Inverse Trigonometric Functions

Domain of ${\text{sec}}^{-1}x$ is $\mathrm{x}\in (-\infty ,-1]\cup [1,\infty )$

\(2[x]+1 \leq-1\) or \(2[x]+1 \geq 1\)

\(\Rightarrow[x] \leq-1\) or \([x] \geq 0\)

\(\Rightarrow \mathrm{x} \in(-\infty, 0)\) or \(x \in[0, \infty)\)

\(\Rightarrow x \in(-\infty, \infty)\)

$$\begin{gathered} 16\left({\left({\sec }^{-1}x\right)}^2+{\left({cosec}^{-1}x\right)}^2\right)............\text{(i)} \\ \text{we know that,} se{c}^{-1}x+\cos e{c}^{-1}x=\frac{\mathrm{\pi }}{2} \\ \text{and} se{c}^{-1}x\in \left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\} \\ \text{now, put in (i), we get} \\ 16\left({\left({\sec }^{-1}x\right)}^2+{\left(\frac{\mathrm{\pi }}{2}-se{c}^{-1}x\right)}^2\right) \\ let se{c}^{-1}x=y \\ 16\left(2y^2-\pi y+\frac{{\mathrm{\pi }}^2}{4}\right)=0 \\ \text{Now}\mathit{\text{, }}\text{it}\text{is}\text{maximum}\text{at} y\mathit{=}\mathrm{\pi } \\ \text{maximum}\text{value}\mathit{=}\mathit{16}\mathit{[}\mathit{2}{\pi }^{\mathit{2}}\mathit{-}{\pi }^{\mathit{2}}\mathit{+}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{]} \\ \mathit{=}\mathit{20}{\pi }^{\mathit{2}} \\ \text{and}\text{minimum}\text{at}\mathit{ }y\mathit{=}\frac{\pi }{\mathit{4}} \\ \mathit{ }\mathit{(}y\mathit{=}\mathit{-}\frac{b}{\mathit{2}a}\mathit{ }\mathit{\text{for the quadratic in}}\mathit{ }y\mathit{)} \\ \text{minimum}\text{value}\mathit{ }\mathit{=}\mathit{16}\mathit{[}\frac{\mathit{2}\mathit{\times }{\pi }^{\mathit{2}}}{\mathit{16}}\mathit{-}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{+}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{]} \\ \mathit{=}\mathit{2}{\pi }^{\mathit{2}} \\ \text{And}\text{sum}\text{of}\text{maximum}\mathit{\text{ \& }}\text{minimum}\mathit{\text{ value}}\mathit{=} \\ \mathit{ }\mathit{20}{\mathrm{\pi }}^2+2{\mathrm{\pi }}^2=22{\mathrm{\pi }}^2 \end{gathered}$$

Let ${\cos }^{-1}x=A$ and ${\sin }^{-1}y=B$. Then $x=\cos A$ and $y=\sin B$.
given condition is $A-B=\alpha$.
The expression is $E=x^2+y^2+2xy\sin \alpha$.
Substitute $x, y$, and $\alpha$ :

$E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B\sin (A-B)$

Using the identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$ :

$E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B(\sin A\cos B-\cos A\sin B)$

$E={\cos }^{2}A+{\sin }^{2}B+2\sin A\cos A\sin B\cos B-2{\cos }^{2}A{\sin }^{2}B$

$E={\cos }^{2}A\left(1-{\sin }^{2}B\right)+{\sin }^{2}B\left(1-{\cos }^{2}A\right)+2\sin A\cos A\sin B\cos B$

$$\begin{gathered} E={\cos }^{2}A{\cos }^{2}B+{\sin }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B \\ E=(\cos A\cos B+\sin A\sin B{)}^2=(\cos (A-B){)}^2={\cos }^2\alpha \end{gathered}$$

The minimum value of ${\cos }^2\alpha \text{ is }0$

At \(\alpha=\frac{\pi}{2}\)

For \(f(x)=\cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)\) to be defined, we need \(-1\leq \dfrac{4x+2[x]}{3}\leq 1\).

So \(-3\leq 4x+2[x]\leq 3\).

Let \([x]=n\), where \(n\in \mathbb{Z}\). Then \(n\leq x<n+1\).

The inequality becomes \(-3\leq 4x+2n\leq 3\), hence \(\dfrac{-3-2n}{4}\leq x\leq \dfrac{3-2n}{4}\).

Now check possible integer values of \(n\).

For \(n=-1\), \(-1\leq x<0\) and \(-\dfrac{1}{4}\leq x\leq \dfrac{5}{4}\), giving \(-\dfrac{1}{4}\leq x<0\).

For \(n=0\), \(0\leq x<1\) and \(-\dfrac{3}{4}\leq x\leq \dfrac{3}{4}\), giving \(0\leq x\leq \dfrac{3}{4}\).

No other integer value of \(n\) gives a common interval.

Thus the domain is \(\left[-\dfrac{1}{4},0\right)\cup\left[0,\dfrac{3}{4}\right]=\left[-\dfrac{1}{4},\dfrac{3}{4}\right]\).

So \(\alpha=-\dfrac{1}{4}\) and \(\beta=\dfrac{3}{4}\).

