If _ ^ / 3 2 k d =1- 1 2 ,(k>0) , then the value of k is:
Mathematics · JEE Main · NTA Exams — Integral Calculus
If \[\int_{\theta}^{\pi / 3} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=1-\frac{1}{\sqrt{2}},(k>0)\], then the value of k is:
- 2
- 1
- 4
- \(\frac{1}{2}\)
Answer
(A) 2
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