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If(a^2-b^2)sinA+2abcosA=(a^2+b^2), find the value of tanA.

If(a^2-b^2)sinA+2abcosA=(a^2+b^2), find the value of tanA.

Grade:10

2 Answers

Vikas TU
14149 Points
7 years ago
There can be many methods for this:
one is:

Divide both sides by cos(A) to get: 

(a^2 - b^2)tan(A) + 2ab = (a^2 + b^2)sec(A) 

Square both sides: 

(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * sec^2(A) 

Since 1 + tan^2(A) = sec^2(A): 

(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * (1 + tan^2(A)) 

((a^2 - b^2)^2 - (a^2 + b^2)^2)tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 - (a^2 + b^2)^2 = 0 

(-4a^2b^2)tan^2(A) + 4ab(a^2 - b^2)tan(A) + 2a^2b^2 - a^4 - b^4 = 0 

tan^2(A) + (b^2 - a^2)/(ab) * tan(A) - 1/2 + (a^2)/(4b^2) + (b^2)/(4a^2) = 0 

Let u = tan(A) 

u = [(a^2 - b^2)/(ab) +/- sqrt((b^2 - a^2)^2/(a^2b^2) - 4((a^2)/(4b^2) + (b^2)/(4a^2) - 1/2))]/2 

u = ((a^2 - b^2)/(ab) +/- sqrt((b^2/(a^2) - 2 + (a^2)/(b^2) - (a^2)/(b^2) - (b^2)/(a^2) + 2)/2 

u = tan(A) = (a/b - b/a +/- sqrt(0))/2 

tan(A) = (a/b - b/a)/2 = (a^2 - b^2)/(2ab) 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
Divide both sides by cos(A) to get: 
(a^2 - b^2)tan(A) + 2ab = (a^2 + b^2)sec(A) 
Square both sides: 
(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * sec^2(A) 
Since 1 + tan^2(A) = sec^2(A): 
(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * (1 + tan^2(A)) 
((a^2 - b^2)^2 - (a^2 + b^2)^2)tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 - (a^2 + b^2)^2 = 0 
(-4a^2b^2)tan^2(A) + 4ab(a^2 - b^2)tan(A) + 2a^2b^2 - a^4 - b^4 = 0 
tan^2(A) + (b^2 - a^2)/(ab) * tan(A) - 1/2 + (a^2)/(4b^2) + (b^2)/(4a^2) = 0 
Let u = tan(A) 
u = [(a^2 - b^2)/(ab) +/- sqrt((b^2 - a^2)^2/(a^2b^2) - 4((a^2)/(4b^2) + (b^2)/(4a^2) - 1/2))]/2 
u = ((a^2 - b^2)/(ab) +/- sqrt((b^2/(a^2) - 2 + (a^2)/(b^2) - (a^2)/(b^2) - (b^2)/(a^2) + 2)/2 
u = tan(A) = (a/b - b/a +/- sqrt(0))/2
tan(A) = (a/b - b/a)/2 = (a^2 - b^2)/(2ab)
 
Thanks and regards,
Kushagra

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