The value of -3-2cosec21-2tan-1-3-2sec21-2tan-1-

The correct option is D (α+β)(α^2+β^2) α ^ 3 2 cosec^ 2 ( 1 2 tan^ 1 α nbsp;β nbsp; nbsp;) +β ^ 3 2 sec^ 2 ( 1 2 tan^ 1 ( β nbsp; ...

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Composition of Inverse Trigonometric Functions and Trigonometric Functions

The value of ...

Question

The value of $\frac{α^{3}}{2} c o s e c^{2} (\frac{1}{2} t a n^{- 1} \frac{α}{β}) + \frac{β^{3}}{2} s e c^{2} (\frac{1}{2} t a n^{- 1} (\frac{β}{α}))$

(α - β) (α^{2} + β^{2})

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(α + β) (α^{2} - β^{2})

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(α + β) (α^{2} + β^{2})

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n o n e o f t h e s e

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Solution

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Similar questions

\frac{α^{3}}{2} c o s e c^{2} \frac{1}{2} ({tan}^{- 1} \frac{α}{β}) + \frac{β^{2}}{2} {sec}^{2} (\frac{1}{2} {tan}^{- 1} \frac{β}{α})

α

β

x^{2} - 4 x + 1 = 0

(α > β),

f (α, β) = \frac{β^{3}}{2} {cosec}^{2} (\frac{1}{2} {tan}^{- 1} \frac{β}{α}) + \frac{α^{3}}{2} {sec}^{2} (\frac{1}{2} {tan}^{- 1} \frac{α}{β})

L M V T

\frac{β - α}{1 + β^{2}} < {tan}^{- 1} β - {tan}^{- 1} α < \frac{β - α}{1 + α^{2}}, β - α < 0

D =

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α + β & α^{2} + β^{2} & 2 α β α + β & 2 α β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} =

∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & α & β 0 & β & α \end{matrix} ∣ ∣ ∣ ∣

D_{2}

D_{2} = \frac{D}{D_{1}}

α, β

y^{2} - 2 y - 7 = 0

α^{2} + β^{2}

α^{3} + β^{3}

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