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प्रश्न
Prove the following trigonometric identities.
`(1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)^2`
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उत्तर
We have to prove `(1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)^2`
We know that, `sin^2 theta + cos^2 theta = 1`
Multiplying both numerator and denominator by `(1 - sin theta)` we have
`(1 - sin theta)/(1 + sin theta) = ((1 - sin theta)(1 - sin theta))/((1 + sin theta)(1 - sin theta))`
`= (1 - sin theta)^2/(1 - sin^2 theta)`
`= ((1 - sin theta)/cos theta)^2`
`= (1/cos theta - sin theta/cos theta)^2`
`= (sec theta - tan theta)^2`
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Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
