SSC CHSL Model Paper 15

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Question : 71 of 100

Marks: +1, -0

If

\cot\θ=3/4,

then

(\sin \θ + \cos \θ - \tan\θ)

is equal to :

$-1/20$
$2/15$
$1/20$
$1/15$

Solution:

\cot\θ=\BC/\AB={3k}/{4k}

i.e. BC = 3k, AB = 4k

∴\AC=√{\AB^2+\BC^2}

=√{(4k)^2+(3k)^2}

=√{25k^2}=5k

∴\sin\θ+\cos\θ-\tan\θ

=\AB/\AC+\BC/\AC-\AB/\BC

={4k}/{5k}+{3k}/{5k}-{4k}/{3k} =1/15

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