JEE Main 2024TrigonometryTrigonometric Ratio and IdentitesEasyQuestionIf tan15∘+1tan75∘+1tan105∘+tan195∘=2a\tan 15^\circ + {1 \over {\tan 75^\circ }} + {1 \over {\tan 105^\circ }} + \tan 195^\circ = 2atan15∘+tan75∘1+tan105∘1+tan195∘=2a, then the value of (a+1a)\left( {a + {1 \over a}} \right)(a+a1) is :OptionsA5−3235 - {3 \over 2}\sqrt 3 5−233B4−234 - 2\sqrt 3 4−23C2D4Check AnswerHide SolutionSolutiontan15∘+tan15∘−tan15∘+tan15∘\tan 15^\circ + \tan 15^\circ - \tan 15^\circ + \tan 15^\circ tan15∘+tan15∘−tan15∘+tan15∘ =2tan15∘= 2\tan 15^\circ=2tan15∘ =2(2−3)=2a⇒a=2−3= 2\left( {2 - \sqrt 3 } \right) = 2a \Rightarrow a = 2 - \sqrt 3=2(2−3)=2a⇒a=2−3 ∴\therefore∴ 1a+a⇒(2+3)+(2−3)=4{1 \over a} + a \Rightarrow \left( {2 + \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) = 4a1+a⇒(2+3)+(2−3)=4