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Question

If (a2b2)sinθ+2abcosθ=a2+b2, find tanθ.

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Solution

If, (a2b2)sinθ+2abcosθ=a2+b2
Let, x=a2b2 and y=a2+b2
Then, 4a2b2=(a2+b2)2(a2b2)2=y2x2
xsinθ+2abcosθ=y
(xsinθ+2abcosθ)2=y2
x2sin2θ+4a2b2cos2θ+4abxsinθcosθ=y2
x2sin2θ+(y2x2)cos2θ+4abxsinθcosθ=y2
x2sinθ+y2cos2θx2cos2θ+4abxsinθcosθ=y2
x2sin2θ+y2cos2θy2x2cos2θ+4abxsinθcosθ=0
x2sin2θ+y2(cos2θ1)x2cos2θ+4abxsinθcosθ=0
x2sin2θy2sin2θx2cos2θ+4abxsinθcosθ=0
sin2θ(y2x2)x2cos2θ+4abxsinθcosθ=0
4a2b2sin2θ+x2cos2θ4abxsinθcosθ=0
(2absinθxcosθ)2=0
2absinθ=xcosθ
sinθcosθ=x2ab
tanθ=a2b22ab
cosecθ=(a2b2)24a2b2a2b2
=a42a2b2+b4+4a2b2a2b2
a4+2a2b2+b4a2b2=(a2+b2)(a2b2)

1033644_1060511_ans_b3ffce46126f487b90ffb93c93d7b738.PNG

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