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Question
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
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Solution
We have to prove `(1 + tan^2 theta)(1 - sin theta)(1 + sin theta) = 1`
We know that
`sin^2 theta + cos^2 theta = 1`
`sec^2 theta - tan^2 theta = 1`
So
`(1 + tan^2 theta)(1 - sin theta) = (1 + tan^2 theta){(1 - sin theta)(1 + sin theta)}`
` = (1 + tan^2 theta)(1 - sin^2 theta)`
`= sec^2 theta cos^2 theta`
` = 1/cos^2 theta cos^2 theta`
= 1
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Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
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We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
