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Answer: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$

Hint: First we will convert $\sin ^{-1}\left(\frac{3}{5}\right)$ into $\tan ^{-1}$.

Given: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$

Explanation:

Let us solve for $\sin ^{-1}\left(\frac{3}{5}\right)$

Let $\sin ^{-1}\left(\frac{3}{5}\right)=\theta$        ............(1)

\begin{aligned} &\sin \theta=\left(\frac{3}{5}\right)=\frac{P}{H} \\ &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}

Now,

\begin{aligned} &\tan \theta=\frac{P}{B} \\ &\tan \theta=\frac{3}{4} \\ &\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}

From equation (1)

$\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$       .............(2)

L.H.S:

$2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$

from Equation (2)

$=2 \tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$                                    $\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$

$=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$                        $\left[\begin{array}{c} -1

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \end{aligned}

$\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$                                                    $\left[\begin{array}{l} x \cdot y>-1 \\ \frac{24}{7} \times \frac{17}{31}=\frac{408}{217}>-1 \end{array}\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{\frac{625}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}

$\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}

Hence it is Prove that $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$

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