RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

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Question : 18 of 120

Marks: +1, -0

If sin θ – cos θ

= 7 /13

and 0 < θ < 90°, then the value of sin θ + cos θ is.

$17 /13$
$13 /17$
$1 /13$
$1 /17$

Solution:

(a)

(sin θ + cos θ)^2 + (sin θ– cos θ)^2= (sin^2 θ+ cos^2 θ) + 2 sin θ. cos θ+ (sin^2 θ+ cos^2 θ) – 2 sin θ. cos θ.

= 1 + 1 = 2

So, (sin θ+ cos θ)^2 + (sin θ– cos θ)^2 = 2

or,

(sin θ + cos θ)^2 +[7 /13]^2=2

or,

(sin θ – cos θ)^2 = 2 –49/169= 289/ 169

sin θ + cos θ =√{[17/13]^ 2}= 17 /13

.

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