)2

(Hypotenuse facet) = 32 + 22

Hypotenuse facet  =  √13

sin A  =  Reverse facet / Hypotenuse facet

sin A  =  3/√13

cos A  =  Adjoining facet / Hypotenuse facet

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

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