Question

please solve the following attachment...​

Answer 1

Step-by-step explanation:

Hey there..Here's the Solution.

1. Prove:- √1-cos∅/√1+cos∅ = Cosec∅-Cot∅

2. Prove:- 1+sin∅/1-sin∅ = Cot²∅/(cosec∅-1)²

→Solⁿ1

=> Squaring both sides

=> 1-cos∅/1+cos∅ = (Cosec∅-Cot∅)²

=> Taking RHS.

=> (Cosec∅-cot∅)²= cosec²∅-2cosec∅cot∅+

cot²∅

[ converting into sin and cos]

=> 1/Sin²∅ - 2×1/sin×cos/sin + cos²∅/sin²

=> 1/Sin²∅-2cos∅/sin²∅+ cos²∅/sin²∅

[ since all denominators are sin²∅ we will simply write the numerators with their signs]

=> 1-2cos∅+cos²∅/sin²∅

Converting numerator into (a-b)² and sin²∅ into 1-cos²∅ as sin²∅+cos²∅=1

= (1-cos∅)²/1-cos²∅

Converting 1-cos²∅ into (a+b)(a-b)

=> (1-cos∅)²/(1+cos∅)(1-cos∅)

Cancelling square in numerator and 1-cos∅ in denominator

= 1-cos∅/1+cos∅ = LHS

Hence Proved!!!

→Solⁿ2

=> taking RHS

=> cot²∅/(cosec∅-1)²=cot²∅/cosec²∅ -2cosec∅+1

[ converting into Sin and cos]

=> cos²∅/sin²∅/1/sin²∅-2/sin∅+1

=> cos²∅/sin²∅/1+sin²∅/sin²∅-2/sin∅

=> cos²∅/sin²∅/1+sin²∅-2sin/sin²∅

[Since Cos²∅= 1-sin²∅ and 1+sin²-2sin=(a-b)²]

=> 1-sin²∅/sin²∅ × Sin²∅/(1-sin∅)²

[sin²∅ and sin²∅ get cancelled]

= 1-sin²∅/(1-sin∅)²

[since 1-sin²∅= a²-b²= (a+b)(a-b)]

=> (1+sin∅)(1-sin∅)/(1-sin∅)²

= 1+sin∅/1-sin∅=LHS

Hence Proved!!!!

I HOPE IT HELPS YOU.... :) :)