Question

if(a²-b²)sinA + 2abcosA=a²+b²,then prove that tanA=a²-b²/2ab​

Answer 1

Step-by-step explanation:

Given If (a²-b²)sin A + 2ab cos A=a²+b²,then prove that tan A=a²-b²/2 ab

• Given (a^2 – b^2)sin A + 2ab cos A = a^2 + b^2
•   Now a^2 + b^2 – (a^2 – b^2)sin A = 2ab cos A
• Squaring both sides we get
• (a^2 + b^2)^2 + (a^2 – b^2)^2 sin^2 A – 2(a^2 + b^2)(a^2 – b^2)sin A = 4a^2b^2 cos^2 A
• (a^2 + b^2) + (a^2 – b^2) sin^2 A – 2(a^2 + b^2)(a^2 – b^2) sinA – 4a^2b^2(1 – sin^2 A) = 0
• [a^2 – b^2)^2 + 4a^2b^2] sin^2A – 2(a^2 + b^2)(a^2 – b^2)sinA + [(a^2 + b^2)^2 – 4a^2b^2] = 0
• (a^2 + b^2)sin A – (a^2 – b^2)]^2 = 0
• So (a^2 + b^2) sin A – (a^2 – b^2) = 0
• Or sin A = a^2 – b^2 / a^2 + b^2
• Or cosec A = a^2 + b^2 / a^2 – b^2
• We know that cot^2 A = cosec^2 A – 1
•      So cot^2 A = (a^2 + b^2 / a^2 – b^2)^2 – 1
• So cot^2 A = (a^2 + b^2)^2 – (a^2 – b^2)^2 / (a^2 – b^2)^2
•                 = (a^2 + b^2 – a^2 + b^2)(a^2 + b^2 + a^2 – b^2) / (a^2 – b^2)^2
•                  = (2b)^2 (2a)^2 / (a^2 – b^2)^2
•                  = 4a^2b^2 / (a^2 - b^2)^2
• So cot A = √4 a^2b^2 / (a^2 – b^2)^2
•                = 2ab / a^2 – b^2
• Or tan A = a^2 – b^2 / 2ab (since cot A = 1/tan A)

Reference link will be

https://brainly.in/question/54471

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