SSC CGL 7 06 2019 Shift 2 Paper

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Question : 56 of 100

Marks: +1, -0

The value of

\ {1}/{\sec x-\tan x}-\ {1}/{\cos x}, 0° < x< 90°,

is equal to:

2secx
tanx
2cosx
cotx

Solution:

\ {1}/{\sec x-\tan x}-\ {1}/{\cos x}

\⇒\ {\cos x}/{1-\sin x}-\ {1}/{\cos x}

\⇒ \ {\cos x(1+\sin x)}/{(1-\sin x)(1+\sin x)}-\ {\cos x}/{\cos ^{2} x}

\⇒\ {\cos x+\sin x \cos x-\cos x}/{\cos ^{2} x}

\⇒\tan x

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