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Question
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
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Solution
LHS = `(sec A - 1)/(sec A + 1)`
= `(1/cos A - 1)/(1/cos A + 1)`
= `((1 - cos A)/cos A)/((1 + cos A)/cos A)`
= `(1 - cos A)/(1 + cos A)`
= RHS
Hence proved.
RELATED QUESTIONS
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
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`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
If x = a cos3θ and y = b sin3θ, prove that `(x/a)^(2/3) + (y/b)^(2/3) = 1`.
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`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
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
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
