$2 t a n^{- 1} (cos x) = t a n^{- 1} (2 c o s e c x)$ then $x = ?$

π / 4

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π / 6

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π / 2

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no solution

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Solution

The correct option is D $π / 4$
Let $k = 2 t a n^{- 1} (c o s x) = t a n^{- 1} (2 c o s e c x)$
$t a n \frac{k}{2} = c o s x$ and $t a n k = 2 c o s e c x$
$t a n k = \frac{2 t a n \frac{k}{2}}{1 - t a n^{2} \frac{k}{2}} = \frac{2 \times c o s x}{1 - c o s^{2} x} = \frac{2 c o s x}{s i n^{2} x} = \frac{2}{s i n x}$
$\Rightarrow t a n x = 1$
$\Rightarrow x = \frac{π}{4}$

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Similar questions

Solution of the equation $2 t a n^{- 1} (c o s x) = t a n^{- 1} (2 c o s e c x)$ is

2 {tan}^{- 1} (cos x) = {tan}^{- 1} (2 c o s e c x)

2 {tan}^{- 1} (cos x) = {tan}^{- 1} ({cosec}^{2} x),

ϵ (0, π)

x

{tan}^{- 1} (\frac{cos x}{1 + sin x}) = \frac{π}{4} - \frac{x}{2}, x \in (- \frac{π}{2}, \frac{π}{2})

{tan}^{- 1} (\frac{cos x}{1 + sin x}) = \frac{π}{4} - \frac{x}{2}, x ϵ (- \frac{π}{2}, \frac{π}{2})

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