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प्रश्न
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
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उत्तर
LHS= `(sectheta- tan theta)/(sec theta + tan theta)`
= `(1/cos theta-sin theta/cos theta)/(1/cos theta+ sin theta/cos theta)`
=`((1-sin theta)/cos theta)/((1+ sin theta)/cos theta)`
=`(1-sin theta)/(1+ sin theta)`
=`((1-sin theta) (1+ sin theta))/( (1+ sin theta )(1+ sin theta)) {"Dividing the numerator and
denominator by"(1 + cos theta)}`
=`((1-sin^2 theta))/((1+ sin theta)^2)`
=`cos^2 theta/(1+ sin theta)^2`
= 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)`
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