RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

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Question : 19 of 120

Marks: +1, -0

If

{x cosec^2 30 . sec^2 45°}/{8cos^2 45°. sin^2 60 °} = tan^2 60° – tan^2 30º, then x = ?

1
– 1
2
0

Solution:

(a)

{x×2^ 2( √2)^2}/{8×[1/√2]^2 ×[√3/2]^2}=( √3)^2-[1/√ 3]^2

or,

{x×4× 2 }/{8×1/2×3/4} = 3 -1/3⇒{8x}/3= {9 -1}/3

or,

8 /3 x=8/3

x = 1

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