Therefore \(12(\alpha+\beta)=12\left(-\dfrac{1}{4}+\dfrac{3}{4}\right)=12\cdot\dfrac{1}{2}=6\).

$$\begin{gathered} f\left(x\right)=16\left({\left({\sec }^{-1}x\right)}^2+{\left({cosec}^{-1}x\right)}^2\right) \\ \Rightarrow f\left(x\right)=16\left({\left({\sec }^{-1}x\right)}^2+{\left(\frac{\mathrm{\pi }}{2}-{\sec }^{-1}x\right)}^2\right) \\ \Rightarrow f\left(x\right)=32\left({\left({\sec }^{-1}x-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right) \\ \Rightarrow f{\left(x\right)}_{min}=32\times \frac{{\mathrm{\pi }}^2}{16}=2{\mathrm{\pi }}^2 \\ \Rightarrow \mathrm{f}{\left(\mathrm{x}\right)}_{\max }=32\left({\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right)=20{\mathrm{\pi }}^2 \\ \Rightarrow \mathrm{f}{\left(\mathrm{x}\right)}_{\min }+\mathrm{f}{\left(\mathrm{x}\right)}_{\max }=22{\mathrm{\pi }}^2 \end{gathered}$$

$$\begin{gathered} 16\left({\left({\sec }^{-1}x\right)}^2+{\left({cosec}^{-1}x\right)}^2\right)............\text{(i)} \\ \text{we know that,} se{c}^{-1}x+\cos e{c}^{-1}x=\frac{\mathrm{\pi }}{2} \\ \text{and} se{c}^{-1}x\in \left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\} \\ \text{now, put in (i), we get} \\ 16\left({\left({\sec }^{-1}x\right)}^2+{\left(\frac{\mathrm{\pi }}{2}-se{c}^{-1}x\right)}^2\right) \\ let se{c}^{-1}x=y \\ 16\left(2y^2-\pi y+\frac{{\mathrm{\pi }}^2}{4}\right)=0 \\ \text{Now}\mathit{\text{, }}\text{it}\text{is}\text{maximum}\text{at} y\mathit{=}\mathrm{\pi } \\ \text{maximum}\text{value}\mathit{=}\mathit{16}\mathit{[}\mathit{2}{\pi }^{\mathit{2}}\mathit{-}{\pi }^{\mathit{2}}\mathit{+}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{]} \\ \mathit{=}\mathit{20}{\pi }^{\mathit{2}} \\ \text{and}\text{minimum}\text{at}\mathit{ }y\mathit{=}\frac{\pi }{\mathit{4}} \\ \mathit{ }\mathit{(}y\mathit{=}\mathit{-}\frac{b}{\mathit{2}a}\mathit{ }\mathit{\text{for the quadratic in}}\mathit{ }y\mathit{)} \\ \text{minimum}\text{value}\mathit{ }\mathit{=}\mathit{16}\mathit{[}\frac{\mathit{2}\mathit{\times }{\pi }^{\mathit{2}}}{\mathit{16}}\mathit{-}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{+}\frac{{\pi }^{\mathit{2}}}{\mathit{4}}\mathit{]} \\ \mathit{=}\mathit{2}{\pi }^{\mathit{2}} \\ \text{And}\text{sum}\text{of}\text{maximum}\mathit{\text{ \& }}\text{minimum}\mathit{\text{ value}}\mathit{=} \\ \mathit{ }\mathit{20}{\mathrm{\pi }}^2+2{\mathrm{\pi }}^2=22{\mathrm{\pi }}^2 \end{gathered}$$

Let \(A=\tan^{-1}(1-\alpha)\) and \(B=\tan^{-1}(1-\beta)\).

Given \(A+B=\frac{\pi}{4}\), so \(\tan(A+B)=1\).

\(\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)}=1\)

\(\frac{2-\alpha-\beta}{\alpha+\beta-\alpha\beta}=1\)

\(2-\alpha-\beta=\alpha+\beta-\alpha\beta\)

\(2=2\alpha+2\beta-\alpha\beta\)

Using \(\beta=\frac{1}{3\alpha}\),

\(2=2\alpha+\frac{2}{3\alpha}-\frac{1}{3}\)

Multiplying by \(3\alpha\),

\(6\alpha=6\alpha^2+2-\alpha\)

\(6\alpha^2-7\alpha+2=0\)

\((3\alpha-2)(2\alpha-1)=0\)

\(\alpha=\frac{2}{3}\) or \(\alpha=\frac{1}{2}\)

If \(\alpha=\frac{2}{3}\), then \(\beta=\frac{1}{2}\);

if \(\alpha=\frac{1}{2}\), then \(\beta=\frac{2}{3}\).

So, \(\alpha+\beta=\frac{7}{6}\)

\(6(\alpha+\beta)=6\cdot\frac{7}{6}=7\)

$$\begin{gathered} \text{ Given, }\cot \left({\cos }^{-1}x\right)=\sec \left({\tan }^{-1}\frac{a}{\sqrt{b^2-a^2}}\right) \\ \Rightarrow cot\left({\cot }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=\sec \left({\sec }^{-1}\frac{b}{\sqrt{b^2-a^2}}\right) \\ \Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}} \\ \Rightarrow \frac{b}{\sqrt{2b^2-a^2}}=x \end{gathered}$$

$$\begin{gathered} \text{ Given, }\cot \left({\cos }^{-1}x\right)=\sec \left({\tan }^{-1}\frac{a}{\sqrt{b^2-a^2}}\right) \\ \Rightarrow cot\left({\cot }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=\sec \left({\sec }^{-1}\frac{b}{\sqrt{b^2-a^2}}\right) \\ \Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}} \\ \Rightarrow \frac{b}{\sqrt{2b^2-a^2}}=x \end{gathered}$$

\(A=\sin^{-1}\alpha,\quad B=\sin^{-1}\beta,\quad C=\sin^{-1}\gamma\)

\(A+B+C=\pi\)

\(\gamma=\sin C=\sin(A+B)\)

\(=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\)

\((\alpha+\beta+\gamma)(\alpha+\beta-\gamma)=3\alpha\beta\)

