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let theta = x
then, √[(1-cos²x )sec²x]
=√ sin²x sec²x
=sinx × secx
= sinx /cos x
= tanx
then, √[(1-cos²x )sec²x]
=√ sin²x sec²x
=sinx × secx
= sinx /cos x
= tanx
Answered by
1
▪▪ Root [ sin^2theta . sec^2theta ]
▪▪ Sin theta × sec theta
▪▪ Sin theta × 1/cos theta
▪▪ sin theta / cos theta
▪▪ tan theta.......... Hence, proved .
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