Question

CosA/1+sinA=secA-tanA

Answer 1

Answer:

Step-by-step explanation:

CosA/1+sinA=secA-tanA

taking LHS

multiply and divide it by ( 1- sinA)

We get

= CosA(1-SinA)/(1+sinA)(1-SinA)

= CosA(1-SinA)/(1-Sin^2A)                                          ;1-Sin^2A= Cos^2A

= CosA(1-SinA)/Cos^2A

=SecA-TanA

and divide we get RHS

Best of Luck For Examination!

Answer 2

$solvng \: l.h.s.\\ \frac{ \cos \: a}{1 + \sin \: a } \\ = > \frac{ \cos \: a }{1 + \sin \: a} \times \frac{1 - \sin \: a }{1 - \sin \: a} \\ = > \frac{ \cos \: a(1 - \sin \: a ) }{1 - { \sin }^{2} a} \\ = > \frac{ \cos \: a(1 - \sin \: a) }{ { \cos}^{2} a} \\ = > \frac{1 - \sin \: a}{ \cos \: a } \\ = > \frac{1}{ \cos \: a } - \frac{ \sin \: a }{ \cos \: a} \\ = > \sec \: a - \tan \: a = r.h.s.$

hope this helps.

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