Prove that #cosB/(1-sinB)=secB+tanB#?

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Answer 1 · 2016-09-10T14:05:11

Please see below.

#cosB/(1-sinB)#

= #cosB/(1-sinB)xx(1+sinB)/(1+sinB)#

= #(cosB(1+sinB))/(1-sin^2B)#

= #(cosB(1+sinB))/cos^2B#

= #(1+sinB)/cosB#

= #1/cosB+sinB/cosB#

= #secB+tanB#