\(\mathrm{LHS}=\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1}\left\{\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \times \frac{7}{24}}\right.\)
\(\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \text { if } x y<1\right] \)
\(= \tan ^{-1}\left\{\frac{48+77}{264-14}\right\}=\tan ^{-1}\left\{\frac{125}{250}\right\}=\tan ^{-1}\left\{\frac{1}{2}\right\}=\mathrm{RHS} \) 
Hence proved.

We have: \(f(x)=x^{ 2 }-3x+2\) ....(1)
\(\therefore \) \(f(f(x))=(f(x))^{ 2 }-3f(x)+2\)  [Using (1)]
\(=(x^{ 2 }-3x+2)^{ 2 }-3(x^{ 2 }-3x+2)+2\)
\(=(x^{ 4 }+9x^{ 2 }+4-6x^{ 3 }-12x+4x^{ 2 })+(-3x^{ 2 }+9x-6)+2=x^{ 4 }-6x^{ 3 }+10x^{ 2 }-3x.\)

Let's consider \(\tan ^{-1}(1)=x\). Then, \(\tan x=1=\tan \left(\frac{\pi}{4}\right)\). \(\therefore \tan ^{-1}(1)=\frac{\pi}{4}\)
Let's assume,\(\cos ^{-1}\left(-\frac{1}{2}\right)=y\).
Then, \(\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)\)
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)
Let's again assume that \(\sin ^{-1}\left(-\frac{1}{2}\right)=z\).
Then, \(\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\).
\(\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
\(\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
\(=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \)
\(=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly
\(\lim _{ x\rightarrow 1 }{ f(x) } =\lim _{ x\rightarrow 1 }{ \left( { 2x }+3 \right) } =2(1)+3=5\)
\(f(1)=2(1)+3=5.\)
\(Thus\quad \lim _{ x\rightarrow 1 }{ f(x) } =f(1)\)
Hence,f is continuous at x = 1.

A = {1, 2, 3 ...13, 14}
R = {(x, y) : 3x - y = 0}
∴ R = {(1, 3), (2, 6), (3, 9),(4, 12)}
R is not reflexive since (1, 2), (2, 2)...(14, 14) ∉ R
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R.[3(3) - 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) - 9 ≠ 0]
Hence R is neither reflexive, nor symmetric, nor transitive. 

Let \(y=f(x)=\frac{2 x-1}{5} \Rightarrow 5 y=2 x-1\)
\(\Rightarrow 2 x-5 y+1 \Rightarrow x=\frac{5 y+1}{2}\)
\(\Rightarrow f^{-1}(y)=\frac{5 y+1}{2}\)
\(\left[\because f\right.\) is invertible, therefore \(\left.y=f(x) \Rightarrow x=f^{-1}(y)\right]\) or \(f^{-1}(x)=\frac{5 x+1}{2}\)

\(L.H.S.={ sin }^{ -1 }\frac { 5 }{ 13 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(\because { sin }^{ -1 }x={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\({ sin }^{ -1 }\frac { 5 }{ 13 } ={ tan }^{ -1 }\left( \frac { 5/13 }{ \sqrt { 1-25/169 } } \right) \)
\(={ tan }^{ -1 }\frac { (5/13) }{ \sqrt { \frac { 144 }{ 169 } } } ={ tan }^{ -1 }\frac { 5 }{ 12 } \)
\(and\quad { cos }^{ -1 }x={ tan }^{ -1 }\left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) \)
\(\therefore \quad { cos }^{ -1 }\frac { 3 }{ 5 } ={ tan }^{ -1 }\left( \frac { \sqrt { 1-\frac { 9 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \sqrt { \frac { 16 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
\({ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 4 }{ 3 } ={ tan }^{ -1 }\left[ \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 2 } }{ 1-\frac { 5\times 4 }{ 12\times 3 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left| \frac { \frac { 15+48 }{ 36 } }{ \frac { 36-20 }{ 36 } } \right| \)
\(={ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =R.H.S.\)

