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Question

If(a2b2)sinθ+2abcosθ=a2+b2findtanθ

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Solution

(a2b2)sinθ+2abcosθ=a2+b2(dividingbycosθ)

= (a2b2)sinθcosθ+2abcosθcosθ=a2+b2cosθ

= (a2b2)tanθ+2ab=(a2+b2)secθ

= squaring both side we get

= (a2b2)2tan2θ+4a2b2+4ab(a2b2)tanθ=(a2+b2)2sec2θ

= (a2b2)2tan2θ+4a2b2+4ab(a2b2)tanθ=(a2+b2)2(1+tan2θ)

= [(a2b2)2(a2+b2)2]tan2θ+4a2b2+4ab(a2b2)tanθ(a2+b2)2=0

= [(a2b2+a2+b2)(a2b2a2b2)]tan2θ+4ab(a2b2)tanθ+4a2b2(a4+b4+2a2b2)=0

4a2b2tan2θ+4ab(a2b2)tanθ+2a2b2a4b4=0

+tan2θ(a2b2ab)tanθ12+a24b2+b24a2=0

tan2θ+(b2a2ab)tanθ12+a24b2+b24a2=0

puttanθ=u,then

u2+(b2a2ab)u12+a24b2+b24a2=0

u=(b2a2ab)±(b2a2ab)24(a24b2+b24a212)2

=a2b2ab±4(b2a2)24(a4+b4)2a2b24a2b22

=a2b2ab±4b4+4a48a2b24a44b4+8a2b24a2b22

tanθ=a2b22ab

1001633_1044605_ans_4e6c2dfec2d14752ae12b6ada2de6e0d.png

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