The expression ( 1 - + )^2 ( 1 - ) ^3 \, cose c^2 ( - ) ( + ), 0° < θ < 90°, is equal to:

Option A. 2tanθ. Answer & Solution Answer: Option A Solution: & ( 1 - + )^2 ( 1 - ) ^3 cose c^2 ( - ) ( + ) & put = 30^ & = ( 1 - 30^ + 30^ )^2 ( 1 - 30^ ) ^330^ cose c^230^ ( 30^ - 30^ ) ( 30^ + 30^ ) & = ( 1 - 1/2 + 3 2 )^2 ( 1 - 3 2 ) ( 2 3 )^3 ( 2 )^2 ( 2 3 - 1 3 ) ( 1 3 + 3 ) & = ( 1 + 3 2…

The expression ( 1 - + )^2 ( 1 - ) ^3 \, cose c^2 ( - ) ( + ), 0° < θ…

Answer: Option A

Solution:

\frac{( 1 - sin θ + cos θ ) ^{2} ( 1 - cos θ ) sec ^{3} θ cose c ^{2} θ}{( sec θ - tan θ ) ( tan θ + cot θ )} put θ = 3 0^{\circ} = \frac{( 1 - sin 30 ^{\circ} + cos 30 ^{\circ} ) ^{2} ( 1 - cos 30 ^{\circ} ) sec ^{3} 30 ^{\circ} cose c ^{2} 30 ^{\circ}}{( sec 30 ^{\circ} - tan 30 ^{\circ} ) ( tan 30 ^{\circ} + cot 30 ^{\circ} )} = \frac{( 1 - \frac{1}{2} + \frac{3}{2} ) ^{2} ( 1 - \frac{3}{2} ) ( \frac{2}{3} ) ^{3} ( 2 ) ^{2}}{( \frac{2}{3} - \frac{1}{3} ) ( \frac{1}{3} + 3 )} = \frac{( \frac{1 + 3}{2} ) ^{2} ( \frac{2 - 3}{2} ) ( \frac{2}{3} ) ^{3} ( 2 ) ^{2}}{( \frac{1}{3} ) ( \frac{1 + 3}{3} )} = \frac{\frac{4 + 2 3}{4} \times \frac{2 - 3}{2} \times \frac{8}{3 3} \times 4}{( \frac{1}{3} ) ( \frac{4}{3} )} = \frac{[ 2 ^{2} - ( 3 ) ^{2} ] \times \frac{8}{3 3}}{\frac{4}{3}} = \frac{\frac{8}{3 3}}{\frac{4}{3}} = \frac{2}{3} Option A: 2 tan 3 0^{\circ} = \frac{2}{3} Option B: cot 3 0^{\circ} = 3 Option C: sin 3 0^{\circ} = \frac{1}{2} Option D: 2 cos 3 0^{\circ} = 2 \times \frac{3}{2} = 3 Only option A is answer .