If ^2 - 3 + 2 ^2 = 1, where 0° < < 90° then what is the value of (cos2 - sin3 + cosec2 )?

Option B. - 3 + 4 3 6. Answer & Solution Answer: Option B Solution: & ^2 - 3 + 2 ^2 = 1 & ^2 - 3 + 2 = ^2 & ^2 - 3 + 2 = 1 - ^2 & 2 ^2 - 3 + 1 = 0 & 2 ^2 - 2 - + 1 = 0 & 2 ( - 1 ) - 1 ( - 1 ) = 0 & ( - 1 ) ( 2 - 1 ) = 0 & = 1\,\, ( not valid ) & and = 1/2 & = 30^ & = 30^ & 2 - 3 + cosec 2 & = 60^ - 90…

If ^2 - 3 + 2 ^2 = 1, where 0° < < 90° then what is the value of (cos…

Answer: Option B

Solution:

\frac{sin ^{2} ϕ - 3 sin ϕ + 2}{cos ^{2} ϕ} = 1 \Rightarrow sin^{2} ϕ - 3 sin ϕ + 2 = cos^{2} ϕ \Rightarrow sin^{2} ϕ - 3 sin ϕ + 2 = 1 - sin^{2} ϕ \Rightarrow 2 sin^{2} ϕ - 3 sin ϕ + 1 = 0 \Rightarrow 2 sin^{2} ϕ - 2 sin ϕ - sin ϕ + 1 = 0 \Rightarrow 2 sin ϕ (sin ϕ - 1) - 1 (sin ϕ - 1) = 0 \Rightarrow (sin ϕ - 1) (2 sin ϕ - 1) = 0 ∴ sin ϕ = 1 (not valid) and sin ϕ = \frac{1}{2} \Rightarrow sin ϕ = sin 3 0^{\circ} ∴ ϕ = 3 0^{\circ} ∴ cos 2 ϕ - sin 3 ϕ + cosec 2 ϕ = cos 6 0^{\circ} - sin 9 0^{\circ} + cosec 6 0^{\circ} = \frac{1}{2} - 1 + \frac{2}{3} = \frac{2}{3} - \frac{1}{2} = \frac{( 4 - 3 )}{2 3} \times \frac{3}{3} = \frac{4 3 - 3}{6} Alternative : Put ϕ = 3 0^{\circ} L .H .S . = \frac{sin ^{2} 30 ^{\circ} - 3 sin 30 ^{\circ} + 2}{cos ^{2} 30 ^{\circ}} = \frac{\frac{1}{4} - \frac{3}{2} + 2}{\frac{3}{4}} = \frac{1 - 6 + 8}{3} = \frac{3}{3} = 1 = R .H .S . ∴ cos 2 ϕ - sin 3 ϕ + cosec 2 ϕ = cos 6 0^{\circ} - sin 9 0^{\circ} + cosec 6 0^{\circ} = \frac{1}{2} - 1 + \frac{2}{3} = - \frac{1}{2} + \frac{2}{3} = \frac{( - 3 + 4 )}{2 3} \times \frac{3}{3} = \frac{- 3 + 4 3}{6}