\((\alpha+\beta)^2-\gamma^2=3\alpha\beta\)

\(\gamma=\sin(A+B)\)

\((\sin A+\sin B)^2-\sin^2(A+B)=3\sin A\sin B\)

\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)

\(\sin A\sin B\bigl(1+\cos(A+B)\bigr)\)

\(\sin A\sin B\bigl(1+\cos(A+B)\bigr)=3\sin A\sin B\)

Since \(\alpha,\beta\ne0\)

\(\sin A\sin B\ne0\)

\(1+\cos(A+B)=3\)

\(\cos(A+B)=\dfrac12\)

\(A+B=\pi-C\)

\(\cos(A+B)=\cos(\pi-C)=-\cos C\)

\(-\cos C=\dfrac12\)

\(\cos C=-\dfrac12\)

\(C\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)

\((\alpha+\beta)^2-\gamma^2=3\alpha\beta\)

\(\alpha^2+\beta^2-\alpha\beta=\gamma^2\)

\(\gamma=\sin(A+B)\)

\(\sin^2(A+B)=\alpha^2+\beta^2-\alpha\beta\)

\(\sin^2(A+B)=(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})^2\)

\(2\sqrt{(1-\alpha^2)(1-\beta^2)}=1\)

\((1-\alpha^2)(1-\beta^2)=\dfrac14\)

\(\cos(A+B)=\sqrt{1-\alpha^2}\sqrt{1-\beta^2}-\alpha\beta\)

\(=\dfrac12-\alpha\beta\)

\(C=\pi-(A+B)\)

\(\gamma=\sin(A+B)\)

\(\gamma^2=\alpha^2+\beta^2-\alpha\beta\)

\(\alpha^2+\beta^2=1+\alpha^2\beta^2-\dfrac14\)

\(\gamma^2=\dfrac34\)

\(\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma=\pi\)

all three quantities are positive, so \(\gamma>0\)

\(\gamma=\dfrac{\sqrt3}{2}\).

$$\begin{gathered} {\cot }^{-1}\left\{\beta +\frac{\left(1+{\beta }^2\right)}{(\alpha -\beta )}\right\}+{\cot }^{-1}\left\{\gamma +\frac{\left(1+{\gamma }^2\right)}{(\beta -\gamma )}\right\}+{\cot }^{-1}\left\{\alpha +\frac{\left(1+{\alpha }^2\right)}{(\gamma -\alpha )}\right\} \\ ={\cot }^{-1}\left(\frac{\alpha \beta +1}{\alpha -\beta }\right)+{\cot }^{-1}\left(\frac{\beta \gamma +1}{\beta -\gamma }\right)+{\cot }^{-1}\left(\frac{\alpha \gamma +1}{\gamma -\alpha }\right) \\ ={\tan }^{-1}\left(\frac{\alpha -\beta }{1+\alpha \beta }\right)+{\tan }^{-1}\left(\frac{\beta -\gamma }{1+\beta \gamma }\right)+\pi +{\tan }^{-1}\left(\frac{\gamma -\alpha }{1+\gamma \alpha }\right) \\ =\left({\tan }^{-1}\alpha -{\tan }^{-1}\beta \right)+\left({\tan }^{-1}\beta -{\tan }^{-1}\gamma \right)+\left(\pi +{\tan }^{-1}\gamma -{\tan }^{-1}\alpha \right) \\ =\pi \end{gathered}$$

$$\begin{gathered} \text{ Domain of }{Sec}^{-1}\left(2\right[x]+1) \\ 2\left[x\right]+1\ge 1\text{ or }2\left[x\right]+1\le -1 \\ 2\left[x\right]\ge 0 \text{ or } 2\left[x\right]\le -2 \\ x\ge 0 \text{ or } \left[x\right]\le -1 \\ \therefore x\in (-\infty ,0)\cup [0,\infty ) \end{gathered}$$

For domain of \(f(x)=\sin ^{-1}\left(\frac{x+[x]}{3}\right)\)

\(-1 \leq \frac{x+[x]}{3} \leq 1\)

\(\Rightarrow-3 \leq x+[x] \leq 3\)

\(\Rightarrow-3-[x] \leq x \leq 3-[x]\)

if \([x]=-1 \Rightarrow x \in[-1,0)\)

if \([x]=0 \Rightarrow x \in[0,1)\)

if \([x]=1 \Rightarrow x \in[1,2)\)

If \([x]=-2\), then \(x \in[-2,-1)\) and \(x+[x]=x-2 \in[-4,-3)\), which is outside \([-3,3]\)

If \([x]=2\), then \(x \in[2,3)\) and \(x+[x]=x+2 \in[4,5)\), which is outside \([-3,3]\)

Hence \(x \in[-1,2)\)

\(\Rightarrow \alpha=-1\) and \(\beta=2\)

\(\Rightarrow \alpha^2+\beta^2=5\)

$$\begin{gathered} \text{ Domain of }{Sec}^{-1}\left(2\right[x]+1) \\ 2\left[x\right]+1\ge 1\text{ or }2\left[x\right]+1\le -1 \\ 2\left[x\right]\ge 0 \text{ or } 2\left[x\right]\le -2 \\ x\ge 0 \text{ or } \left[x\right]\le -1 \\ \therefore x\in (-\infty ,0)\cup [0,\infty ) \end{gathered}$$

Since \(x>0\),

\(\tan^{-1}(x),\tan^{-1}(2x)\in\left(0,\frac{\pi}{2}\right)\)

and their sum is \(\frac{\pi}{4}\).

