If b sin θ = a, then sec θ + tan θ = ?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 06 Dec 2022 Shift 1)

Answer 1

Given:

b sin θ= a

Concept used:

sinθ = Perpendicular/ Hypotenuse

secθ = Hypotenuse/ Base

tanθ = Perpendicular/ Base

Calculation:

bsinθ = a

sinθ = a/b

So, Perpendicular = a & Hypotenuse = b

(Hypotenuse)²= (Perpendicular)²+ (Base)²

b²= a²+ Base²

base²= b²- a²

\(base = {√{b^2-a^2} }\)

secθ = Hypotenuse/ base= b/\({ √{b^2-a^2} }\)

tanθ= perpendicular/ base= a/\({ √{b^2-a^2} }\)

So, secθ+ tanθ = b/\({ √{b^2-a^2} }\)+ (a/\({ √{b^2-a^2} }\))

secθ + tanθ = (b+ a)/(\({ √{b^2-a^2} }\))

∵ (b + a) = √(b + a) × √(b + a)

⇒ secθ + tanθ = \( {√{(b+a)(b+a)} }\)/ \( { √{(b-a)(b+a)} }\)

secθ+ tanθ= \(√ {\frac{{b + a}}{{b - a}}} \)

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