Q53 of 268 Page 5

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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