RRB NTPC 17 January 2017 Paper

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Question : 55 of 120

Marks: +1, -0

If sinx =

4/5

then cosec x + cot x =

$31 /12$
$35 /12$
2
$1 /2$

Solution:

(3) sin x =

4/5

, cosec x + cot x = ?

\⇒ \cosec x + \cot x = \ { 1 } /{ \sin x } + \ { \cos x } /{ \sin x } =

{ 1 + \cos x }/ { \sin x }

.......(i)
Now,

\sin ^ { 2 } x + \cos ^ { 2 } x = 1

⇒ ( { 4 } { 5 } ) ^ { 2 } + \cos ^ { 2 } x = 1

⇒ \cos ^ { 2 } x = 1 - { 16 }/ { 25 } = { 9 }/ { 25 }

\⇒ \cos x = { 3 } /{ 5 } , \sin x = { 4 }/ { 5 }

Substituting the values in equation(i), we get

\ { 1 + { 3 }/{ 5 } } /{ { 4 }/ { 5 } } = 2

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