Question

prove that

$\frac{ \sin \: a \tan \: a }{1 - \cos \: a} = 1 + \sec \: a$

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Answer 1

Answer:

$\frac{ \sin \: a \: \tan \: a }{1 - \cos \: a } = \frac{ { \sin \: a }^{2} }{ \cos \: a \: (1 - \cos \: a) } \\ = \frac{1 - { \cos}^{2} \: a}{ \cos \: a \: ( 1 - \cos \: a) } \\ = \frac{1 + \cos \: a }{ \cos \: a } \\ = \frac{ \sin \: a \: \tan \: a }{1 - \cos \: a} = 1 + \sec \: a$

hence proved

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Answer 2

Answer:

In this type of question first you have to change tan \a\a

in \bold{ \frac{sina}{cosa} }

cosa

sina

form then you have to use the suitable identity.

\bold{ tan \: a \: = \frac{sin \: a}{cos \: a} }tana=

cosa

sina

Identity used :-

\bold{{ \sin}^{2} a = 1 - {cos}^{2} a }sin

2

a=1−cos

2

a

Dear!! User kindly refer to attachment.

Step-by-step explanation:

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