x=cot theta+tan theta y=sec theta-cos theta then (x^2y) ^2/3-(xy^2) ^2/3=
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Answer:
The question should be:
prove that (x
2
y)
3
2
−(xy
2
)
3
2
=1
cotθ+tanθ=
tanθ
1+tan
2
θ
=
tanθ
sec
2
θ
⇒x=cotθ+tanθ=
sinθcosθ
1
secθ−cosθ=
cosθ
1cos
2
θ
=
cosθ
sin
2
θ
⇒y=secθ−cosθ=
cosθ
sin
2
θ
∴x
2
=
sin
2
θcos
2
θ
1
,y
2
=
cos
2
θ
sin
4
θ
∴x
2
y=
sin
2
θcos
2
θ
1
×
cosθ
sin
2
θ
=sec
3
θ
∴xy
2
=
sinθcosθ
1
×
cos
2
θ
sin
4
θ
=tan
3
θ
∴(x
2
y)
3
2
−(xy
2
)
3
2
=(sec
3
θ)
3
2
−(tan
3
θ)
3
2
=sec
2
θ−tan
2
θ
=1
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