If fθ=tansin-123+cos2θ, then

A
$f (\frac{π}{4}) = 1$
B
$f (\frac{π}{4}) = \sqrt{2}$
C
$\frac{d (f (θ))}{d (cosθ)} at θ = \frac{π}{4} is - 2$
D
$f' (\frac{π}{4}) = \sqrt{2}$

Solution:

$\begin{array}{l} f (θ) = \tan (\sin^{- 1} \sqrt{\frac{2}{3 + \cos 2 θ}}) \\ = \tan (\tan^{- 1} \sqrt{\frac{2}{1 + \cos 2 θ}}) \\ = \sqrt{\frac{2}{1 + \cos 2 θ}} = secθ \\ \Rightarrow f (\frac{π}{4}) = \sqrt{2} \\ f' (θ) = secθtanθ \\ \Rightarrow f' (\frac{π}{4}) = \sqrt{2} \\ {\frac{d (f (θ))}{d (cosθ)}|}_{θ = \frac{π}{4}} = {(\frac{- 1}{\cos^{2} θ})}_{θ = \frac{π}{4}} = - 2 \end{array}$

$If f (θ) = \tan (\sin^{- 1} \sqrt{\frac{2}{3 + \cos 2 θ}}), then$

Solution:

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