Question

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

Answer 1

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Answer 2

Solution:

Given LHS = $\frac{1+secA-tanA}{1+secA+tanA}$

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

/* By Trigonometric identity:

sec²A - tan²A = 1 */

= $\frac{[(secA+tanA)(secA-tanA)+(secA-tanA)]}{(1+secA+tanA)}$

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

= $\frac{(secA-tanA)(secA+tanA+1)}{(secA+tanA+1)}$

After cancellation, we get

= $secA-tanA$

= $\frac{1}{cosA}-\frac{sinA}{cosA}$

= $\frac{(1-sinA)}{cosA}$

= RHS

Therefore,

$\frac{1+secA-tanA}{1+secA+tanA}$=$\frac{(1-sinA)}{cosA}$

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