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SSC CGL 2022 PYQ — Question 74

SSC CGL - 06 Dec 2022 - Shift 1 • Quantitative Aptitude • Q74

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

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Correct answer: A

Explanation

The correct answer is Option 1 i.e. b+ab−a\sqrt{{b+a} \over b-a}

Given:

b sin θ  = a

sin θ  = a/b

Concept Used:

sin θ  = PHP\over H

sec θ = HBH \over B

tanθ = PBP \over B

Explanation:

According to the given question,

P = a, H = b

So, B = H2−P2\sqrt{H^2 - P^2} = b2−a2\sqrt{b^2 - a^2}

Now, sec θ + tan θ = bb2−a2+ab2−a2 {b \over \sqrt{b^2 - a^2}} + {a \over \sqrt{b^2 - a^2}} 

⇒ b+ab2−a2{b + a} \over \sqrt{b^2 - a^2} = b+ab+ab−ab+a{\sqrt{b +a} \sqrt{b + a}} \over {\sqrt{b - a} \sqrt{b + a}} = b+ab−a\sqrt{{b+a} \over b-a}

Hence the value of sec θ + tan θ  is b+ab−a\sqrt{{b+a} \over b-a}