Math, asked by subususmi11gmailcom, 9 months ago

Prove the trigonometric identity:
If cos A + COS²A = 1. prove that sin²A - sin²A = 1​

Answers

Answered by Samriddhi2410
3

Answer:

Step-by-step explanation:

Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA= 2/sin²A-cos²A=2/1-2cos ²A

LHS = Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA

= {(Sin A+cosA)² + (Sin A - cosA)²} /(sin A -cosA)(sinA+cosA)

={ (sin²A+cos²A + 2sinA.cosA) + (sin²A+cos²A - 2sinA.cosA)} / (sin² A - cos²A)

= {(1+2sinA.cosA) + (1- 2sinA.cosA) / (sin² A - cos²A)

[sin²θ+cos²θ=1]

= 1 + 1 / (sin² A - cos²A)

LHS = 2 / (sin² A - cos²A) = 2/(1- cos²A - cos²A) = 2/ (1- 2cos²A) = RHS

[sin²θ = 1 - cos²θ]

HOPE THIS ANSWER WILL HELP YOU..

Answered by hrockyreddy2345
0

Answer:

Step-by-step explanation:

Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA= 2/sin²A-cos²A=2/1-2cos ²A

LHS = Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA

= {(Sin A+cosA)² + (Sin A - cosA)²} /(sin A -cosA)(sinA+cosA)

={ (sin²A+cos²A + 2sinA.cosA) + (sin²A+cos²A - 2sinA.cosA)} / (sin² A - cos²A)

= {(1+2sinA.cosA) + (1- 2sinA.cosA) / (sin² A - cos²A)

[sin²θ+cos²θ=1]

= 1 + 1 / (sin² A - cos²A)

LHS = 2 / (sin² A - cos²A) = 2/(1- cos²A - cos²A) = 2/ (1- 2cos²A) = RHS

[sin²θ = 1 - cos²θ]

Similar questions