tan \(\theta\) = 3,
tan 3\(\theta\) = \(\frac { 3\tan { \theta } -\tan ^{ 3 }{ \theta } }{ 1-3\tan ^{ 2 }{ \theta } } \)
\(=\frac { 9-27 }{ 1-27 } =\frac { -18 }{ -26 } =\frac { 9 }{ 13 } \)

(i) sin (A + 60°) + sin (A - 60°) = sin A.
LHS = sin (A + 60°) + sin (A - 60°)
\(= \sin A \cos 60^{\circ}+\cos A \sin 60^{\circ} +\sin A \cos 60^{\circ}-\cos A \sin 60^{\circ}\)
= 2 sin A cos 60° = 2 sin A \(\left( \frac { 1 }{ 2 } \right) \) = sin A = RHS  
Hence proved

Since A and B are acute angles, A and B lies in the I quadrant
\(\cos A=\frac{13}{14} \Rightarrow \sin A=\sqrt{1-\cos ^2 A}\)
\(=\sqrt{1-\frac{169}{196}}=\frac{\sqrt{27}}{14}\)
\(=\frac{3 \sqrt{3}}{13}\)
\(\cos B =\frac{1}{7} \Rightarrow \sin B=\sqrt{1-\cos ^2 B} \)
\(=\sqrt{1-\frac{1}{49}} \)
\(=\sqrt{\frac{48}{49}}=\frac{4 \sqrt{3}}{7} \)
cos( A - B) = cos A cos B + sin A sin B
= \(\left( \frac { 13 }{ 14 } \right) \frac { 1 }{ 7 } +\frac { 4\sqrt { 3 } }{ 7 } \left( \frac { \sqrt { 27 } }{ 14 } \right) =\frac { 13 }{ 98 } +\frac { 49 }{ 98 } =\frac { 1 }{ 2 } \)
\(\therefore \cos { \left( A-B \right) } =\frac { 1 }{ 2 } \)
\(\Rightarrow A-B=\frac { \pi }{ 3 } \ \left[ \because \cos { { 60 }^{ o }=\frac { 1 }{ 2 } } \right] \)
Hence Proved.

\(\tan ^{-1} 2 x+\tan ^{-1} 3 x=\pi / 4\)
\(\tan^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}\right)=\frac{\pi}{4}\)
\(\frac{5 x}{1-6 x^2}=\tan \pi / 4=1 \text { if } 6 x^2<1 \)
\(5 x=1-6 x^2 \quad x^2<1 / 6 \)
\(6 x^2+5 x-1=0 \quad \frac{-1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}} \)
\(6 x^2+6 x-x-1=0 \)
\(x=-1, \frac{1}{6} \text { and } \frac{-1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}} \)
\(x=1 / 6\)

Let B = \(\begin{vmatrix} 1 &-3 &2 \\4 &-1&2\\3&5&2 \end{vmatrix}\)
Minor of 1 = M₁₁ = \(\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}=-2-10=-12\)
Minor of -3 = M₁₂ = \(\begin{vmatrix}4 &2 \\ 3 & 2 \end{vmatrix}=8-6=2\)
Minor of 2 = M₁₃ = \(\begin{vmatrix} 4 & -1 \\ 3 & 5\end{vmatrix}=20+3=23\)
Minor of 4 = M₂₁ = \(\begin{vmatrix} -3 & 2 \\5 & 2 \end{vmatrix}=-6+10=-16\)
Minor of -1 = M₂₂ = \(\begin{vmatrix}1 & 2 \\ 3 & 2 \end{vmatrix}=2-6=-4\)
Minor of 2 = M₂₃ = \(\begin{vmatrix} 1& -3 \\3 &5 \end{vmatrix}=5+9=14\)
Minor of 3 = M₃₁ = \(\begin{vmatrix} -3 &2 \\ -1 & 2 \end{vmatrix}=-6+2=-4\)
Minor of 3 = M₃₂ = \(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=2-8=-6\)
Minor of 2 = M₃₃ = \(\begin{vmatrix} 1 & -3 \\4 & -1 \end{vmatrix}=-1+12=11\)
Co-factor of 1 = A₁₁ = (-1)¹⁺¹ M₁₁ = -12
Co-factor of -3 = A₁₂= (-1)¹⁺² M₁₂ = -2
Co-factor of 2 = A₁₃= (-1)¹⁺³ M₁₃ = 23
Co-factor of 4 = A₂₁ = (-1)²⁺¹ M₂₁ = 16
Co-factor of -1 = A₂₂ = (-1)²⁺² M₂₂ = -4
Co-factor of 2 = A₂₃ = (-1)2+3 M₂₃ = -14
Co-factor of 3 = A₃₁ = (-1)³⁺¹ M₃₁ = -4
Co- factor of 5 = A₃₂ = (-1)³⁺² M₃₂ = 6
Co-factor of 2 = A₃₃ = (-1)³⁺³ M₃₃ = 11

