Answer
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Hint: Revise all the formulas of trigonometry and all the properties of inverse trigonometric functions. The inverse trigonometric functions are the inverse functions of the trigonometric functions for example inverse of sine, cosine, tangent etc.
Complete step by step solution:
We have to solve $\tan (1/2{{\sin }^{-1}}3/4)$ for that let us assume that
$1/2{{\sin }^{-1}}3/4=\theta $
Now by cross multiplication
$1/2{{\sin }^{-1}}3/4=\theta $ becomes
${{\sin }^{-1}}3/4=2\theta $ ---- (1)
Now by multiplying sin on both sides of the equation(1)
$\sin ({{\sin }^{-1}}3/4)=\sin 2\theta $
$\sin 2\theta =3/4$ ---- (2)
Now by using the formula $\sin 2\theta =2\tan \theta /(1+{{\tan }^{2}}\theta )$ on equation (2)
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$
Now by cross multiplication
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$ becomes
$4(2\tan \theta )=3(1+{{\tan }^{2}}\theta )$ ----- (3)
By solving the brackets of equation (3)
$8\tan \theta =3+3{{\tan }^{2}}\theta $
$3{{\tan }^{2}}\theta -8\tan \theta +3=0$ ------- (4)
To find the roots of a quadratic equation we use the formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So by using the above formula on equation (4)
$\tan \theta =-(-8)\pm \sqrt{64-(4 \times 3 \times 3}/2 \times 3$
So, $\tan \theta =4\pm \sqrt{7}/3$ ----- (5)
Now by taking tan inverse on both sides of equation (5)
$\theta ={{\tan }^{-1}}\left[ 4\pm \sqrt{7}/3 \right]$
$\tan \theta =4\pm \sqrt{7}/3$
As $\theta =1/2{{\sin }^{-1}}3/4$
So, $\tan \theta =4\pm \sqrt{7}/3$
Since ,
$\begin{align}
& -\pi /2\le {{\sin }^{-1}}3/4\le \pi /2 \\
& -\pi /4\le 1/2{{\sin }^{-1}}3/4\le \pi /4 \\
\end{align}$
Therefore, $\tan (-\pi /4)\le \tan 1/2({{\sin }^{-1}}3/4)\le \tan \pi /4$
$-1\le \tan (1/2{{\sin }^{-1}}3/4)\le 1$
Since, $4+\sqrt{7}/3>1$ so it is ignored
Therefore, $\tan (1/2{{\sin }^{-1}}3/4)=4-\sqrt{7}/3$.
Note:
There is a restriction on $\sin \theta $ i.e. $-\pi /2\le {{\sin }^{-1}}\theta \le \pi /2$. So all the values which are greater than one should be ignored. Always use the correct trigonometric formula to solve a particular equation as using the wrong formula leads towards the wrong answer.
Complete step by step solution:
We have to solve $\tan (1/2{{\sin }^{-1}}3/4)$ for that let us assume that
$1/2{{\sin }^{-1}}3/4=\theta $
Now by cross multiplication
$1/2{{\sin }^{-1}}3/4=\theta $ becomes
${{\sin }^{-1}}3/4=2\theta $ ---- (1)
Now by multiplying sin on both sides of the equation(1)
$\sin ({{\sin }^{-1}}3/4)=\sin 2\theta $
$\sin 2\theta =3/4$ ---- (2)
Now by using the formula $\sin 2\theta =2\tan \theta /(1+{{\tan }^{2}}\theta )$ on equation (2)
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$
Now by cross multiplication
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$ becomes
$4(2\tan \theta )=3(1+{{\tan }^{2}}\theta )$ ----- (3)
By solving the brackets of equation (3)
$8\tan \theta =3+3{{\tan }^{2}}\theta $
$3{{\tan }^{2}}\theta -8\tan \theta +3=0$ ------- (4)
To find the roots of a quadratic equation we use the formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So by using the above formula on equation (4)
$\tan \theta =-(-8)\pm \sqrt{64-(4 \times 3 \times 3}/2 \times 3$
So, $\tan \theta =4\pm \sqrt{7}/3$ ----- (5)
Now by taking tan inverse on both sides of equation (5)
$\theta ={{\tan }^{-1}}\left[ 4\pm \sqrt{7}/3 \right]$
$\tan \theta =4\pm \sqrt{7}/3$
As $\theta =1/2{{\sin }^{-1}}3/4$
So, $\tan \theta =4\pm \sqrt{7}/3$
Since ,
$\begin{align}
& -\pi /2\le {{\sin }^{-1}}3/4\le \pi /2 \\
& -\pi /4\le 1/2{{\sin }^{-1}}3/4\le \pi /4 \\
\end{align}$
Therefore, $\tan (-\pi /4)\le \tan 1/2({{\sin }^{-1}}3/4)\le \tan \pi /4$
$-1\le \tan (1/2{{\sin }^{-1}}3/4)\le 1$
Since, $4+\sqrt{7}/3>1$ so it is ignored
Therefore, $\tan (1/2{{\sin }^{-1}}3/4)=4-\sqrt{7}/3$.
Note:
There is a restriction on $\sin \theta $ i.e. $-\pi /2\le {{\sin }^{-1}}\theta \le \pi /2$. So all the values which are greater than one should be ignored. Always use the correct trigonometric formula to solve a particular equation as using the wrong formula leads towards the wrong answer.
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