RRB Group D Model Paper 7 with solutions

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Question : 86 of 100

Marks: +1, -0

If 0° < θ < 90° and 2 sec θ = 3

cosec^2

θ, then θ is

$π/6$
$π/4$
$π/3$
$π/5$

Solution:

(3) Given, 22sec θ = 3

cosec^2

θ

2/{cosθ} = 3/{sin^2θ}

⇒

2/{cosθ}=3/{1-cos^2θ}

2

cos^2θ

+ 3cosθ - 2 = 0
2

cos^2θ

+ 4cosθ - cosθ - 2 = 0
2 cosθ (cosθ + 2) - 1(cosθ + 2) = 0
and (2 cosθ - 1)(cosθ + 2) = 0
So, cosθ =

1/2

= cos60° ⇒ θ =

π/3

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