Question

Prove that tan theta + sec theta -1 upon tan theta -sec theta +1 = sec theta + tan theta

Answer 1

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Answer 2

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

Solution:

Given that, we have to prove:

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

Take the LHS

$\frac{tan\ \theta + sec\ \theta -1}{tan\ \theta - sec\ \theta + 1}$

We know that,

$sec^2 \theta -tan^2 \theta = 1$

Therefore,

$\frac{tan\ \theta + sec\ \theta -(sec^2\theta - tan^2\theta)}{tan\ \theta - sec\ \theta + 1}\\\\Expand\ by\ a^2 - b^2 = (a+b)(a-b)\\\\\frac{tan\ \theta + sec\ \theta - (( sec \theta + tan \theta)(sec\ \theta - tan\ \theta))}{tan\ \theta - sec\ \theta + 1}\\\\Take\ sec \theta + tan \theta\ as\ common\\\\\frac{sec \theta + tan \theta(1-(sec\ \theta - tan\ \theta)}{tan\ \theta - sec\ \theta + 1}\\\\\frac{sec \theta + tan \theta( 1 - sec\ \theta + tan \theta)}{tan\ \theta - sec\ \theta + 1}$

$sec\ \theta + tan\ \theta$

Thus,

LHS = RHS

Thus proved

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