Question

prove it :tanA-secA+1/tanA+secA-1=sinA-1/cosA​

Answer 1

Question :

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

Answer :

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

PROOF :

We know that,

${sec}^{2} (a) - {tan}^{2} (a) = 1$

now, here by using this property we can replace the Value of 1 in numerator.

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

$\frac{tan(a) - \sec(a) + ( {tan}(a) + \sec(a))( \tan(a) - \sec(a)) }{ \tan(a) + \sec(a) + 1 }$

Now, by taking , (Tan A - Sec A) common , we get,

$\frac{( \tan(a) - \sec(a))(1 + \tan(a) + \sec(a)) }{(1 + \tan(a) + \sec(a)) }$

Now, by cancelling common terms in both numerator and denominator , we get

$\implies \: \tan(a) - \sec(a)$

We know that,

$\tan(a) = \frac{ \sin(a) }{ \ \cos(a) }$

$\sec(a) = \frac{1}{ \cos(a) }$

NOW, by substituting these values in the eq. we get,,,

$\implies \: \frac{ \sin(a) }{ \cos(a) } - \frac{1}{ \cos(a) }$

$\implies \: \frac{( \sin(a) - 1)}{ \cos(a) }$

Therefore,now LHS=RHS and

hence proved that

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

Answer 2

i hope it is helpful for u