Question

Prove that : secA + tanA - 1/tanA - secA + 1 =1 - sinA/cos A= cos A / 1 -sinA​

Answer 1

We have to prove that

(secA + tanA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA = cosA/(1 - sinA)

solution : we know, sec²A - tan²A = 1

⇒(secA - tanA)(secA + tanA) = 1 ......(1)

LHS = (secA + tanA - 1)/(tanA - secA + 1)

= {secA + tanA - (sec²A - tan²A)}/(tanA - secA + 1) [ from equation (1) . ]

= {secA + tanA - (secA + tanA)(secA - tanA)}/(tanA - secA + 1)

= {(secA + tanA)(1 - secA + tanA)}/(tanA - secA + 1)

= (secA + tanA)

= 1/cosA + sinA/cosA

= (1 + sinA)/cosA .......(2)

again, (1 + sinA)/cosA = (1 + sinA)/sinA × (1 - sinA)/(1 - sinA)

= (1 + sinA)(1 - sinA)/cosA(1 - sinA)

= (1 - sin²A)/cosA(1 - sinA)

= cos²A/cosA(1 - sinA)

= cosA/(1 - sinA) ........(3)

from equations (2) and (3) we get,

(secA + tanA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA = cosA/(1 - sinA)

[ hence proved ]

Answer 2

$\frac{tanA + secA - 1}{1 + tanA - secA} \\ \\ = \frac{secA + tanA - 1}{( {sec}^{2}A - {tan}^{2} A) - (secA - tanA) } \\ \\ = \frac{secA + tanA - 1}{(secA - tanA)(secA + tanA - 1)} \\ \\ = \frac{1}{secA - tanA} \\ \\ = \frac{1}{ \frac{1 - sinA}{cosA} } \\ \\ = \frac{cosA}{1 - sinA}$