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Question

If cosec θsin θ=m and sec θcos θ=n, prove that (m2n)2/3+(mn2)2/3=1.

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Solution

cosec θsinθ=m(Given)
1sinθsinθ=m
cos2θsinθ=m

Similarly,
secθcosθ=n(Given)

1cosθcosθ=n

sin2θcosθ=n

Therefore,
Taking L.H.S.-
(m2n)2/3+(mn2)2/3

=(cos4θsin2θsin2θcosθ)2/3+(cos2θsinθsin4θcos2θ)2/3

=(cos3θ)2/3+(sin3θ)2/3

=cos2θ+sin2θ

=1

= R.H.S.

Hence proved.

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