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.
karthikeya , 9 Years ago
Grade 10
Trigonometry> If(a^2-b^2)sinA+2abcosA=(a^2+b^2), find t...
2 Answers
Vikas TU
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)
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)
Approved Last Activity: 9 Years ago
Kushagra Madhukar
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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