RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

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Question : 19 of 120

Marks: +1, -0

If 3 cos θ= 5 sin θ, then the value of

{5sinθ- 2sec^3θ +2cosθ} /{5sinθ+ 2sec3θ- 2cosθ}

is equal to

$271/ 979$
$376/ 2937$
$542/ 2937$
None of these

Solution:

(a) Given, 3 cosθ = 5 sin θ ⇒ tan θ =

3/5

.
sec θ =

√{1 +tan^2 θ}=√{1 +[3/5]^2}

=√{{25+ 9}/25} =√34/5

In expression, dividing the numerator &denominator by cos θ,

={5 tan θ 2sec^4θ+ 2}/{5tan θ+2sec^4θ- 2}

{5×3/5/-2×[√34/5]^4+2}/{5×3/5+2×[√34/5]^4-2}

{3-2×1156/625+2}/{3+2×156/625-2}={5- 2312/625}{1 +2312/625}

=

813 /2937=271/ 979

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