Question

1 + secA​/secA=sin2A/1-cosA​

Answer 1

Answer:

sorry, butvis the question correct,, I think it should be cos²A in the last portion.

then sin²A +cos² A = 1 will come

isn't it?

make brainlist pls

Answer 2

Question:-

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

Answer:-

• Let us solve LHS part , we can write a LHS as

$\large➦ \frac{1 + \sec \: A}{ \sec \: A} = \frac{ 1 + \frac{1 }{ \cos \:A } }{ \frac{1}{ \cos \:A } }$

$\large ➦ \frac{ \frac{ \cos A \: + 1}{ \cos \: A} }{ \frac{1}{cos \: A} }$

$➦ \: \frac{ \cos \:A \: + 1 }{ \cos \: A} = \frac{ \cos \: A }{1}$

$➦1 + \cos A$

$➦1 + \cos \: A \times \frac{1 - \cos A}{1 - \cos \: A}$

$➦ \frac{(1)^2 - (\cos)^2}{1 - \cos \:A }$

$➦\frac{1 - \cos^{2} \:A }{1 - \cos \: A}$

$➦ \frac{ \sin^{2}A}{1 - \cos \: A}$

$RHS = LHS$