\(\cos 2 A + \cos 4 A= 2 \cos \frac{2 A + 4 A}{2} \cos \frac{2 A-4 A}{2}\)
= 2cos(3A) cos(A)

sin (-105)° = - sin (105°) 
= - sin (60° + 45°)
= - [sin 60° cos 45° + cos 60° sin 45°]
= \(-\left[ \frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } \right] \)
= \(-\left[ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right] \)

sin 75^o =sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
[\(\because\)sin (A+B) = sin A cos B + cos A sinB]
\(=\frac{1}{\sqrt{2}}\times\frac{\sqrt3}{2}+\frac{1}{\sqrt{2}}\times\frac12=\frac{\sqrt3+1}{2\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt6+\sqrt2}{4}\)

Let tan^-1(-1) = y where \(\frac{-\pi}{2}\le y\le \frac{\pi}{2}\)
\(\Rightarrow\) tan y = -1
=\(-\tan\left(\frac{\pi}{4}\right)\)
\(\Rightarrow y=\frac{-\pi}{4}\)

Consider \(\frac{\sin A+\sin B}{\cos A+\cos B}=\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}\)
\(\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{\sin \frac{A+B}{2}}{\cos \frac{A+B}{2}}\)
\(\tan \left(\frac{A+B}{2}\right)=1 / 2\)

\(\text {LHS }=(\cos \alpha-\cos \beta)^2+(\sin \alpha-\sin \beta)^2\)
\(=\left(2 \sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta}{2}\right)^2+\left(2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}\right)^2\)
\(=4 \sin ^2 \frac{\alpha-\beta}{2}\left[\sin ^2 \frac{\alpha+\beta}{2}+\cos ^2 \frac{\alpha+\beta}{2}\right]\)
\(=4 \sin ^2 \frac{\alpha-\beta}{2}=\text { RHS }\)
Hence proved.

\(\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1}\left(\frac{4}{7}\right)\)
\(\tan ^{-1}\left(\frac{x+1+x-1}{1-\left(x^2-1\right)}\right)=\tan ^{-1}\left(\frac{4}{7}\right)\)
\(\frac{2 x}{2-x^2}=\frac{4}{7} \ x^2-1<1\)
\(14 x=8-4 x^2 \ x^2<2\)
\((4 x^2+14 x-8=0\ -\sqrt{2})\)
\(\div 2 \ \ 2 x^2+7 x-4=0\)
\((2 x-1)(x+4)=0\)
\((x=\frac{1}{2} \ \ [\because-\sqrt{2}]\)

Given A = \(\begin{bmatrix}1 &3&3 \\1 &4&3\\1&3&4 \end{bmatrix} \)
\(A_{11}=\text {Cofactor of } 1=16-9=7\)
\(A_{12}=\text {Cofactor of } 3=-(4-3)=-1\)
\(A_{13}=\text {Cofactor of } 3=3-4=-1\)
\(A_{21}=\text {Cofactor of } 1=-(12-9)=-3 \)
\(A_{22}=\text {Cofactor of } 4=4-3=1 \)
\(A_{23}=\text {Cofactor of } 3=-(3-3)=0 \)
\( A_{31}=\text {Cofactor of } 1=9-12=-3 \)
\(A_{32}=\text {Cofactor of } 3=3-3=0 \)
\(A_{33}=\text {Cofactor of } 4=4-3=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right)\)
A^-1 = \(\begin{bmatrix} 7&-3&-3\\-1&1&0\\-1&0&1 \end{bmatrix}\)
\(|A| =1(16-9)-3(4-3)+3(3-4) \)
\(=7-3-3=1 \neq 0\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left(\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right)\left(\begin{array}{ccc} 7 & -3 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 7-3-3 & -3+3+0 & -3+0+3 \\ 7-4-3 & -3+4+0 & -3+0+3 \\ 7-3-4 & -3+3+0 & -3+0+4 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I\)
\(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|A| I\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)