Question

secA-1÷secA+1 = (sinA÷ 1+cosA)whole square
​

Answer 1

$\huge{\underline{\textbf{Solution:-}}}$

$\implies$sec A - 1 \ sec A + 1 

$\bigstar\sf{ Multiplying\:and\:dividing\:by\:sec A - 1}$

$\implies$(sec A - 1)² / (sec²A - 1) 

$\implies$1 × (sec²A + 1 - 2secA) → [sec²A - 1 = tan² A]

$\bigstar\sf{Dividing\:question\:by\:tan^{2}A}$

$\implies$cot²A (tan²A + 2 - 2secA) → [1 / tanA = cotA] 

$\implies$1 + 2cot²A - 2 × 1/cosA × cosA/sinA × cosA/sinA → [simply multiplying] 

$\implies$[cosA / sinA = cotA × 1/cosA = secA] 

$\implies$(1 + cot²A) + cot²A -2cotAcosecA 

$\implies$cosec²A + cot²A - 2cosec²Acot²A = cosec² A

$\implies$(cotA - cosecA)²

______________________________________________________________________________