Question
Write the following in the simplest form:
$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
✓
Answer
$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{1+\Big(\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big)^2}\end{pmatrix}\end{Bmatrix}$ $\Big\{\text{Since},2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{\frac{1+\text{x}+1-\text{x}}{1+\text{x}}}\end{pmatrix}\end{Bmatrix}$
$=2\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{1+\text{x}}{2}$
$=\sqrt{1-\text{x}}\sqrt{1+\text{x}}$
$=\sqrt{1-\text{x}^2}$
Hence, $\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}=\sqrt{1-\text{x}^2}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{1+\Big(\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big)^2}\end{pmatrix}\end{Bmatrix}$ $\Big\{\text{Since},2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{\frac{1+\text{x}+1-\text{x}}{1+\text{x}}}\end{pmatrix}\end{Bmatrix}$
$=2\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{1+\text{x}}{2}$
$=\sqrt{1-\text{x}}\sqrt{1+\text{x}}$
$=\sqrt{1-\text{x}^2}$
Hence, $\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}=\sqrt{1-\text{x}^2}$
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