If sin θ = 6/10, find the value of cos θ + tan θ.
Answer :
Consider ∆ ABC be right angled at B and ∠ACB = θ
It is given that
sin θ = AB/AC
⇒ sin θ = 6/10
Take AB = 6x then AC = 10x
In right angled ∆ ABC
AC2 = AB2 + BC2
(10x)2 = (6x)2 + BC2
BC2 = 100x2 – 36x2 = 64x2
BC2 = (8x)2
⇒ BC = 8x
In right angled ∆ ABC
cos θ = base/hypotenuse
⇒ cos θ = BC/AC
cos θ = 8x/10x = 4/5
In right angled ∆ ABC
tan θ = perpendicular/base
⇒ tan θ = AB/BC
tan θ = 6x/8x = ¾
Here
cos θ + tan θ = 4/5 + ¾
Taking LCM
= (16 + 15)/ 20
= 31/20
= 1 (11/20)
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