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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^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
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= R.H.S.
