**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

AC^{2} = AB^{2} + BC^{2}

(10x)^{2} = (6x)^{2} + BC^{2}

BC^{2} = 100x^{2} – 36x^{2} = 64x^{2}

BC^{2} = (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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