Question

sin⍬+ cos⍬=√2 cos(90°-⍬),show that cot⍬= √2 − 1​

Answer 1

Required Answer:-

Given:

• sin θ + cos θ = √2 cos(90° - θ)

To Prove:

• cot θ = √2 - 1.

Proof:

Given,

➡ sin θ + cos θ = √2 cos(90° - θ)

➡ sin θ + cos θ = √2 sin θ (As cos(90° - θ) = sin θ)

➡ cos θ = √2 sin θ - sin θ

➡ cos θ = sin θ(√2 - 1)

➡ cos θ/sin θ = √2 - 1

➡ 1/tan θ = √2 - 1 (tan θ = sin θ/cos θ)

➡ cot θ = √2 - 1 (cot θ = 1/tan θ)

Hence Proved.

Formula Used:

• cos(90° - θ) = sin θ
• sin θ/cos θ = tan θ
• tan θ = 1/cot θ

Answer 2

Given:-

• sin θ + cos θ = √2 cos(90° - θ)

To Prove:-

• cot θ = √2 - 1.

Proof:-

Given,

➡ sin θ + cos θ = √2 cos(90° - θ)

➡ sin θ + cos θ = √2 sin θ (As cos(90° - θ) = sin θ)

➡ cos θ = √2 sin θ - sin θ

➡ cos θ = sin θ(√2 - 1)

➡ cos θ/sin θ = √2 - 1

➡ 1/tan θ = √2 - 1 (tan θ = sin θ/cos θ)

➡ cot θ = √2 - 1 (cot θ = 1/tan θ)

Hence Proved.

Formula Used:

• cos(90° - θ) = sin θ
• sin θ/cos θ = tan θ
• tan θ = 1/cot θ