Given, lim x → π 2 tan 2 x [ 2 sin 2 x + 3 sin x + 4 − sin 2 x + 6 sin x + 2 ] \lim_{x \to \dfrac{\pi}{2}}\tan^{2}x\left[\sqrt{2\sin^{2}x + 3\sin x + 4} - \sqrt{\sin^{2}x + 6\sin x + 2}\right] lim x → 2 π tan 2 x [ 2 sin 2 x + 3 sin x + 4 − sin 2 x + 6 sin x + 2 ]
= lim x → π 2 tan 2 x [ ( 2 sin 2 x + 3 sin x + 4 ) 2 − ( sin 2 x + 6 sin x + 2 ) 2 ] 2 sin 2 x + 3 sin x + 4 + sin 2 x + 6 sin x + 2 = \lim_{x \to \dfrac{\pi}{2}}\tan^{2}x\dfrac{\left[{\left(\sqrt{2\sin^{2}x + 3\sin x + 4}\right)}^{2} - {\left(\sqrt{\sin^{2}x + 6\sin x + 2}\right)}^{2}\right]}{\sqrt{2\sin^{2}x + 3\sin x + 4} + \sqrt{\sin^{2}x + 6\sin x + 2}} = lim x → 2 π tan 2 x 2 sin 2 x + 3 sin x + 4 + sin 2 x + 6 sin x + 2 [ ( 2 sin 2 x + 3 sin x + 4 ) 2 − ( sin 2 x + 6 sin x + 2 ) 2 ]
= lim x → π 2 tan 2 x [ sin 2 x − 3 sin x + 2 ] 9 + 9 = \lim_{x \to \dfrac{\pi}{2}}\dfrac{\tan^{2}x\left[\sin^{2}x - 3\sin x + 2\right]}{\sqrt{9} + \sqrt{9}} = lim x → 2 π 9 + 9 tan 2 x [ sin 2 x − 3 sin x + 2 ]
= lim x → π 2 tan 2 x ( sin x − 1 ) ( sin x − 2 ) 6 = \lim_{x \to \dfrac{\pi}{2}}\dfrac{\tan^{2}x\left(\sin x - 1\right)\left(\sin x - 2\right)}{6} = lim x → 2 π 6 tan 2 x ( sin x − 1 ) ( sin x − 2 )
= 1 6 lim x → π 2 tan 2 x ( 1 − sin x ) = \dfrac{1}{6}\lim_{x \to \dfrac{\pi}{2}}\tan^{2}x\left(1 - \sin x\right) = 6 1 lim x → 2 π tan 2 x ( 1 − sin x )
= 1 6 lim x → π 2 sin 2 x ( 1 − sin x ) ( 1 − sin x ) ( 1 + sin x ) = 1 12 = \dfrac{1}{6}\lim_{x \to \dfrac{\pi}{2}}\dfrac{\sin^{2}x\left(1 - \sin x\right)}{\left(1 - \sin x\right)\left(1 + \sin x\right)} = \dfrac{1}{12} = 6 1 lim x → 2 π ( 1 − sin x ) ( 1 + sin x ) sin 2 x ( 1 − sin x ) = 12 1
