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Question

If sinθ=x2y2x2+y2, then find the values of cosθ and 1tanθ

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Solution

we know that
sinθ=sideoppositetoangleθhypotenuseorsinθ=Perpendicularhypotenuse
sinθ=x2y2x2+y2PH=x2y2x2+y2ABBC=x2y2x2+y2
side opposite to angle θ=AB=x2y2
hypotenuse AC=x2+y2
In right angled ABC, we have
(AB)2+(BC)2=(AC)2(x2y2)2+(BC)2=(x2+y2)2(BC)2=(x2+y2)2(x2y2)2 (By Pythagoras theorem)
(BC)2=[x2+y2+x2y2][x2+y2(x2y2)]=(2x2)(2y2)
(using identity a2b2=(a+b)(ab))
BC=4x2y2=±2xy
taking positie square root since, side cannot be negative
cosθ=Basehypotenuse=BCAC=2xyx2+y2
and tanθ=PerpendicularBase=ABBC=x2y22xy
so, 1tanθ=1x2y22xy=2xyx2y2
1791327_1814100_ans_7b0f8a67b59c42b3a027a5270e57da9d.PNG

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