Therefore,

\(\tan\left(\tan^{-1}(x)+\tan^{-1}(2x)\right)=\tan\frac{\pi}{4}=1\)

\(\frac{x+2x}{1-2x^2}=1\)

\(\frac{3x}{1-2x^2}=1\)

\(2x^2+3x-1=0\)

\(x=\frac{-3\pm\sqrt{17}}{4}\)

Only \(x=\frac{-3+\sqrt{17}}{4}>0\) is acceptable.

Hence there is exactly one positive real solution.

\(\begin{aligned}
& \sec ^2\left(\tan ^{-1} \alpha\right)+\cos e c^2\left(\cot ^{-1} \beta\right)=36, \quad \alpha+\beta=8 \\
& 1+\alpha^2+1+\beta^2=36 \Rightarrow \alpha^2+\beta^2=34 \\
& \Rightarrow \alpha^2+(8-\alpha)^2=34 \\
& \Rightarrow 2 \alpha^2-16 \alpha+30=0 \\
& \alpha^2-8 \alpha+15=0 \Rightarrow \alpha=5, \beta=3 \\
& \alpha^2+\beta=28
\end{aligned}\)

$$\begin{gathered} {\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1) \\ \text{ Now }-1\le 2x-1\le 1 \\ 0\le x\le 1 \\ \Rightarrow \pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1)\ge \frac{\pi }{2} \\ \text{ and }{\cos }^{-1}x\text{ for }x\in [0,1]\text{ always lies in }\left[0,\frac{\pi }{2}\right] \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS}=\frac{\pi }{2} \\ \Rightarrow {\cos }^{-1}x=\frac{\pi }{2}\Rightarrow x=0 \\ \mathrm{Hence} \mathrm{only} x=0 \text{is the possible solution}. \\ \text{ Sum of all solution}=0. \end{gathered}$$

\(f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)\)

\(\Rightarrow 2 x+3 \neq 0 \text { and }-1 \leq \frac{x-1}{2 x+3} \leq 1\)

\(\Rightarrow \frac{x-1-2 x-3}{2 x+3} \leq 0 \Rightarrow \frac{x+4}{2 x+3} \geq 0\)

Also, \(0 \leq \frac{x-1+2 x+3}{2 x+3} \Rightarrow \frac{3 x+2}{2 x+3} \geq 0\)

\(\Rightarrow x \in\left(-\infty, \frac{-3}{2}\right) \cup\left[\frac{-2}{3}, \infty\right)\)

\(\Rightarrow \text { domain } \Rightarrow x \in(-\infty,-4] \cup\left(\frac{-2}{3}, \infty\right)\)

\(\Rightarrow R-\left(-4, \frac{-2}{3}\right]\)

\(\Rightarrow \alpha=-4, \beta=\frac{-2}{3}\)

\(\Rightarrow 12 \times(-4)\left(\frac{-2}{3}\right)=32\)

\(\begin{aligned}
& \cos ^{-1}\left[\frac{12}{13} \cos x+\frac{5}{13} \sin x\right] \\
& let \cos \alpha=\frac{12}{13}. \\
&\sin \alpha=\frac{5}{13} \\
& \cos ^{-1}[\cos \alpha \cos x+\sin \alpha \sin x] \\
& \cos ^{-1}(\cos (x-\alpha)) \\
&= x-\tan ^{-1} \frac{5}{12}
\end{aligned}\)

Let \(\alpha=\sin^{-1}\!\left(\frac35\right)\)

Then \(\sin\alpha=\frac35,\quad \cos\alpha=\frac45\)

\(\tan\alpha=\frac34\)

Let \(\beta=\cos^{-1}\!\left(\frac{2}{\sqrt5}\right)\)

\(\cos\beta=\frac{2}{\sqrt5},\quad \sin\beta=\frac1{\sqrt5}\)

\(\tan\beta=\frac12\)

\(\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}
=\frac{2\cdot\frac12}{1-\frac14}
=\frac43\)

\(\tan(\alpha-2\beta)
=\frac{\tan\alpha-\tan2\beta}{1+\tan\alpha\tan2\beta}\)

\(=\frac{\frac34-\frac43}{1+\frac34\cdot\frac43}
=\frac{-\frac7{12}}2
=-\frac7{24}\)

Range of ${\text{sec}}^{-1}x$ is $\mathrm{x}\in (-\infty ,-1]\cup [1,\infty )$

\(2[x]+1 \leq-1\) or \(2[x]+1 \geq 1\)

\(\Rightarrow[x] \leq-1\) or \([x] \geq 0\)

\(\Rightarrow \mathrm{x} \in(-\infty, 0)\) or \(x \in[0, \infty)\)

\(\Rightarrow x \in(-\infty, \infty)\)

97

Let \(\theta=\tan ^{-1}(x \sqrt{2})\)

\(\tan \theta=x \sqrt{2}\)

\(\sin \theta=\frac{x \sqrt{2}}{\sqrt{1+2 x^2}}\)

Therefore we get

\(\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\frac{x \sqrt{2}}{\sqrt{1+2 x^2}}\)

Now let \(\phi=\sin ^{-1} \sqrt{1-x^2}\)

\(\cos \phi=\sqrt{1-\sin ^2 \phi}=\sqrt{1-\left(1-x^2\right)}=\sqrt{x^2}=x\)

\(\cot \phi=\frac{\cos \phi}{\sin \phi}=\frac{x}{\sqrt{1-x^2}}\)

So, \(\cot \left(\sin ^{-1} \sqrt{1-x^2}\right)=\frac{x}{\sqrt{1-x^2}}\)

Now, the equation becomes

\(\frac{x \sqrt{2}}{\sqrt{1+2 x^2}}=\frac{x}{\sqrt{1-x^2}}\)

\(2\left(1-x^2\right)=1+2 x^2\)

\(x^2=\frac{1}{4}\)

Since \(x \in(0,1)\),

\(x=\frac{1}{2}\)

Assume \(\sin ^{-1} \mathbf{x}=\theta\)

\( \cos (2 \theta)=\frac{1}{9}\)
\(\sin \theta= \pm \frac{2}{3}\)
as \(m\) and \(n\) are co-prime natural numbers,

\(x=\frac{2}{3}\)
i.e. \(m=2, n=3\)

So, the quadratic equation becomes \(2 x^2-3 x+1=0\)

whose roots are \(\alpha=1, \beta=\frac{1}{2}\)
\(\left(1, \frac{1}{2}\right)\) lies on \(5 x+8 y=9\).

