Question

Sin-1(root over (x/(1+x))
Find the function in simplest form

Answer 1

Step-by-step explanation:

tan^-1(✓x)

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Answer 2

$\displaystyle \sf {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg) = \bf {tan}^{ - 1} \sqrt{x}$

Given :

$\displaystyle \sf The \: function \: : \: {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg)$

To find :

The function in simplest form

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg)$

Step 2 of 2 :

Express the function in simplest form

$\displaystyle \sf {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg)$

$\displaystyle \sf Let\: \: x = {tan}^{2} \theta \: \: such \: that \: \: \: \theta = {tan}^{ - 1} \sqrt{x}$

Thus we get

$\displaystyle \sf {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg)$

$\displaystyle \sf = {sin}^{ - 1} \bigg( \sqrt{ \frac{ {tan}^{2} \theta}{1 + {tan}^{2} \theta} } \bigg)$

$\displaystyle \sf = {sin}^{ - 1} \bigg( \sqrt{ \frac{ {tan}^{2} \theta}{ {sec}^{2} \theta} } \bigg)\: \: \: \bigg[ \: \because \:1 +{tan}^{2} \theta = {sec}^{2} \theta \bigg]$

$\displaystyle \sf = {sin}^{ - 1} \bigg( \frac{ {tan}^{} \theta}{ {sec}^{} \theta} \bigg)$

$\displaystyle \sf = {sin}^{ - 1} \bigg( \frac{ sin \theta }{ cos \theta \times {sec}^{} \theta} \bigg)$

$\displaystyle \sf = {sin}^{ - 1} ( sin \theta )$

$\displaystyle \sf = \theta$

$\displaystyle \sf = {tan}^{ - 1} \sqrt{x}$

$\boxed{ \: \: \displaystyle \sf \therefore \: {sin}^{ - 1} \bigg( \sqrt{ \frac{x}{1 + x} } \bigg) = \bf {tan}^{ - 1} \sqrt{x} \: \: }$

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