SSC CHSL 9 Jan 2017 Shift 3 Paper

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Question : 57 of 100

Marks: +1, -0

[{SinA}/{(1+CosA)}]+[({1+CosA})/{SinA}]

is equal to

2secA
2cosecA
2tanA
2cotA

Solution:

Expression :

[{SinA}/{(1+CosA)}]

+[({1+CosA})/{SinA}]

Taking L.C.M, we get :
=

{(sin^2A)+(1+cosA)^2}/{sinA(1+cosA)}

=

{sin^2 A+cos^2 A+2cosA+1}/{sinA(1+cosA)}

Using

(sin^2A + cos^2A = 1)

=

{2+2cosA}/{sinA(1+cosA)}

={2(1+cosA)}/{sinA(1+cosA)}

=

2/{sinA}

= 2cosecA

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