RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 4 of 120

Marks: +1, -0

I f

4x =secθand 4/X= tan θ

, then

8 [x^2-1/ x^2]

is

$1 /16$
$1/ 8$
$1 /2$
$1 /4$

Solution:

Given 4x

= sec θ ⇒ x =\text"secθ"/\text"4"

4/x

= tan θ
⇒

1/x

={tanθ}/4

∴ 8[x^2=-1/x^2]=8 [{sec^2θ}/16- {tan^2θ} /16]=1/2

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