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`(tan A + tanB )/(cot A + cot B) = tan A tan B`
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LHS = `(tan A + tanB )/(cot A + cot B) `
=`(tan A + tan B)/(1/ tan A + 1/ tanB)`
=` (tan A + tan B)/( (tan A+tan B)/ (tan A tan B)`
=`(tan A tan B ( tan A + tan B))/((tan A + tan B ))`
= ЁЭСбЁЭСОЁЭСЫЁЭР┤ ЁЭСбЁЭСОЁЭСЫЁЭР╡
= RHS
Hence, LHS = RHS
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Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
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sin2 A cot2 A + cos2 A tan2 A = 1
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Prove the following identity :
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Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot 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`
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