Q. 53

# If (a2 – b2) sin θ + 2ab cosθ = a2 + b2, then prove that .

Taking (a2 – b2) sin θ + 2ab cos θ = a2 + b2

We know that

Then, substituting the above values in the given equation, we get

Now, substituting, , we hwve

( a2 – b2)2t – 2ab(1 – t2) = (a2 + b2)(1+t2)

Simplify, we get

(a2 + 2ab + b2)t2 – 2(a2 – b2)t + (a2 –2ab +b2)=0

(a+b)2 t2 – 2(a2 – b2)t + (a – b)2 = 0

(a+b)2 t2 –2 (a – b)(a+b)t + (a – b)2 =0

[(a+b)t – (a – b)]2 = 0 [ (a b)2 = (a2 + b2 – 2ab)]

[(a+b)t – (a – b)] = 0

(a+b)t = (a – b)

We know that,

Hence Proved

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