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Question
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Solution 1
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ) + (cosec2θ + sec2θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`
= (1) + (1 + cot2θ + 1 + tan2θ) + (2) + (2)
= 7 + tan2θ + cot2θ
= R.H.S
Solution 2
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ ) + 1 + cot2θ + 2 sin θ x `1/sin θ` + 1 + tan2 θ + 2cos θ. `1/cos θ`
= 1 + 1 + 1 + 2 + 2 + tan2 θ + cot2θ
= 7 + tan2 θ + cot2θ
= RHS
Hence proved.
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`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
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If cot θ = `40/9`, find the values of cosec θ and sinθ,
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`
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