Assume \(\sin ^{-1} \mathbf{x}=\theta\)

\( \cos (2 \theta)=\frac{1}{9}\)
\(\sin \theta= \pm \frac{2}{3}\)
as \(m\) and \(n\) are co-prime natural numbers,

\(x=\frac{2}{3}\)
i.e. \(m=2, n=3\)

So, the quadratic equation becomes \(2 x^2-3 x+1=0\)

whose roots are \(\alpha=1, \beta=\frac{1}{2}\)
\(\left(1, \frac{1}{2}\right)\) lies on \(5 x+8 y=9\).

$$\begin{gathered} \cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)=\cos \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{33}{56}\right) \\ =\cos \left({\tan }^{-1}\left(\frac{3/4+5/12}{1-\frac{3}{4}·\frac{5}{12}}\right)+{\tan }^{-1}\left(\frac{33}{56}\right)\right) \\ =\cos \left({\tan }^{-1}\frac{56}{33}+{\cot }^{-1}\left(\frac{56}{33}\right)\right) \\ =\cos \pi /2=0 \end{gathered}$$

\(\begin{aligned}
& \cos ^{-1}\left[\frac{12}{13} \cos x+\frac{5}{13} \sin x\right] \\
& let \cos \alpha=\frac{12}{13}. \\
&\sin \alpha=\frac{5}{13} \\
& \cos ^{-1}[\cos \alpha \cos x+\sin \alpha \sin x] \\
& \cos ^{-1}(\cos (x-\alpha)) \\
&= x-\tan ^{-1} \frac{5}{12}
\end{aligned}\)

Let ${\theta }_1=2$ radians. Since $\frac{\pi }{2}<2<\pi ,\sec 2$ is negative.
Consider the first term's argument:

$$\begin{gathered} E_1=\frac{\sqrt{1+{\tan }^{2}2}-1}{\tan 2}=\frac{\left|\sec 2\right|-1}{\tan 2} \\ =\frac{-\sec 2-1}{\tan 2}=\frac{-\frac{1}{\cos 2}-1}{\frac{\sin 2}{\cos 2}}=\frac{-(1+\cos 2)}{\sin 2} \end{gathered}$$

Using half-angle formulas:

$E_1=\frac{-2{\cos }^{2}1}{2\sin 1\cos 1}=-\cot 1$

The first term is ${\cot }^{-1}(-\cot 1)=\pi -{\cot }^{-1}\left(\cot 1\right)=\pi -1$(since 1 is in the principal range $(0,\pi )$).

Let ${\theta }_2=\frac{1}{2}$ radian. Since $0<\frac{1}{2}<\frac{\pi }{2},\sec \left(\frac{1}{2}\right)$ is positive.
Consider the second term's argument:

$$\begin{gathered} E_2=\frac{\sqrt{1+{\tan }^2(1/2)}+1}{\tan (1/2)}=\frac{\sec (1/2)+1}{\tan (1/2)}=\frac{1+\cos (1/2)}{\sin (1/2)} \\ =\frac{2{\cos }^2(1/4)}{2\sin (1/4)\cos (1/4)}=\cot (1/4) \end{gathered}$$

The second term is ${\cot }^{-1}\left(\cot \right(1/4\left)\right)=\frac{1}{4}$ (since $\frac{1}{4}$ is in the principal range $(0,\pi )$).

The full expression is:

$(\pi -1)-\frac{1}{4}=\pi -\frac{5}{4}$

\(\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{4}{4 \mathrm{n}^2+3}\right)\)

\(T_n=\tan ^{-1}\left(\frac{\left(n+\frac{1}{2}\right)-\left(n-\frac{1}{2}\right)}{1+\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)}\right)\)

\(\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\mathrm{n}-\frac{1}{2}\right)\)

\(\mathrm{T}_1+\mathrm{T}_2+\ldots+\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)\)

\(\mathrm{S}_{\infty}=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)\)

Required value =

$$\begin{gathered} {\cot }^{-1}\left(\frac{\alpha \beta +1}{\alpha -\beta }\right)+{\cot }^{-1}\left(\frac{\beta \gamma +1}{\beta -\gamma }\right)+{\cot }^{-1}\left(\frac{\alpha \gamma +1}{\gamma -\alpha }\right) \\ ={\tan }^{-1}\left(\frac{\alpha -\beta }{1+\alpha \beta }\right)+{\tan }^{-1}\left(\frac{\beta -\gamma }{1+\beta \gamma }\right)+\pi +{\tan }^{-1}\left(\frac{\gamma -\alpha }{1+\gamma \alpha }\right) \\ =\left({\tan }^{-1}\alpha -{\tan }^{-1}\beta \right)+\left({\tan }^{-1}\beta -{\tan }^{-1}\gamma \right)+\left(\pi +{\tan }^{-1}\gamma -{\tan }^{-1}\alpha \right) \\ =\pi \end{gathered}$$

