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Trigonometric Equations

If α∈ - 3 π...

Question

If $α \in (- \frac{3 π}{2}, - π)$ then find the value of ${tan}^{- 1} (cot α) - {cot}^{- 1} (tan α) + {sin}^{- 1} (sin α) + {cos}^{- 1} (cos α)$

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Solution

$\begin{matrix} α \in (\frac{- 3}{2}, - π) N o w, {tan}^{- 1} (cot α) - {cot}^{- 1} (tan α) + {sin}^{- 1} (sin α) + {cos}^{- 1} (cos α) = (\frac{π}{2} - α - 2 π) - ((\frac{π}{2} - α) - π) + - π - α + (2 π + α) = - π + π = 0 \end{matrix}$
Hence, solved.

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α ϵ (0, \frac{π}{2})

, then the value of

{tan}^{- 1} (cot α) - {cot}^{- 1} (tan α) + {sin}^{- 1} (sin α) - {cos}^{- 1} (cos α)

α ϵ (- \frac{π}{2}, 0)

, then find the value of

{tan}^{- 1} (cot α) - {cot}^{- 1} (tan α)

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt (1 + x^{2}) + \sqrt (1 + x^{2})} = α

. then prove that

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\frac{m t a n (α - θ)}{{c o s}^{2} θ} = \frac{n t a n θ}{{c o s}^{2} (α - θ)}

, then prove that

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{c o s}^{- 1} \frac{c o s α + c o s β}{1 + c o s α c o s β} = 2 {t a n}^{- 1} (t a n \frac{α}{2} t a n \frac{β}{2})

{cos}^{- 1} α + {cos}^{- 1} β + {cos}^{- 1} γ = 3 π

α (β + γ) + β (γ + α) + γ (α + β)

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