Question

sinA-cosA+1/sinA+cosA-1​

Answer 1

$\huge {\bf {\underline {Answer}}}$$\tt LHS = \frac {sinA - cosA + 1}{sinA + cosA -1} \\ \tt \large =\frac {\frac{sinA}{cosA} - \frac{cosA}{cosA} + \frac{1}{cosA}} {\frac{sinA}{cosA} + \frac{cosA}{cosA} - \frac{1}{cosA}} \\ \tt =\frac{tanA - 1 + secA}{tanA + 1 - secA} \\ \tt =\frac{tanA + secA - 1}{tanA - secA + 1} \\ \tt =\frac{tanA + secA - (sec²A - tan²A)}{tanA - secA + 1} \\ \tt =\frac{tanA + secA - (secA + tanA)(secA - tanA)}{tanA - secA + 1} \\ \tt =\frac{(tanA + secA) [1 - (secA - tanA)]}{tanA - secA + 1} \\ \tt =\frac{(tanA + secA) (1 - secA + tanA)}{tanA - secA + 1} \\ \tt = tanA + secA \\ \tt =\frac{tanA + secA}{1} \\ \tt =\frac{tanA + secA}{sec²A - tan²A} \\ \tt =\frac{tanA + secA}{(secA - tanA)(secA + tanA)} \\ \tt =\frac{1}{secA - tanA} = RHS$

hope it helps dear... ☺️✌

Answer 2

$\huge {\bf {\underline {Answer}}}$

$\tt LHS = \frac {sinA - cosA + 1}{sinA + cosA -1} \\ \tt =\frac {\frac{sinA}{cosA} - \frac{cosA}{cosA} + \frac{1}{cosA}} {\frac{sinA}{cosA} + \frac{cosA}{cosA} - \frac{1}{cosA}} \\ \tt =\frac{tanA - 1 + secA}{tanA + 1 - secA} \\ \tt =\frac{tanA + secA - 1}{tanA - secA + 1} \\ \tt =\frac{tanA + secA - (sec²A - tan²A)}{tanA - secA + 1} \\ \tt =\frac{tanA + secA - (secA + tanA)(secA - tanA)}{tanA - secA + 1} \\ \tt =\frac{(tanA + secA) [1 - (secA - tanA)]}{tanA - secA + 1} \\ \tt =\frac{(tanA + secA) (1 - secA + tanA)}{tanA - secA + 1} \\ \tt = tanA + secA \\ \tt =\frac{tanA + secA}{1} \\ \tt =\frac{tanA + secA}{sec²A - tan²A} \\ \tt =\frac{tanA + secA}{(secA - tanA)(secA + tanA)} \\ \tt =\frac{1}{secA - tanA} = RHS$