If ^2 ^2 - 3 + 2 = 1,\, lies in the first quadrant, then the value of ^2 /2 + ^2 /2 + is:

Option B. 7 3 54. Answer & Solution Answer: Option B Solution: & ^2 ^2 - 3 + 2 = 1 & ^2 = ^2 - 3 + 2 & 1 - ^2 = ^2 - 3 + 2 & 0 = 2 ^2 - 3 + 1 & 2 ^2 - 3 + 1 = 0 & 2 ^2 - 2 - + 1 = 0 & 2 ( - 1 ) - 1 ( - 1 ) = 0 & ( 2 - 1 ) ( - 1 ) = 0 & = 1/2 = 60^ & = 1 = 0^ & ^2 /2 + ^2 /2 + & = ^2 60^ 2 + ^2 60…

If ^2 ^2 - 3 + 2 = 1,\, lies in the first quadrant, then the value of…

Answer: Option B

Solution:

\frac{sin ^{2} θ}{cos ^{2} θ - 3 cos θ + 2} = 1 sin^{2} θ = cos^{2} θ - 3 cos θ + 2 1 - cos^{2} θ = cos^{2} θ - 3 cos θ + 2 0 = 2 cos^{2} θ - 3 cos θ + 1 2 cos^{2} θ - 3 cos θ + 1 = 0 2 cos^{2} θ - 2 cos θ - cos θ + 1 = 0 2 cos θ (cos θ - 1) - 1 (cos θ - 1) = 0 (2 cos θ - 1) (cos θ - 1) = 0 cos θ = \frac{1}{2} = cos 6 0^{\circ} cos θ = 1 = cos 0^{\circ} \Rightarrow \frac{tan ^{2} \frac{θ}{2} + sin ^{2} \frac{θ}{2}}{tan θ + sin θ} = \frac{tan ^{2} \frac{60 ^{\circ}}{2} + sin ^{2} \frac{60 ^{\circ}}{2}}{tan 60 ^{\circ} + sin 60 ^{\circ}} = \frac{tan ^{2} 30 ^{\circ} + sin ^{2} 30 ^{\circ}}{tan 60 ^{\circ} + sin 60 ^{\circ}} = \frac{( \frac{1}{3} ) ^{2} + ( \frac{1}{2} ) ^{2}}{3 + \frac{3}{2}} = \frac{\frac{1}{3} + \frac{1}{4}}{\frac{2 3 + 3}{2}} = \frac{\frac{4 + 3}{12}}{\frac{3 3}{2}} = \frac{7 \times 2}{12 \times 3 3} = \frac{7}{18 3} \times \frac{3}{3} = \frac{7 3}{54}