On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
\(\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

\({ sin }^{ -1 }\frac { 2a }{ 1+{ a }^{ 2 } } =2{ tan }^{ -1 }a\)
\(and\quad { sin }^{ -1 }\frac { 2b }{ 1+{ b }^{ 2 } } =2{ tan }^{ -1 }b\)
\(as\left[ 2{ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \right] \)
\(2{ tan }^{ -1 }a+2{ tan }^{ -1 }b=2ta{ n }^{ -1 }x\)
\({ tan }^{ -1 }a+{ tan }^{ -1 }b={ tan }^{ -1 }x\)
\({ tan }^{ -1 }\left( \frac { a+b }{ 1-ab } \right) =ta{ n }^{ -1 }x\)
\(x=\frac { a+b }{ 1-ab } \)  
Hence Proved.

\(f(x)=x^{ 2 },x\in Ra\)
\(f(1.1)=(1.1)^{ 2 }z\)
\(=1.21\)
\(f(1)=(1)^{ 2 }=1\)
\(\frac { f(1.1)-f(1) }{ 1.1-1 } =\frac { 1.21 }{ 1.1-1 } =\frac { 0.21 }{ 0.1 } \)
\(=2.1\)

\({ tan }^{ -1 }\left[ \frac { cos\quad x }{ 1+sin\quad s } \right] \) \(\quad \because \) \(\begin{cases} cos\quad x={ cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } \\ and\quad 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { { cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] Divide\quad by\quad cos\frac { x }{ 2 } ,\quad we\quad get\)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { x }{ 2 } }{ 1+tan\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) \right] =\frac { \pi }{ 4 } -\frac { x }{ 2 } \)

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

a₃₂ = -11
Alternative Method:
Given \(A =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|\)
⇒ a₃₂ \(=\left| \begin{matrix} 5 & 8 \\ 2 & 1 \end{matrix} \right|\)
⇒ a₃₂ = 5-16 = -11

We have, \(\left[ \begin{matrix} 2x+y & x-y \\ x-z & x+y+z \end{matrix} \right] =\left[ \begin{matrix} 10 & -1 \\ 2 & 8 \end{matrix} \right] \)
\(\Rightarrow\) 2x + y = 10, x - y = - 1
x - z = 2 and x + y + z = 8
\(\therefore\) 2(y - 1) + y = 10  \(\Rightarrow\)  2y + y + 2 = 10
\(\Rightarrow\)3y = 12  \(\Rightarrow\)  y = 4
\(\therefore\)  x = 3
3 - z,  \(\Rightarrow\)  z = 1
\(\therefore\)  x = 3, y = 4, z = 1

\(\frac{d y}{d x}=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}\)

Given a * b = 1cm
(a, b) (2*3)*6 = {1cm(2,3)}*6
= 6*6 = 1cm (6, 6) = 6

For the function \(f\left( x \right) ={ x }^{ 3 }-{ 6x }^{ 2 }+ax+b\), it is given that f(1) = f(3) = 0.
Find the values of a and b and hence verify Rolle's Theorem on [1, 3].

(i) Commutativity.
(a,b) * (c,d) = (a+c, b+d)
= (c+a, d+b)
[∵ Addition is commutative in N]
= (c,d) * (a,b).
Hence ' * ' is commutative
(ii) Associativity.
[(a*b)*(c,d)]*(e,f)] = (a+c, b+d)*(e,f)
= [(a+c)+e,(b+d)+f)
= (a+(c+e),b+(d+f)
[∵ Addition is associative in N]
= (a,b)*(c+e,d+f)
= (a,b)*[(c,d)*(e,f)]
Hence, '*' is associative.
(iii) Let (x, y) be the identity element.
Then (a,b)*(x,y) = (a,b)
\(\Rightarrow \) (a+x, b+y) = (a,b)
\(\Rightarrow \) a+x = a, b+y = b
\(\Rightarrow \) x = 0, y = 0.
But \(0\notin N\).
Hence, identity element does not exist. 

(i) Let \(x_{ 1 },x_{ 2 }y\in R\).
Now  \(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \)  \(3-4_{ x1 }=3-4_{ x2 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\Rightarrow f\) is one-one.
Let \(y\in R\). Let \(y=f(x_{ 0 })\).
Then \(3-4x_{ 0 }=y\Rightarrow x_{ 0 }=\frac { 3-y }{ 4 } \).
Now \(y\in R\Rightarrow \frac { 3-y }{ 4 } \in R\Rightarrow x_{ 0 }\in R\)
\(f(x_{ 0 })=3-4x_{ 0 }=3-4\frac { 3-y }{ 4 } =3-3+y=y\).
\(\because \) For each \(y\in R\) , there exists \(x_{ 0 }\in R\) such that 
\(f(x_{ 0 })=y\)
\(\because \)  \(f\) is onto
Hence, 'f' is ne-one and onto or bijective.
(ii) Here  f(1) = 1 + 1 = 2,
f(-1) = 1 + 1 = 2.
Now \(1\neq -1\) but f(1) = f(-1)
\(\because \)   \(f\) is onto
Also range of \(f\) is \([1,\infty )\neq R\)
\(\because \)  \(f\) is onto.
Hence, 'f' os not bijective

