Question

If x = a secA + b tanA and y = a tanA + b secA,prove that x^2 - y^2 = a^2 - b^2

Answer 1

Answer:

given x=a sec A+btan A ...[1]

y=a tan A+bsec A...[2]

squaring and sub (1)-(2)on both sides

there fore x^2-y^2=a^2 sec^2A+b^2 tan^2A-(a^2tan^2 A+b^2 sec ^2A)

where sec A=1+tan A

a^2(sec^2A-tan^2A)+b^2(tan^2A-sec^2A) where tan A-sec A=-1

a^2(1)+b^2(-1)

there fore x^2-y^2=a^2-b^2

Lhs=Rhs

Hence proved

a^21+ tan^2a+b^2tan^2A