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Question
` tan^2 theta - 1/( cos^2 theta )=-1`
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Solution
LHS= `tan^2 theta - 1/(cos^2 theta)`
=` (sin^2 theta )/( cos^2 theta) - 1/(cos^2 theta)`
=`(sin ^2 theta-1)/(cos^2 theta)`
=` (-cos^2 theta )/(cos^2 theta)`
= -1
= RHS
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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