If 11+tanθ+11−tanθ{1 \over 1+ tan\theta} + {1 \over 1- tan\theta}1+tanθ1+1−tanθ1 = 4, 0° < θ < 90°, then what is the value of cosec θ + sec θ + sin θ? A3+426{3+4 \sqrt{2}\over \sqrt{6}}63+42 B833{8 \sqrt{3}\over 3}383 C332{3\sqrt{3}\over 2}233 D4+326{4+3 \sqrt{2}\over \sqrt{6}}64+32 Show answer & explanationCorrect answer: AExplanationThe correct answer is option 1 i.e. 3+426{3+4 \sqrt{2}\over \sqrt{6}}63+42. 11+tanθ+11−tanθ{1 \over 1+ tan\theta} + {1 \over 1- tan\theta}1+tanθ1+1−tanθ1 = 4 ⇒ 1−tanθ+1+tanθ1−tan2θ{1- tan\theta+1+ tan\theta \over 1- tan^2\theta} 1−tan2θ1−tanθ+1+tanθ = 4 ⇒ 21−tan2θ{2\over 1- tan^2\theta} 1−tan2θ2 = 4 ⇒ 1 - tan2θ = 121\over221 ⇒ tan2θ = 121\over221 ⇒ tanθ = 121\over\sqrt221 = PBP\over BBP ⇒ H2 = 12 + (2)2(\sqrt2)^2(2)2 ⇒ H = 3\sqrt33 ⇒ cosec θ + sec θ + sin θ ⇒ 3\sqrt33 + 32\sqrt3\over\sqrt223 + 131\over\sqrt331 = 3+426{3+4 \sqrt{2}\over \sqrt{6}}63+42