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Properties Derived from Trigonometric Identities

Prove that ...

Question

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

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Solution

$t a n^{- 1} (\frac{c o s x}{1 + s i n x}) = t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{c o s^{2} \frac{x}{2} - s i n^{2} \frac{x}{2}}{c o s^{2} \frac{x}{2} + s i n^{2} \frac{x}{2} + 2. c o s \frac{x}{2} . s i n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠$
$= t a n^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(c o s \frac{x}{2} - s i n \frac{x}{2}) (c o s \frac{x}{2} + s i n \frac{x}{2})}{{(c o s \frac{x}{2} + sin \frac{x}{2})}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$= t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{c o s \frac{x}{2} - s i n \frac{x}{2}}{c o s \frac{x}{2} + s i n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠ = t a n^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{c o s \frac{x}{2}}{c o s \frac{x}{2}} - \frac{s i n \frac{x}{2}}{c o s \frac{x}{2}}}{\frac{c o s \frac{x}{2}}{c o s \frac{x}{2}} + \frac{s i n \frac{x}{2}}{c o s \frac{x}{2}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$
$= t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1 - t a n \frac{x}{2}}{1 + t a n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠ = t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{t a n \frac{π}{4} - t a n \frac{x}{2}}{1 + t a n \frac{π}{2} t a n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠$
$= t a n^{- 1} [t a n (\frac{π}{4} - \frac{x}{2})]$
$= \frac{π}{4} - \frac{x}{2}$

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Similar questions

{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}{p} - \frac{x}{q}, - \frac{π}{2} < x < \frac{3 π}{2}

.
Find the value of p and q

{tan}^{- 1} (\frac{cos x}{1 - sin x}), - \frac{π}{2} < x < \frac{π}{2}

in the simplest form.

x \in (- \frac{π}{2}, \frac{3 π}{2}),

{tan}^{- 1} (\frac{cos x}{1 + sin x})

Q. Prove that 2

t a n^{- 1} \frac{1}{2} - t a n^{- 1} \frac{1}{7} = \frac{π}{4}

s i n^{- 1} x = s i n^{- 1} (3 x - 4 x^{3}), x ϵ [\frac{- 1}{2}, \frac{1}{2}]

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