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Question
`(1-cos^2theta) sec^2 theta = tan^2 theta`
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Solution
LHS = `(1-cos^2 theta)sec^2 theta`
=`sin^2 theta xx sec^2 theta (∵ sin^2 theta + cos^2 theta = 1)`
= `sin^2 theta xx 1/(cos^2 theta)`
=`(sin^2 theta)/(cos^2 theta)`
=`tan^2 theta`
=RHS
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= `cos^2theta xx square .....[1 + tan^2theta = square]`
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