Question

x=tanA+sinA and y=tanA,prove that [x+y]2. - [x+y]2 = 1
[ x-y]. [ 2]

Answer 1

Given: x = tanA + sinA and y = tanA – sinA
Therefore,
x + y = tanA + sinA + tanA – sinA = 2 tanA
x – y = tanA + sinA – tanA + sinA = 2 sinA
LHS = (x+y/x-y)2 – (x+y/2)2
Substitute the value of x+y  and  x-y
LHS = (2 tanA /2 sinA)2 – (2 tanA/2)2
        = ((sinA/cosA)/sinA)2 – tan2A
         = (1/cosA)2 – tan2A
         = sec2A – tan2A
         = 1 + tan2A – tan2A
          = 1
         = RHS
Hence proved.
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