(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

 \(\left| \frac { i\left( 2+i \right) ^{ 2 } }{ \left( 1+i \right) ^{ 2 } } \right| =\frac { \left| i \right| \left| \left( 2+i \right) ^{ 3 } \right| }{ \left| \left( 1+i \right) ^{ 2 } \right| } =\frac { \left( \sqrt { 4+1 } \right) ^{ 3 } }{ \left( \sqrt { 2 } \right) ^{ 2 } } \) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)
= \(\frac { \left( \sqrt { 5 } \right) ^{ 3 } }{ 2 } =\frac { 5\sqrt { 5 } }{ 2 } \).

\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi +\frac { \pi }{ 4 } \right) \right) \) \(\because \frac { 5\pi }{ 4 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
= \({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 4 } \right) \right) \)
= \( \frac {- \pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) 

Comparing the given equation of the plane with \(\vec { r } .\vec { n } \) = p, we have \(\vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \).
We know that the perpendicular distance from the given point with position vector u to the plane \(\vec { r } .\vec { n } \)= p is given by \(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } \). Therefore, substi \(\vec { u } \)= (2, 5, -3) = \(2\hat { i } +5\hat { j } -3\hat { k } \) and \(\ \vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \) in the formula, we get
\(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } =\frac { |(2\hat { i } +5\hat { j } -3\hat { k } ).(6\hat { i } -3\hat { j } +2\hat { k } )-5| }{ |6\hat { i } -3\hat { j } +2\hat { k } | } \) = 2 unit.

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24

Here A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0, A = 0 either A or C is 0.
Hence, the given equation represents a parabola.

Let sin^-1 \(\left( -\frac { 1 }{ 2 } \right) \) = y. Then sin y = -\(\frac{1}{2}\)
The range of the principal value of sin^-1x is \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and hence, Let us find y \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) Such that sin y = -\(\frac{1}{2}\). Clearly, y = -\(\frac{\pi}{6}\)
Thus, the principal value of sin^-1\(\left( -\frac { 1 }{ 2 } \right) \) is -\(\frac{\pi}{6}\). This corresponds to -30^o.

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 3
∴ \(\rho \)(A) ≤ min(3, 3) = 2
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right| =1\left| \begin{matrix} 4 & -6 \\ 1 & -1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} \right| \)
[Expanded along R₁]
= 1(-4+6)+2(-2+30)+3(2-20)
= 1(2)+2(28)+3(-18)
= 2+56-54 = 58-54 = 4 ≠ 0
∴ \(\rho \)(A) = 3

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

We consider \(\frac { 1+i }{ 1-i } =\frac { \left( 1+i \right) \left( 1+i \right) }{ \left( 1-i \right) \left( 1+i \right) } =\frac { 1+2i }{ 1+1 } =\frac { 2i }{ 2 } =i\)
and \(\frac { 1-i }{ 1+i } =\left( \frac { 1+{ i } }{ 1-i } \right) ^{ -1 }=\frac { 1 }{ i } =-i\)
Therefore,\(\left( \frac { 1+i }{ 1-i } \right) ^{ 3 }-\left( \frac { 1-i }{ 1-i } \right) ^{ 2 }\)= i³-(-i)³ = - i - i = -2i