Question

Prove that tan A -1+sec A / tan A +1 -sec A = sec a + tan A

Answer 1

To prove:

$\sf \dfrac{tan \: a - 1 + sec \: a}{tan \: a \: + 1 - sec \: a} = sec \: a + tan \: a$

Identities to be used in doing :

• 1 + tan² A = sec² A
• a² - b² = ( a + b) ( a - b)

Using identity

$\tt \because 1 + {tan}^{2} a = {sec}^{2} a \\ \therefore \tt1 = {sec}^{2} a - {tan}^{2} a$

$\tt\frac{tan\: a - ({sec}^{2}a - {tan}^{2}a) + sec \: a }{tan \: a + 1 - sec \: a} \\ \\ \bf using \: identity \\ \\ \bf {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ \implies \tt \frac{tan \: a + sec \: a - \{(sec \: a + tan \: a)(sec \: a - tan \: a)\}}{tan \: a + 1 - sec \: a}$

Now , Taking Tan A + Sec A common

$\implies \ \tt\frac{tan \: a + sec \: a(1 - (sec \: a - tan \: a)}{tan \: a + 1 - sec \: a} \\ \\ \implies \tt \frac{tan \: a + sec \: a(1 - sec \: a + tan \: a)}{tan \: a + 1 - sec \: a} \\ \\ \tt \implies \frac{tan \: a + sec \: a \cancel{(tan \: a + 1 - sec \: a)}} {\cancel{tan \: a + 1 - sec \: a}} \\ \\ \implies \boxed{\boxed{\tt tan \: a + sec \: a}}$

Hence Proved