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

Ans: 2 methods to solve this question:

1st Method:

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

\therefore \frac{\frac{\sin\theta}{\cos\theta} + \frac{1}{cos\theta} -1}{\frac{\sin\theta}{\cos\theta} - \frac{1}{cos\theta} - 1}

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

Multiplying numerator & denominator by (sin θ + cos θ + 1), we get:

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

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

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

We know that sin2 θ + cos2 θ = 1

\therefore \frac{(\sin^2\theta - \cos^2\theta) + 2\sin\theta + (\sin^2\theta + \cos^2\theta) } {2 \sin\theta\cos\theta}

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

=       \frac{(1 + \sin\theta)}{\cos\theta}         =         RHS             Hence Proved!

2nd Method:

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

We know that, 1 + tan2 θ = sec2 θ

\therefore      1 =  sec2 θ – tan2 θ

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

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

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

=       tan θ + sec θ

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

=       \frac{(1 + \sin\theta)}{\cos\theta}         =         RHS             Hence Proved!

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