\(Let\ y=log(log\ x)\)
\(\frac { dy }{ dx } =\frac { 1 }{ log\quad x } .\frac { 1 }{ x } =\frac { 1 }{ x\quad log\quad x } \)
\( and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { \frac { d }{ dx } (x\quad log\quad x) }{ { \left( x\quad log\quad x \right) }^{ 2 } } \)
\(=-\frac { x.\frac { 1 }{ x } +log\quad x.1 }{ { \left( x\quad log\quad x \right) }^{ 2 } } \)
\( =-\frac { 1+log\quad x }{ { \left( x\quad log\quad x \right) }^{ 2 } }\)

We have
\(2 A-B=2\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]-\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2 & 4 & 6 \\ 4 & 6 & 2 \end{array}\right]+\left[\begin{array}{ccc} -3 & 1 & -3 \\ 1 & 0 & -2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2-3 & 4+1 & 6-3 \\ 4+1 & 6+0 & 2-2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 5 & 3 \\ 5 & 6 & 0 \end{array}\right]\)

(i) We have : \(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \) .
\(\therefore \ A\prime =\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] =A.\)
Hence, A is a symmetric matrix.
(ii) We have: \(A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \)
\(\therefore \ A\prime =\left[ \begin{matrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{matrix} \right] \)
\(=(-1)\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \)
\(=(-1)A=-A.\)
Hence, A is a skew-symmetric matrix.

Area of \(\Delta ABC={1\over 2}\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|\)
\(={1\over2}\begin{vmatrix} a&a+b&1\\b&c+a&1\\c&a+b&1 \end{vmatrix}\)
\(={1\over2}\begin{vmatrix}a+b+C&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1 \end{vmatrix}\)
\({1\over2}(a+b+c)\begin{vmatrix}1&b+c&1\\1&c+a&1\\1&a+b&1 \end{vmatrix}\)
\(={1\over2}(a+b+c)(0)=0\)

We write
\( \tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)= \tan ^{-1}\left[\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \)
\(= \tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}\right] \)
\(= \tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]=\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right] \)
\( =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2} \)

\(cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta & cos\theta sin\theta \\ -cos\theta sin\theta & { cos }^{ 2 }\theta \end{bmatrix}+\begin{bmatrix} { sin }^{ 2 }\theta & -sin\theta cos\theta \\ sin\theta cos\theta & { sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta & cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta & { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).

0

Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
\(a * b=\left\{\begin{array}{ll} a+b, & \text { if } a+b<6 \\ a+b-6 & \text { if } a+b \geq 6 \end{array}\right.\)
An element e ∈ X is the identity element for the operation *, if a * e = a = e * a ∀ a ∈ X
For a ∈ X we observed that
a * 0 = a + 0 =a   [a ∈ X ⇒ a + 0 < 6]
0 * a = 0 + a = a        [a ∈ X ⇒ 0 + a < 6]
a * 0 = a = 0 * a ∀ a ∈ X
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a.
ie \(\left\{\begin{array}{ll} a+b=0=b+a & \text { if } a+b<6 \\ a+6-6=0=b+a-6 & \text { if } a+b \geq 6 \end{array}\right.\)
i.e.,
a = −b or b = 6 − a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ −b.

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

Since f(x) is continuous at x=0 and x=5,
\(\therefore\) at x = 3, LHL = RHL
\(\Rightarrow \lim _{ { x\rightarrow 3 }^{ - } }{ f\left( x \right) =\lim _{ { x\rightarrow 0 }^{ + } }{ f\left( x \right) } } \)
\(\lim _{ x\rightarrow 3 }{ (1)=\lim _{ x\rightarrow 3 }{ ax+b } } \)
\(1=a\times 3+b\)
\(\Rightarrow\) 3a + b = 1
Similarly, at x = 5,
LHL=RHL
\(\lim _{ { x\rightarrow 5 }^{ - } }{ f\left( x \right) = } \lim _{ { x\rightarrow 5 }^{ + } }{ f\left( x \right) } \)
\(\lim _{ x\rightarrow 5 }{ (ax+b)=\lim _{ x\rightarrow 5 }{ (7) } } \)
\(\Rightarrow a(5)+b=7\)
\(\Rightarrow 5a+b=7\)
Solving equations (i) and (ii), we get
a=3 and b=-8

Given that sin[ cot ^-1(x + 1)] = cos(tan^-1x)....(i)
We know that,
\({ cot }^{ -1 }(A)={ sin }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x + 1
Applying this identity in equation (i), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos({ tan }^{ -1 }x)\) ...(ii)
Also, we know that
\({ tan }^{ -1 }A={ cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x
Applying this identity in equation (ii), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos\left( { cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
\(\Rightarrow \frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } =\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)
\(\left[ \because { sin }^{ -1 }(sin\theta )=\theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
and cos^-1 (cos \(\theta \)) = \(\theta \) \(\forall \theta \) [0, \(\pi \)]]
Squaring and Reciprocating both side, we have
1+(1 + x)² = 1 + x²
\(\Rightarrow\) 1 + 1 + x² + 2x = 1 + x²
\(\Rightarrow\) 1 + 2x = 0
\(\Rightarrow\)  \(x=-\frac { 1 }{ 2 } \quad \)

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

As sec \(\left( { tan }^{ -1 }\frac { y }{ 3 } \right) \) \(=\sqrt { 1+{ tan }^{ 2 }\left( { tan }^{ -1 }\frac { y }{ 3 } \right) } \)
\(=\sqrt { 1+\frac { { y }^{ 2 } }{ 9 } } =\frac { \sqrt { 9+{ y }^{ 2 } } }{ 3 } \)