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Question
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
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Solution
We know that `sin^2 theta + cos^2 theta = 1`
Multiplying both numerator and the denominator by `(1 - sin theta)`, we have
`cos theta/(1 + sin theta) = (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`
`= (cos theta(1 - sin theta))/(1 - sin^2 theta)`
`= (cos theta (1 - sin theta))/cos^2 theta`
`= (1 - sin theta)/cos theta`
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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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