If cot-1-cos - 1-2 - tan-1-cos - 1-2 - -x- then sin x equals

Cot^ 1[(cos α )^1/2 ] tan^ 1[(cos α )^1/2 ]=x nbsp; ⇒tan^ 1[1√(cos α) ] tan^ 1 [√(cos α) ] = x ⇒tan^ 1[1√(cos α) √(cos α)1 + √(cosα)√(cosα) ] = x ⇒tan^ ...

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Standard XII

Mathematics

Inverse Function

If cot-1[cos ...

Question

If ${cot}^{- 1} [{(cos α)}^{1 / 2}] - {tan}^{- 1} [{(cos α)}^{1 / 2}] = x,$ then $sin x$ equals

{tan}^{2} (\frac{α}{2})

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tan α

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cot (\frac{α}{2})

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{cot}^{2} (\frac{α}{2})

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Solution

The correct option is A ${tan}^{2} (\frac{α}{2})$
${cot}^{- 1} [{(cos α)}^{1 / 2}] - {tan}^{- 1} [{(cos α)}^{1 / 2}] = x$

$\Rightarrow {tan}^{- 1} [\frac{1}{\sqrt{cos α}}] - {tan}^{- 1} [\sqrt{cos α}] = x$

$\Rightarrow {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{1}{\sqrt{cos α}} - \sqrt{cos α}}{1 + \frac{\sqrt{cos α}}{\sqrt{cos α}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = x$

$\Rightarrow {tan}^{- 1} [\frac{1 - cos α}{2 \sqrt{cos α}}] = x$

$\Rightarrow tan x = \frac{1 - cos α}{2 \sqrt{cos α}}$

$∴ sin x = \frac{1 - cos α}{1 + cos α}$

$= \frac{2 {sin}^{2} \frac{α}{2}}{2 {cos}^{2} \frac{α}{2}} = {tan}^{2} (\frac{α}{2})$

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Inverse Function

Standard XII Mathematics

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If cot−1[(cos α)1/2]−tan−1[(cos α)1/2]=x, then sinx equals

If ${cot}^{- 1} [{(cos α)}^{1 / 2}] - {tan}^{- 1} [{(cos α)}^{1 / 2}] = x,$ then $sin x$ equals