MCQ
જો $2tan^{-1}(cosx) = tan^{-1}(cosec^2x)$ તો $x =$
- A$\frac{\pi}{2}$
- B$\pi$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{3}$
$\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{1}{\sin ^{2} x} \Rightarrow 2 \cos x=1 \Rightarrow x=\frac{\pi}{3}$
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$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad k \quad, \quad x=0$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} ,\,\,\, x>0$
એ $x=0$ આગળ સતત હોય તો $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ ની કિમંત મેળવો.