JEE Main 20199 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002If ∫0π3tanθ2k secθ dθ=1−12,(k>0),\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),0∫3π2ksecθtanθdθ=1−21,(k>0), then value of k is :AOption A. 4BOption B. 12{1 \over 2}21COption C. 1DOption D. 2CheckSkip · show solution
