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Question

If x=3 sin θ and y=4 cos θ, find the value of 16x2+9y2.


Solution

x=3 sin θ
Squaring both sides
   x2=9 sin2 θ
sin2 θ=x29     ...(i)
And y=4 cos θ
Squaring both sides
y2=16 cos2 θ
cos2θ=y216    ...(ii)
On adding eq. (i) and eq. (ii)
sin2θ+cos2θ=x29+y216
                     1=x29+y216  (Since, sin2θ+cos2θ=1)
                     1=16x2+9y2144
                     16x2+9y2=144
                     16x2+9y2=144
                     16x2+9y2=12

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