Question

please solve my question correctly correctly correctly....
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Answer 1

Hi !

Solution :- given sin²¢ = cos³¢

We have proved , cot^6¢ -co²t¢ = 1

= cot^6¢ = 1 + cot²¢

= cot^6¢ = cosec²¢

Now , from LHS .

cot^6¢ = cos^6¢/sin^6¢

=( cos³¢)²/sin^6¢

But , cos³¢ = sin²¢ ( given )

so, (sin²¢)²/sin^6¢

= sin⁴¢/sin^6¢ = 1/sin²¢ = cosec²¢

Here proved , LHS = RHS .

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Hope it's helpful