Question

Please proved that question

Answer 1

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Answer 2

Given,

asin^2¢ + bcos^2¢ = c

asin^2¢ + b(1-sin^2¢) = c

asin^2¢ + b - bsin^2¢ = c

sin^2¢(a-b) = c - b

sin^2¢ = (c - b) / (a - b)............equation 1.

Now, because sin^2¢ = 1 - cos^¢, so by equation 1....

1 - cos^2¢ = (c - b) / (a - b)

cos^2¢ = 1 - (c - b) / (a - b)

cos^2¢ = (a - b - c + b) / (a - b)

cos^2 ¢ = (a - c) / (a - b)........equation 2.

Now, if we divide equation 1 by equation 2...

sin^2¢ / cos^2¢ = (c - b) / (a - c)

tan^2¢ = (c - b) / (a - c)

tan ¢ = √{ (c - b) / (a - c) }.......●

Hope it was helpful.