If cosθ−sinθ=3cos(90°−θ),0°<θ<90°cos θ - sin θ = \sqrt3 cos (90° - θ), 0° < θ<90° cosθ−sinθ=3cos(90°−θ),0°<θ<90° then find the value if tanθ - cot θ. A3−23(1+3)\frac {3-2\sqrt3}{(1+\sqrt3)}(1+3)3−23 B−3−23(1+3)-\frac {3-2\sqrt3}{(1+\sqrt3)}−(1+3)3−23 C−3+23(1−3)-\frac {3+2\sqrt3}{(1-\sqrt3)}−(1−3)3+23 D3+23(1−3)\frac {3+2\sqrt3}{(1-\sqrt3)}(1−3)3+23 Show answer & explanationCorrect answer: BExplanationThe correct answer is Option 2 i.e −3−23(1+3)-\frac {3-2\sqrt3}{(1+\sqrt3)}−(1+3)3−23 ⇒ cosθ−sinθ=3cos(90°−θ),0°<θ<90°cos θ - sin θ = \sqrt3 cos (90° - θ), 0° < θ<90° cosθ−sinθ=3cos(90°−θ),0°<θ<90° ⇒ cosθ−sinθ=3Sinθcos θ - sin θ = \sqrt3 Sin θ cosθ−sinθ=3Sinθ ⇒cosθ=sinθ(3+1)⇒cotθ=3+1⇒ cos θ = sin θ (\sqrt3 + 1) \\ ⇒ cot θ = \sqrt3 +1 ⇒cosθ=sinθ(3+1)⇒cotθ=3+1 ⇒tanθ=13+1⇒ tan \theta = \frac{1}{\sqrt3 +1 }⇒tanθ=3+11 ⇒ tanθ - cot θ = 13+1−(3+1) \frac{1}{\sqrt3 +1 } - (\sqrt3 +1) 3+11−(3+1) On further simplification one gets, ⇒ tanθ - cot θ = −3−23(1+3)-\frac {3-2\sqrt3}{(1+\sqrt3)}−(1+3)3−23.