Find the value of $sin ({sec}^{- 1} (\frac{u}{2}))$ $sin(sec^(-1)(u/2))$ ?

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Answer 1 · 2017-03-30T05:16:58

$sin ({sec}^{- 1} (\frac{u}{2})) = \frac{1}{u} \sqrt{u^{2} - 4}$ $sin(sec^(-1)(u/2))=1/usqrt(u^2-4)$

Let ${sec}^{- 1} (\frac{u}{2}) = x$ $sec^(-1)(u/2)=x$ , then

$sec x = \frac{u}{2}$ $secx=u/2$ and $cos x = \frac{2}{u}$ $cosx=2/u$

Therefore $sin x = \sqrt{1 - {cos}^{2} x}$ $sinx=sqrt(1-cos^2x)$

= $\sqrt{1 - {(\frac{2}{u})}^{2}}$ $sqrt(1-(2/u)^2)$

= $\sqrt{1 - \frac{4}{u^{2}}}$ $sqrt(1-4/u^2)$

= $\sqrt{\frac{u^{2} - 4}{u^{2}}}$ $sqrt((u^2-4)/u^2)$

= $\frac{1}{u} \sqrt{u^{2} - 4}$ $1/usqrt(u^2-4)$

Now $sin x = \frac{1}{u} \sqrt{u^{2} - 4}$ $sinx=1/usqrt(u^2-4)$ , but $x = {sec}^{- 1} (\frac{u}{2})$ $x=sec^(-1)(u/2)$

Hence $sin ({sec}^{- 1} (\frac{u}{2})) = \frac{1}{u} \sqrt{u^{2} - 4}$ $sin(sec^(-1)(u/2))=1/usqrt(u^2-4)$

Find the value of sin(sec^(-1)(u/2))sin(sec−1(u2))sin(sec^(-1)(u/2))?