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प्रश्न
Prove the following:
`(tan theta + 1/costheta)^2 + (tan theta - 1/costheta)^2 = 2((1 + sin^2theta)/(1 - sin^2theta))`
उत्तर
L.H.S. = `(tan theta + 1/costheta)^2 + (tan theta - 1/costheta)^2`
= `(sintheta/costheta + 1/costheta)^2 + (sintheta/costheta - 1/costheta)^2`
= `((sintheta+ 1)^2)/cos^2theta + ((sintheta - 1)^2)/cos^2theta`
= `((sintheta + 1)^2 + (sintheta - 1)^2)/cos^2theta`
= `(sin^2theta + 2sintheta + 1 + sin^2theta - 2sintheta + 1)/cos^2theta`
= `2((sin^2theta+1)/(cos^2theta))`
= `2((1 + sin^2theta)/(1 - sin^2theta))`
= R.H.S.
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