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Question
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
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Solution
`Given that
`x = a sec theta + b tan theta`
`y = a ta theta + b sec theta`
We have to prove `x^2 - y^2 = a^2 - b^2`
We know that `sec^2 theta - tan^2 theta = 1`
So,
`x^2 - y^2`
`= (a sec theta + b tan theta)^2 - (a tan theta + b sec theta)^2`
`= (a^2 sec^2 theta + 2 ab sec theta + b^2 tan^2 theta) - (a^2 tan^2 theta + 2 ab sec theta tan theta + b^2 + sec^2 theta)`
`= a^2 (sec^2 theta - tan^2 theta) - b^2 (sec^2 theta - tan^2 theta)`
`= a^2 - b^2 `
Hence proved.
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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
