RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 6 of 120

Marks: +1, -0

I f

sin θ+cosθ=√2 cos(90 - θ)

, then cot θis

$√2 +1$
0
$√ 2$
$√2 -1$

Solution:

sin θ + cos θ = √2 cos(90 - θ) = √2 sin θ

Squaring both side

sin^2 θ + cos^2 θ + 2sin θ cos θ = 2sin^2 θ

cos^2 θ = sin^2 θ – 2 cos θ sin θ

Divide by

sin^2 θ

cot ^2 θ = 1 – 2 cot θ

cot ^2 θ + 2cot θ – 1 = 0

cot θ =

{-2 ± √{4 + 4}}/2

cot θ

= √2 - 1 or cot θ = -√ 2 - 1

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