\(\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { a-b }{ a+b } \right) } =\frac { \pi }{ 4 } \)
Alternative Method :
\(\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { a-b }{ a+b } \right) } =\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } -\tan ^{ -1 }{ \left( \frac { \frac { a }{ b } -1 }{ \frac { a }{ b } +1 } \right) } \)
By taking \(\frac { a }{ b } =tan\quad \theta \)
\({ tan }^{ -1 }(tan\quad \theta )-{ tan }^{ -1 }\left( \frac { tan\quad \theta -tan\left( \frac { \pi }{ 4 } \right) }{ 1+tan\quad \theta \quad tan\left( \frac { \pi }{ 4 } \right) } \right) \)
\(={ tan }^{ -1 }(tan\quad \theta )-{ tan }^{ -1 }\left[ tan\left( \theta -\frac { \pi }{ 4 } \right) \right] \)
\(=\theta -\theta +\frac { \pi }{ 4 } \ \left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
\(=\frac { \pi }{ 4 } \)

Given, R = {(x, y): 3x² -7xy + 4y² = 0; x, y ∊ N}
= {(x, y) : 3x² - 3xy - 4xy + 4y² = 0; x, y∊N}
= {(x, y) : 3x(x - y) - 4y (x - y) = 0; x, y∊N}
= {(x, y) : (x - y)(3x -4y)=0; x, y∊ N}
Here, we get those ordered pairs (x, y) which satisfy the equation defined in R.
Reflexive Since, xRx ⇒ (x - x)(3x - 4x) = 0
⇒ 0 = 0, which is true, so R is reflexive.
Symmetric Now, xRy ⇒ (x - y)(3x - 4 y) = 0
and yRx =>(y - x)(3y - 4x) = 0
Clearly, xRy =# yRx
So, R is not symmetric.
Transitive Now, xRy ⇒ (x - y)(3x - 4y) = 0
yRz ⇒ (y - z)(3y - 4z) = 0
and xRz ⇒ (x-z) (3x-4z) = 0
Here, (x - y) (3x - 4y) = 0 and (y - z)(3y - 4z) = 0
⇏ (x-z) (3x-4z) = 0
So, R is not transitive.

\(\text { We have } \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\)
\(\text { Or }\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}\)
\(\text { i.e. }\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\text { Therefore } \quad \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}=1\)
\(\text { or } \quad 6 x^{2}+5 x-1=0 \text { i.e., }(6 x-1)(x+1)=0\)
\(\text { which gives } \ x=\frac{1}{6} \text { or } x=-1 .\)
Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative\(x=\frac{1}{6}\) is the only solution of the given equation

Getting Sin^-1 \(\left( \frac { 8 }{ 17 } \right) =tan^{ -1 }\frac { 8 }{ 15 } \)
and \(sin^{ -1 }\left( \frac { 3 }{ 5 } \right) =tan^{ -1 }\frac { 3 }{ 4 } \)
LHS = \(tan^{ -1 }\frac { 8 }{ 15 } +tan^{ -1 }\frac { 3 }{ 4 } \)
\(=tan^{ -1 }\left( \frac { \frac { 8 }{ 15 } +\frac { 3 }{ 4 } }{ 1-\frac { 8 }{ 15 } \times \frac { 3 }{ 4 } } \right) \)
\(=tan^{ -1 }\left( \frac { 77 }{ 36 } \right) \)
\(=cos^{ -1 }\left( \frac { 36 }{ 85 } \right) \)
= RHS

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

Cosine functions respected to any interval \(\left[-\pi, 0]\right.\)

By the properties, gof is ene-one \(\Rightarrow\) is one-one.

A function f : A ➝ B is said to be onto, if for every b ∈ B, there exists an element a in A such that f(a) = b

As sin^-1 x = \(\frac { \pi }{ 2 } \), sin^-1 y = \(\frac { \pi }{ 2 } \), sin^-1z = \(\frac { \pi }{ 2 } \)
⇒ x = 1, y = 1, z = 1
∴ x + y² + z³ = 1 + 1 + 1 = 3

y = f(x)
⇒ 5x + 4
⇒ \(\frac{y-4}{5}\)
∴ f^-1(y) = \(\frac{y-4}{5}\)
or f^-1(x) = \(\frac{x-4}{5}\)

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c) 

(c) Given, f(x)= y = cos^-1 x
We know that the domain of f(x) is equivalent to the range of value of x for which f(x) exist.
Let f(x) = \(\theta\)
\(\Rightarrow\) cos^-3 x = \(\theta\) \(\Rightarrow\) x = cos \(\theta\)
Since, the range of cos \(\theta\) is [-1, 1]
\(\therefore\) The range of x is [-1, 1].
Hence, the domain of f(x) is [-1, 1]
Thus, Assertion (A) is correct.
The principal value branch of cos^-1 x is [0, \(\pi\)].
\(\therefore\) Reason is wrong.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.