Let \(t=\tan\theta\)

\(\frac{6t}{9+t^2}=\frac{2\cdot(2t/3)}{1+(2t/3)^2}\)

\(\sin^{-1}\left(\frac{6t}{9+t^2}\right)=2\tan^{-1}\left(\frac{t}{3}\right)\)

because \(\tan^{-1}\left(\frac{t}{3}\right)\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right]\)

Hence the equation becomes

\(\theta=\tan^{-1}(2t)-\tan^{-1}\left(\frac{t}{3}\right)\)

\(\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)\)

\(\theta=\tan^{-1}\left(\frac{5t/3}{1+\frac{2t^2}{3}}\right)\)

\(\theta=\tan^{-1}\left(\frac{5t}{3+2t^2}\right)\)

Taking tangent on both sides

\(t=\frac{5t}{3+2t^2}\)

\(t(3+2t^2)=5t\)

\(2t(t^2-1)=0\)

\(t=0,\pm1\)

\(t=0 \Rightarrow \theta=0\)

\(t=1 \Rightarrow \theta=\frac{\pi}{4}\)

\(t=-1 \Rightarrow \theta=-\frac{\pi}{4}\)

Thus there are \(3\) real solutions

\(\begin{aligned}
& \tan ^{-1}\left(\frac{\alpha-\beta}{1+\alpha \beta}\right)+\tan ^{-1}\left(\frac{\beta-\gamma}{1+\beta \gamma}\right)+ \pi +\tan ^{-1}\left(\frac{\gamma-\alpha}{1+\alpha \gamma}\right) \\
& =\tan ^{-1} \alpha-\tan \beta+\tan ^{-1} \beta-\tan ^{-1} \gamma+ \pi +\tan ^{-1} \gamma-\tan ^{-1} \alpha \\
& =\pi .
\end{aligned}\)

Let ${\sin }^{-1}\frac{2}{\sqrt{13}}=\theta \& {\cos }^{-1}\frac{3}{\sqrt{10}}=\varphi$

$\sin \theta =\frac{2}{\sqrt{13}} \& \cos \varphi =\frac{3}{\sqrt{10}}$

$\tan \theta =\frac{2}{3} \& \tan \varphi =\frac{1}{3}$

$$\begin{gathered} \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta } \\ =\frac{12}{5} \end{gathered}$$

$$\begin{gathered} \tan 2\varphi =\frac{2\tan \varphi }{1-{\tan }^2\varphi } \\ =\frac{3}{4} \end{gathered}$$

$\tan \left(2\theta -2\varphi \right)=\frac{\tan 2\theta -\tan 2\varphi }{1+\tan 2\theta \tan 2\varphi }$

$=\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot \frac{3}{4}}$

$=\frac{33}{56}$

Since \(\dfrac{3\pi}{2}<5<2\pi\)

the principal value of \(\sin^{-1}(\sin5)\) is

\(a=5-2\pi\)

\(\pi<5<2\pi\)

\(b=2\pi-5\)

\(a=-(2\pi-5)\)

and \(b=2\pi-5\)

\(a^2+b^2
=(2\pi-5)^2+(2\pi-5)^2
=2(2\pi-5)^2\)

Since \(\frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}}\), we get:

\(\sin^{-1}\left(\frac{1}{2}\right)<\sin^{-1}\left(\frac{6}{11}\right)<\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\).

So, \(\frac{\pi}{6}<\sin^{-1}\left(\frac{6}{11}\right)<\frac{\pi}{4}\)

Multiplying by \(3\):

\(\frac{\pi}{2}<\alpha<\frac{3\pi}{4}\), hence \(\cos\alpha<0\).

So, Statement II is true.

Since \(\frac{4}{9}<\frac{1}{2}\), we get:

\(\cos^{-1}\left(\frac{4}{9}\right)>\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)

Also, \(\frac{4}{9}>0\), so \(\cos^{-1}\left(\frac{4}{9}\right)<\frac{\pi}{2}\)

Thus, \(\frac{\pi}{3}<\cos^{-1}\left(\frac{4}{9}\right)<\frac{\pi}{2}\)

Multiplying by \(3\):

\(\pi<\beta<\frac{3\pi}{2}\)

Now, \(\frac{\pi}{2}<\alpha<\frac{3\pi}{4}\) and \(\pi<\beta<\frac{3\pi}{2}\)

So, \(\frac{3\pi}{2}<\alpha+\beta<\frac{9\pi}{4}\)

In this interval, \(\cos(\alpha+\beta)>0\),

so Statement I is true.

Assume \(\sin ^{-1} \mathrm{x}=\theta\)

$$
\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}
$$

as \(m\) and \(n\) are co-prime natural numbers,

\(x=\frac{2}{3}\)

i.e. \(m=2, n=3\)
So, the quadratic equation becomes \(2 x^2-3 x+1=0\) whose roots are \(\alpha=1, \beta=\frac{1}{2} \left(1, \frac{1}{2}\right)\) lies on \(5 x+8 y=9\).

