Question

Prove that 1+sina/1-sina=seca+tana​

Answer 1

✯ The question will be To prove :

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

Solution

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Taking LHS that is

$\frac{1 + \sin(a) }{1 - \sin(a) }$

Multiple and divide Nr and Dr with

1+ sin a because it becomes 1 and there will be no affect on the fraction after multiplying by 1

$\frac{1 + \sin(a) }{1 - \sin(a) } \times \frac{1 + \sin(a) }{1 + \sin(a) }$

$\frac{( {1 + \sin(a) )}^{2} }{1 - { \sin(a)) }^{2} }$

And we know that1- sin²a = cos²a .So ,

$\frac{( {1 + \sin(a)) }^{2} }{ { \cos(a) }^{2} }$

Now writing 1 and sin²a separately.

$( { \frac{1}{ \cos(a) } + \frac{ \sin(a) }{ \cos(a) } )}^{2}$

We know that

• 1/cos a. = sec a
• sin a/cos a= tan a

So replacing these values we got

LHS = ( seca + tan a )² = RHS

hence proved.