Mathematics 2015 Jamb Examination

We can rearrange the terms to group the parts of a perfect square trinomial: \[ a^2 - 4a + 4 - b^2 \] The first three terms form the square of a binomial: \[ (a - 2)^2 - b^2 \] This is now a difference of two squares, which follows the pattern \( X^2 - Y^2 = (X + Y)(X - Y) \). Here, \( X = (a - 2) \) and \( Y = b \): \[ ((a - 2) + b)((a - 2) - b) \] \[ (a - 2 + b)(a - 2 - b) \]

Since \( \tan x = \dfrac{\text{opposite}}{\text{adjacent}} = \frac{2}{3} \), we find the hypotenuse \( h \) using Pythagoras' theorem: \[ h = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \] Now, find \( \sin x = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{2}{\sqrt{13}} \). To find \( 2\sin x \): \[ 2\sin x = 2 \times \dfrac{2}{\sqrt{13}} = \dfrac{4}{\sqrt{13}} \] Rationalizing the denominator: \[ \dfrac{4}{\sqrt{13}} \times \dfrac{\sqrt{13}}{\sqrt{13}} = \dfrac{4\sqrt{13}}{13} \]