Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

L.H.S = \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 1 }{ 7 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 7 } \right) \left( \frac { 1 }{ 13 } \right) } \right) ={ tan }^{ -1 }\left( \cfrac { \frac { 13+7 }{ 91 } }{ \frac { 91-1 }{ 91 } } \right) \)

\(=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac {90}{91}}\right)=\tan ^{-1}\left(\frac{20}{\not 91} \times \frac{\not 91}{90}\right) \)
\(=\tan ^{-1}\left(\frac{\not 20^{2}}{\not 90^{9}}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\mathrm{RHS} \)
Hence proved.

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

We know that tan^-1 x + tan^-1 y = tan^-1\(\frac{x+y}{1-xy}\), xy < 1
Thus, tan⁻¹\(\frac{1}{2}+tan^{-1}\frac{1}{3}={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 3 } \right) } ={ tan }^{ -1 }(1)=\frac { \pi }{ 4 } \)

 \(tan\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right) \)
Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =x\)
\(\Rightarrow \frac { 1 }{ 2 } =cosx\)
\(\Rightarrow cosc=cos\frac { \pi }{ 3 } \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
Let \({ sin }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\)
\(\Rightarrow \left( \frac { -1 }{ 2 } \right) =siny\)
\(\Rightarrow siny=\frac { -1 }{ 2 } =-sin\frac { \pi }{ 6 } =\left( \frac { -\pi }{ 6 } \right) \)
\(\Rightarrow y=\frac { -\pi }{ 6 } \)
\(\therefore { tan }^{ -1 }\left( cos^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \({ tan }^{ -1 }\left( \frac { \pi }{ 3 } -\left( \frac { -\pi }{ 6 } \right) \right) ={ tan }^{ -1 }\left( \frac { \pi }{ 3 } +\frac { \pi }{ 6 } \right) \)
= \(tan\left( \frac { 2\pi +\pi }{ 0 } \right) =tan\left( \frac { 3\pi }{ 6 } \right) =tan\left( \frac { \pi }{ 2 } \right) \)
= \(\infty \)

By definition, the domain of yx = cos^-1 x is -1. This leads to \(-1\le\frac{2+sinx}{3}\le1\) which is same as -3\(\le\)2+sinx\(\le\)3
so, -5\(\le sin\ x\le1 \) reduces to -1\(\le sin\ x\le1 \), which gives
-sin^-1(1)\(\le x\le sin^-1(1) or -\frac{\pi}{2}\le x\le \frac{\pi}{2}\)
Thus, the domain of cos-1\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

\({ sin }^{ -1 }\left( \frac { sinx+cosx }{ \sqrt { 2 } } \right)\)
\(\begin{aligned} & =\sin ^{-1}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right) \\ & =\sin ^{-1}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right) \\ & =\sin ^{-1}\left(\sin \left(x+\frac{\pi}{4}\right)\right)=\pi+\frac{\pi}{4} \end{aligned}\)

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 }\)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 }\)
\(\begin{aligned} & \Rightarrow\left(\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}}{\frac{(x-2)(x+2)}{(x+2)}}\right)=\tan \frac{\pi}{4}=1 \\ & \Rightarrow \quad \frac{x^2+x-2+x^2-x-2}{\left(x^2-4\right)-\left(x^2-1\right)}=1 \end{aligned}\)
\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 }\)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } }\)

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

Let sec^-1 \(\frac{5}{4}=\theta\).
Then, sec \(\theta=\frac{5}{4}\) and hence, cos \(\theta=\frac{4}{5}\)
Also, sin \(\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-{ \left( \frac { 4 }{ 5 } \right) }^{ 2 } } =\frac { 3 }{ 5 } \), which gives \(\theta=sin^{-1}(\frac{3}{5})\)
Thus, sec^-1\(\left( \frac { 5 }{ 4 } \right) ={ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) and\ { sin }^{ -1 }\frac { 3 }{ 5 } +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =2{ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
We know that sin^-1\(\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) =2{ sin }^{ -1 }x,\quad if|x|\le \frac { 1 }{ \sqrt { 2 } } \)
Since \(\frac { 3 }{ 5 } <\frac { 1 }{ \sqrt { 2 } } \), we have 2\({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) ={ sin }^{ -1 }\left( 2\times \frac { 3 }{ 5 } \sqrt { { 1-\left( \frac { 3 }{ 5 } \right) }^{ 2 } } \right) ={ sin }^{ -1 }\left( \frac { 24 }{ 25 } \right) \)
Hence, \(sin\left[ { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) \right] =sin\left( { sin }^{ -1 }\left( \frac { 24 }{ 25 } \right) \right) =\frac { 24 }{ 25 } ,\ since\frac { 24 }{ 25 } \in [-1,1]\)