Question

prove that tanO+secO-1/tanO-secO+1 = secO+tanO​

Answer 1

Step-by-step explanation:

tanO+secO-1/tanO-secO+1

=(secO+tanO)-(sec²O-tan²O)/(tanO-secO+1)

=(secO+tanO)(1-secO+tanO)/(tanO-secO+1)

=secO+tanO

Answer 2

Solution →

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

LHS →

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

We know that →

• ${ \sec }^{2} \theta - { \tan }^{2} \theta = 1$

So put the value of 1→

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

$= \frac{tan \: \theta \: + sec \:\theta \: - (sec \: \theta + tan \:\theta)(sec \: \theta - tan \: \theta) }{tan \: \theta - sec \:\theta + 1 }$

Now,

$tan \: \theta + sec \: \theta \: is \: common$

So,

$= \frac{(tan \: \theta + sec \: \theta)(1 - sec \: \theta + tan \: \theta)}{(1 - sec \: \theta + tan \: \theta)}$

$( 1 - sec \: \theta + tan \: \theta) \: will \: be \: cancelled$

$\boxed{\red{= tan \: \theta + sec \: \theta }}$ RHS

Hence Proved !!

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