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Question
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
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Solution
We have,
tan α = n tan β
⇒ `tan β = tan α/n`
⇒ `cot β = n/tan α`
sin α = m sin β
⇒ `sin β = sin α /m`
⇒ `cosec β = m/sin α`
Since, cosec2 β - cot2 β = 1
⇒ `m^2/sin^2 α - n^2/tan^2 α = 1`
⇒ `m^2/sin^2 α - (n^2cos^2α )/sin^2 α = 1`
⇒ m2 - n2cos2 α = sin2 α
⇒ m2 - n2cos2 α = 1 - cos2 α
⇒ m2 - 1 = (n2 - 1)cos2 α
⇒ cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Hence proved.
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`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
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9 sec2 A − 9 tan2 A is equal to
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`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Choose the correct alternative:
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
