RRB NTPC Mains or Stage 2 Model Paper 1 with solutions

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Question : 39 of 120

Marks: +1, -0

If cot

x = 5/ 12

, then sin

x + 1 /{cot x }sec x + x = ?

$173/ 13$
$77 /13$
$75 /13$
$17/ 13$

Solution:

(2) cot x =

5/12

∴tan

x =12/5

∴ sec x = √{1+ tan^2 x}

= √{1+[12/5]^2}

= √{1+144/25}=√{{25 +144}/25}

=√{169/25}

=13/5

∴ sin x ={tanx}/{secx} =12/5×5/13

=12/13

∴sinx + tanx + secx
=

12/13+12/5+13/5

[∵ 1/ {cot x}=tan x]

=

12/13+25/5 =12/13+ 5

={12 +65}/13=77/13

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