\(16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right) \ldots(i)\)

we know that,\(\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\)

and \(\sec ^{-1} x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}\)

now ,put in(i),we get

\(16\left(\left(\sec ^{-1} x\right)^2+\left(\frac{\pi}{2}-\sec ^{-1} x\right)^2\right)\)

let \(\sec ^{-1} x=y\)

\(16\left(2 y^2-\pi y+\frac{\pi^2}{4}\right)=0\)

Now, it is maximum at \(y=\pi\)

maximum value \(=16\left[2 \pi^2-\pi^2+\frac{\pi^2}{4}\right]\)

\(=20 \pi^2\)

and minimum at \(y=\frac{\pi}{4}\)

\(\left(y=-\frac{b}{2 a}\right.\) for the quadratic in \(\left.y\right)\)

minimum value \(=16\left[\frac{2 \times \pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}\right]\)

\(=2 \pi^2\)

And sum of maximum &minimum value= \(20 \pi^2+2 \pi^2=22 \pi^2\)

Let \(\alpha=\tan^{-1}\frac{5}{12}\)

\(\cos\alpha=\frac{12}{13}\), \(\sin\alpha=\frac{5}{13}\)

\(\frac{12}{13}\cos x+\frac{5}{13}\sin x=\cos\alpha\cos x+\sin\alpha\sin x\)

\(=\cos(x-\alpha)\)

\(\cos^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)=\cos^{-1}(\cos(x-\alpha))\)

\(\frac{\pi}{2}\le x\le\frac{3\pi}{4}\) and \(0<\alpha<\frac{\pi}{2}\)

\(0<x-\alpha<\pi\)

Hence \(x-\alpha\) already lies in the principal range of \(\cos^{-1}\), namely \([0,\pi]\).

\(\cos^{-1}(\cos(x-\alpha))=x-\alpha\)

\(=x-\tan^{-1}\frac{5}{12}\)

\(\cot^{-1}(\cot(-11))\) must be taken in principal range \((0,\pi)\).

Since \(-11+4\pi\in(0,\pi)\),

\(\cot^{-1}(\cot(-11))=4\pi-11\)

\(\cos^{-1}\dfrac1{\sqrt2}=\dfrac{\pi}{4}\)

\(10\sin\left(2\cos^{-1}\dfrac1{\sqrt2}\right)=10\sin\dfrac{\pi}{2}=10\)

Also, \(\sin(2\tan^{-1}2)=\dfrac{2\tan(\tan^{-1}2)}{1+\tan^2(\tan^{-1}2)}=\dfrac{4}{5}\)

So \(10\sin(2\tan^{-1}2)=8\).

Total value \(=4\pi-11+10+8=4\pi+7\)

\(\begin{aligned}
& \tan ^{-1}\left(\frac{\alpha-\beta}{1+\alpha \beta}\right)+\tan ^{-1}\left(\frac{\beta-\gamma}{1+\beta \gamma}\right)+ \pi +\tan ^{-1}\left(\frac{\gamma-\alpha}{1+\alpha \gamma}\right) \\
& =\tan ^{-1} \alpha-\tan \beta+\tan ^{-1} \beta-\tan ^{-1} \gamma+ \pi +\tan ^{-1} \gamma-\tan ^{-1} \alpha \\
& =\pi .
\end{aligned}\)

\(\begin{aligned}
& \sec ^2\left(\tan ^{-1} \alpha\right)+\cos e c^2\left(\cot ^{-1} \beta\right)=36, \quad \alpha+\beta=8 \\
& 1+\alpha^2+1+\beta^2=36 \Rightarrow \alpha^2+\beta^2=34 \\
& \Rightarrow \alpha^2+(8-\alpha)^2=34 \\
& \Rightarrow 2 \alpha^2-16 \alpha+30=0 \\
& \alpha^2-8 \alpha+15=0 \Rightarrow \alpha=5, \beta=3 \\
& \alpha^2+\beta=28
\end{aligned}\)

\(16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right) \ldots(i)\)

we know that,\(\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\)

and \(\sec ^{-1} x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}\)

now ,put in(i),we get

\(16\left(\left(\sec ^{-1} x\right)^2+\left(\frac{\pi}{2}-\sec ^{-1} x\right)^2\right)\)

let \(\sec ^{-1} x=y\)

\(16\left(2 y^2-\pi y+\frac{\pi^2}{4}\right)=0\)

Now, it is maximum at \(y=\pi\)

maximum value \(=16\left[2 \pi^2-\pi^2+\frac{\pi^2}{4}\right]\)

\(=20 \pi^2\)

and minimum at \(y=\frac{\pi}{4}\)

\(\left(y=-\frac{b}{2 a}\right.\) for the quadratic in \(\left.y\right)\)

minimum value \(=16\left[\frac{2 \times \pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}\right]\)

\(=2 \pi^2\)

And sum of maximum &minimum value= \(20 \pi^2+2 \pi^2=22 \pi^2\)

$$\sin^{-1}\!\left(\frac{\sqrt3}{2}x+\frac12\sqrt{1-x^{2}}\right),\qquad -\frac12<x<\frac1{\sqrt2}.$$

Let

$$x=\sin\theta \quad\Rightarrow\quad \sqrt{1-x^2}=\cos\theta,$$

with $\theta=\sin^{-1}x\in\left(-\frac\pi6,\frac\pi4\right)$ (from the given range of $x$).

Then the expression inside $\sin^{-1}$ becomes

$$\frac{\sqrt3}{2}\sin\theta+\frac12\cos\theta = \sin\theta\cos\frac\pi6+\cos\theta\sin\frac\pi6 = \sin\left(\theta+\frac\pi6\right).$$

So the whole expression is

$$\sin^{-1}\Big(\sin\big(\theta+\tfrac\pi6\big)\Big).$$

Now check the range of $\theta+\frac\pi6$:

$$\theta\in\left(-\frac\pi6,\frac\pi4\right) \implies \theta+\frac\pi6\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac\pi2,\frac\pi2\right].$$

On this interval, $\sin^{-1}(\sin y)=y$. Thus

$$\sin^{-1}\left(\frac{\sqrt3}{2}x+\frac12\sqrt{1-x^{2}}\right) = \theta+\frac\pi6 = \sin^{-1}x+\frac\pi6.$$

