Prove that: (1 + sec θ - tan θ) / (1 + sec θ + tan θ) = (1

Q) Prove that : $\frac{1 + \sec \theta - \tan \theta} {1 + \sec \theta + \tan \theta} = \frac{1 - \sin \theta} {\cos \theta}$

Ans:

Let’s start from simplifying the LHS:

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

= $\frac{(1 + \frac{1}{\cos \theta} - \frac{\sin \theta)}{\cos \theta} )} {(1 + \frac{1}{\cos \theta} + \frac{\sin \theta)}{\cos \theta} )}$

= $\frac{\cos \theta + 1 - \sin \theta} {\cos \theta + 1 + \sin \theta}$

Since, we need to get demnominator in simplified form, hence let’s multiply nominator and denominator by cos θ + sin θ – 1, we get:

LHS = $\frac{\cos \theta - \sin \theta + 1} {\cos \theta + \sin \theta + 1} \times \frac{\cos \theta + \sin \theta - 1}{\cos \theta + \sin \theta -1}$

= $\frac{(\cos \theta + 1 - \sin \theta)\times (\cos \theta + \sin \theta - 1 )}{(\cos \theta + 1 + \sin \theta) \times ((\cos \theta + \sin \theta - 1)}$

= $\frac{(\cos \theta + (1 - \sin \theta))\times (\cos \theta - (1 - \sin \theta))}{((\cos \theta + \sin \theta) + 1) \times ((\cos \theta + \sin \theta) - 1)}$

= $\frac{(\cos^2 \theta - (1 - \sin \theta)^2)}{((\cos \theta + \sin \theta)^2 - 1)}$

= $\frac{(\cos^2 \theta - (1 + \sin^2 \theta - 2 \sin \theta))}{((\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta) - 1)}$

= $\frac{(\cos^2 \theta - 1 - \sin^2 \theta + 2 \sin \theta)}{((\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta) - 1)}$

We know that sin2 θ + cos2 θ = 1

∴ LHS = $\frac{(\cos^2 \theta - (\cos^2 \theta + \sin^2 \theta) - \sin^2 \theta + 2 \sin \theta)}{((1 + 2 \sin \theta \cos \theta) - 1)}$

= $\frac{(\cos^2 \theta - \cos^2 \theta - \sin^2 \theta - \sin^2 \theta + 2 \sin \theta)}{(1 + 2 \sin \theta \cos \theta - 1)}$

= $\frac{(\cancel{\cos^2 \theta} - \cancel{\cos^2 \theta} - \sin^2 \theta - \sin^2 \theta + 2 \sin \theta)}{(\cancel{1} + 2 \sin \theta \cos \theta - \cancel{1})}$