Question

Plz solve this problem of trigonometryi am an aakashian​

Answer 1

Step-by-step explanation:

$\bf L.H.S = \tt \dfrac{sec\: \theta + tan \: \theta - 1}{tan \: \theta - sec \: \theta + 1}$

$: \implies \tt \dfrac{\frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1}{ \frac{sin \: \theta}{cos \: \theta} - \frac{1}{cos \: \theta} + 1 } \: = \dfrac{1 + sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta}$

$: \implies \tt\dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta + (cos \: \theta - 1)} \: \times \: \dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta - (cos \: \theta - 1)}$

$: \implies \tt\dfrac{ sin^{2} \: \theta + cos^{2} \: \theta + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: (cos \: \theta - 1)}{sin^{2} \: \theta - (cos \: \theta - 1)^{2} }$

$: \implies \tt\dfrac{1 + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - sin^{2} \: \theta - cos^{2} \: \theta + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2} \: \theta}$

$: \implies \tt\dfrac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: \dfrac{1 + sin \: \theta}{cos \: \theta}$

$: \implies\tt\dfrac{1 + sin \: \theta}{cos \: \theta} \: \times \: \dfrac{1 - sin \: \theta}{1 - sin \: \theta}$

$: \implies\tt\dfrac{1 + sin^{2} \: \theta}{cos \: (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos^{2} \: \theta}{cos \: \theta (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos \: \theta}{1 - sin \: \theta} \: = \: \bf{ R.H.S}$

$\huge\bigstar \:\underline{\red{\sf Hence, Proved}} \: \bigstar$