So the correct option is:

$$\boxed{{\displaystyle \frac{\pi }{6}+{\sin }^{-1}x}}$$

\(-1 \leq\left|\frac{2-|x|}{4}\right| \leq 1\)

\(\Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1\)

\(-4 \leq 2-|x| \leq 4\)

\(-6 \leq-|x| \leq 2 \)

\(-2 \leq|x| \leq 6\)

\( |x| \leq 6 \)
\( \Rightarrow x \in[-6,6] .......{1}\)
Now, \(3-x \neq 1\)
And \(x \neq 2\)

\(\text { and } 3-x>0\)

\(x<3 \ldots .(3)\)

\(\text { From (1), (2) and (3) }\)

\(\Rightarrow x \in[-6,3)-\{2\}\)

\( \alpha=6\)

\(\beta=3\)

\(\gamma=2\)

\(\alpha+\beta+\gamma=11\)

\(\alpha=\sin^{-1}\frac{3}{5}\), \(\beta=\sin^{-1}\frac{5}{13}\)

\(\cos\alpha=\frac{4}{5}\), \(\cos\beta=\frac{12}{13}\)

\(\sin(\alpha+\beta)=\frac{3}{5}\cdot\frac{12}{13}+\frac{4}{5}\cdot\frac{5}{13}=\frac{56}{65}\)

\(\cos(\alpha+\beta)=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}\)

\(\alpha+\beta=\sin^{-1}\frac{56}{65}=\cos^{-1}\frac{33}{65}\)

\(\gamma=\sin^{-1}\frac{33}{65}\)

Since \(\cos\gamma=\frac{56}{65}\),

\((\alpha+\beta)+\gamma=\frac{\pi}{2}\)

\(\cos\left(\alpha+\beta+\gamma\right)=\cos\frac{\pi}{2}=0\)

$$\begin{gathered} {\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1) \\ \text{ Now }-1\le 2x-1\le 1 \\ 0\le x\le 1 \\ \Rightarrow \pi +{\sin }^{-1}x+{\sin }^{-1}(2x-1)\ge \frac{\pi }{2} \\ \text{ and }{\cos }^{-1}x\text{ for }x\in [0,1]\text{ always lies in }\left[0,\frac{\pi }{2}\right] \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS}=\frac{\pi }{2} \\ \Rightarrow {\cos }^{-1}x=\frac{\pi }{2}\Rightarrow x=0 \\ \mathrm{Hence} \mathrm{only} x=0 \text{is the possible solution}. \\ \text{ Sum of all solution}=0. \end{gathered}$$

\(\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right), \frac{-1}{2}<x<\frac{1}{\sqrt{2}}\)

Let \( \sin ^{-1}(x)=\theta \frac{-\pi}{6}<\theta<\frac{\pi}{4}\)

\(\Rightarrow \mathrm{x}=\sin \theta,\)

\(\Rightarrow \sin ^{-1}\left(\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta\right)\)

\(\Rightarrow \sin ^{-1}\left(\sin \left(\theta+\frac{\pi}{6}\right)\right)=\theta+\frac{\pi}{6}\)

\(\Rightarrow \sin ^{-1}(\mathrm{x})+\frac{\pi}{6}\)

$$\begin{gathered} {\cos }^{-1}\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right) \\ \text{where}, \frac{\pi }{2}\le x\le \frac{3\pi }{4} \\ \text{let}, \cos \theta =\frac{12}{13}, \text{then} \sin \theta =\frac{5}{13} \\ \text{and} \tan \theta =\frac{5}{12} \\ \theta ={\tan }^{-1}\frac{5}{12} \\ Now, {\cos }^{-1}\left(\cos \theta .\cos x+\sin \theta .\sin x\right) \\ {\cos }^{-1}\left(\cos \right(x-\theta \left)\right) \\ x-\theta \\ x-{\tan }^{-1}\frac{5}{12} \end{gathered}$$

$$\begin{gathered} \text{Let,} \\ A = \cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right) \\ A=\cos \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{33}{56}\right) \\ \text{we know,} \\ {\tan }^{-1}x+{\tan }^{-1}y ={\tan }^{-1}\left(\frac{x+y}{1-xy}\right) \\ \text{Now}, \\ A =\cos \left({\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}·\frac{5}{12}}\right)+{\tan }^{-1}\frac{33}{56}\right) \\ A=\cos \left({\tan }^{-1}\frac{56}{33}+{\cot }^{-1}\frac{56}{33}\right) \\ A = \cos \left(\frac{\mathrm{\pi }}{2}\right)=0 \end{gathered}$$

Sol: $E = \cot^{-1}\left(\frac{-\sec 2 - 1}{\tan 2}\right) - \cot^{-1}\left(\frac{\sec \frac{1}{2} + 1}{\tan \frac{1}{2}}\right)$

$$\begin{array}{l}={\cot }^{-1}(-\left(\frac{1+\cos 2}{\sin 2}\right))-{\cot }^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right)\\ ={\cot }^{-1}(-\cot 1)-{\cot }^{-1}(\cot \frac{1}{4})\\ =\pi -1-\frac{1}{4}=\pi -\frac{5}{4}\end{array}$$

$$\begin{gathered} \cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65}\right)=\cos \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{33}{56}\right) \\ =\cos \left({\tan }^{-1}\left(\frac{3/4+5/12}{1-\frac{3}{4}·\frac{5}{12}}\right)+{\tan }^{-1}\left(\frac{33}{56}\right)\right) \\ =\cos \left({\tan }^{-1}\frac{56}{33}+{\cot }^{-1}\left(\frac{56}{33}\right)\right) \\ =\cos \pi /2=0 \end{gathered}$$