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Question
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
We have `x^2 - y^2 = [( a sec theta + b tan theta )^2 - ( a tan theta + b sec theta )^2]`
= `(a^2 sec^2 theta + b^2 tan^2 theta + 2 ab sec theta tan theta) - (a^2 tan^2 theta + b^2 sec^2 theta + 2 ab tan theta sec theta)`
= `a^2 sec^2 theta + b^2 tan^2 theta - a^2 tan^2 theta - b^2 sec^2 theta`
= `(a^2 sec^2 theta - a^2 tan^2 theta)-( b^2 sec^2 theta - b^2 tan ^2 theta)`
= `a^2 ( sec^2 theta - tan^2 theta )-b^2 ( sec^2 theta - tan^2 theta)`
= `a^2 - b^2 [∵ sec^2 theta - tan^2 theta =1]`
Hence, `x^2 - y^2 = a^2 - b^2`
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