NIMCET 2012 Question Paper with solutions

Given, A - B = $\frac{\mathit{\pi }}{4}$
⇒ tan (A - B) = tan $\frac{\mathit{\pi }}{4}$ = 1
⇒ $\frac{\mathit{t}\mathit{a}\mathit{n}\mathit{A}-\mathit{t}\mathit{a}\mathit{n}\mathit{B}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{A}.\mathit{t}\mathit{a}\mathit{n}\mathit{B}}$ = 1
⇒ tan A - tan B = 1 + tan A . tan B
⇒ 1 - tan A + tan B + tan A . tan B = 0
⇒ 2 = 1 + tan A - tan B + tan A. tan B = 0
⇒ 2 = (1 - tan B) + tan A (1 - tan B)
⇒ 2 = (1 - tan B) (1 + tan A)