Question

Plz solve this problem of trigonometry
i am an aakashian​

Answer 1

Explanation:

`\bf L.H.S = \tt (sec\: \theta + tan \: \theta - 1)/(tan \: \theta - sec \: \theta + 1) \\ \\`

`: \implies \tt ((1)/(cos \: \theta) + (sin \: \theta)/(cos \: \theta) - 1)/( (sin \: \theta)/(cos \: \theta) - (1)/(cos \: \theta) + 1 ) \: = (1 + sin \: \theta - cos \: \theta)/(sin \: \theta + cos \: \theta) \\ \\`

`: \implies \tt( sin \: \theta - (cos \: \theta - 1))/(sin \: \theta + (cos \: \theta - 1)) \: * \: ( sin \: \theta - (cos \: \theta - 1))/(sin \: \theta - (cos \: \theta - 1)) \\ \\`

`: \implies \tt( 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(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(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(2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta))/( 2 \: cos \: \theta - 2 \: cos^(2) \: \theta) \\ \\`

`: \implies \tt\frac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: (1 + sin \: \theta)/(cos \: \theta) \\ \\`

`: \implies\tt(1 + sin \: \theta)/(cos \: \theta) \: * \: (1 - sin \: \theta)/(1 - sin \: \theta) \\ \\`

`: \implies\tt(1 + sin^(2) \: \theta)/(cos \: (1 - sin \: \theta)) \\ \\`

`: \implies\tt(cos^(2) \: \theta)/(cos \: \theta (1 - sin \: \theta)) \\ \\`

`: \implies\tt(cos \: \theta)/(1 - sin \: \theta) \: = \: \bf{ R.H.S}\\ \\`

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