Taking (a2 – b2) sin θ + 2ab cos θ = a2 + b2
We know that
Then, substituting the above values in the given equation, we get
⇒ (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)
Hence Proved