Question

please solve the above, the solution given in the book is very confusing. please do not copy paste since I really need to understand this.​

Answer 1

Answer:

this the answer for your question

Answer 2

Answer:

$Given: {sec}^{2} \theta = 1 + {tan}^{2} \theta \\ \therefore \: {sec}^{2} \theta - {tan}^{2} \theta = 1 \\ \therefore \: ({sec} \theta + {tan} \theta)({sec} \theta - {tan} \theta) = 1 \\ \implies ({sec} \theta + {tan} \theta) = \frac{1}{({sec} \theta - {tan} \theta) } \\ \implies \frac{({sec} \theta + {tan} \theta) }{1} = \frac{1}{({sec} \theta - {tan} \theta) } \\ let \: \frac{({sec} \theta + {tan} \theta) }{1} = \frac{1}{({sec} \theta - {tan} \theta) } = k \: (constant) \\ \therefore \: k = \frac{({sec} \theta + {tan} \theta) }{1} = \frac{1}{({sec} \theta - {tan} \theta) } .....(1) \\ \therefore \: k = \frac{1 + {sin} \theta}{{cos\theta}} = \frac{cos\theta}{1 - {sin} \theta} \\ multiplying \: numerator \: and \: \\ denominator \: by \: (- 1) \: to \: extreme \: \\ right \: ratio \: we \: find : \\k = \frac{1 + {sin} \theta}{{cos\theta}} = \frac{( - 1) \times cos\theta}{ - 1(1 - {sin} \theta) } \\ k = \frac{1 + {sin} \theta}{{cos\theta}} = \frac{ - cos\theta}{ - 1 + {sin} \theta} \\ \: by \: theorem \: on \: equal \: ratios \: we \: have : \\ k = \frac{1 + {sin} \theta- cos\theta}{{cos\theta}- 1 + {sin} \theta} \\ \implies \: k = \frac{ {sin} \theta- cos\theta + 1}{{sin} \theta + {cos\theta}- 1 } .....(2) \\ hence \: from \: (1) \: and \: (2) \: we \: find: \\ \frac{ {sin} \theta- cos\theta + 1}{{sin} \theta + {cos\theta}- 1 } = \frac{1}{({sec} \theta - {tan} \theta) } \\ thus \: proved$