We know that \(\cos^2x+\sin^2x=1\)
\(\Rightarrow\cos x=\pm\sqrt{1-\sin^2x}\)
In the III quadrant, cos x is negative
\(\therefore\cos x=-\sqrt{1-\sin^2x}=-\sqrt{1-\frac{24}{25}}=-\frac15\left[\because\sin^2x=\frac{4(6)}{25}=\frac{24}{25}\right]\)
In the III quadrant, tan x is positive
\(\therefore\tan x=\frac{\sin x}{\cos x}=\frac{-2\sqrt6}{5}\times\frac{-5}{1}=2\sqrt{6}\)
\(cosec x=\frac{1}{\sin x}=\frac{-5}{2\sqrt6}\)
\(\sec x=\frac{1}{\cos x}=-5\) and
\(\cot x=\frac{1}{\tan x}=\frac{1}{2\sqrt6}\)

A + B = 45° \(\Rightarrow\) tan (A + B) = tan 45^°
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=1\)
\(\Rightarrow\) tan A + tan B = 1 - tanA tanB
Adding 1 both sides
\(\Rightarrow\) (1 + tan A) + tan B (1 + t an A) = 2
\(\Rightarrow\) (1 + tanA)(1 + tan B) = 2
take A = B then 2A =  45^°
A = B = \(22{ \frac { 1 }{ 2 } }^{ ° }\) 
\(\left( 1+\tan { 22{ \frac { 1 }{ 2 } }^{ ° } } \right) \left( 1+\tan { 22{ \frac { 1 }{ 2 } }^{ ° } } \right) =2\)
\(\left( 1+\tan { 22{ \frac { 1 }{ 2 } }^{ ° } } \right) =2\)
\(\Rightarrow\)  \(1+\tan { 22{ \frac { 1 }{ 2 } }^{ ° } } =\pm \sqrt { 2 } \)
\(\Rightarrow\)\(\ \tan { 22{ \frac { 1 }{ 2 } }^{ ° } } =\pm \sqrt { 2 } -1 \)
Hence Proved.

\(\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\)
\(\mathrm{A}=85^{\circ}, \mathrm{B}=5^{\circ}\)
\(\tan \left(85^{\circ}-5^{\circ}\right)=\frac{\tan 85^{\circ}-\tan 5^{\circ}}{1+\tan 85^{\circ} \tan 5^{\circ}}\)
\(\tan 80^{\circ}=\frac{\tan 85^{\circ}-\tan 5^{\circ}}{1+\tan 85^{\circ} \tan \left(90^{\circ}-5\right)}\)
\(=\frac{\tan 85^{\circ}-\tan 5^{\circ}}{1+\tan 85^{\circ} \cot 85^{\circ}}\)
\(=\frac{\tan 85^{\circ}-\tan 5^{\circ}}{1+\tan 85^{\circ} \frac{1}{\tan 85^{\circ}}}\)
\(2 \tan 80^{\circ}=\tan 85^{\circ}-\tan 5^{\circ}\)
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.

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

A lies in I quadrant and B lies in the IlI quadrant

\(\cos A=\sqrt{1-\sin ^2 A}\)
\(=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\cos B=-\frac{12}{13}\)
\(\sin B=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13}\)
sin (A + B)
 = sin A cos B - cos A sin B = \(\left( \frac { 3 }{ 5 } \right) \left( -\frac { 12 }{ 13 } \right) -\left( \frac { 4 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) \)
= \(\frac { -36 }{ 65 } +\frac { 20 }{ 65 } =\frac { -16 }{ 65 } \)

A lies in I quadrant and B lies in the IlI quadrant

\(\cos A=\sqrt{1-\sin ^2 A}\)
\(=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\cos B=-\frac{12}{13}\)
\(\sin B=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13}\)
tan (A - B)
\(=\frac{\sin (A-B)}{\cos (A-B)} \)
\(=\frac{\frac{-16}{\frac{35}{63}}}{\frac{-63}{63}}=\frac{16}{63} \)

\(\ cosec { \left( { 1125 }^{ o } \right) } =\ cosec { \left( 3\times { 360 }^{ o }+{ 45 }^{ o } \right) } \)
\(=\ cosec { { 45 }^{ o } } =\sqrt { 2 } \) (I quad)

\(\tan \left(-855^{\circ}\right)=-\tan 855^{\circ}\)
\(=-\tan \left(2 \times 360^{\circ}+135^{\circ}\right)\)
\(=-\tan 135^{\circ}\)
\(=-\tan \left(180^{\circ}-45^{\circ}\right)(\text { II quad })\)
\(=\tan 45^{\circ}=1\)

\(\sec { \left( { 390 }^{ o } \right) } \) (I quad)
\(\sec { \left( { 390 }^{ o } \right) } =\sec { \left( { 360 }^{ o }+{ 30 }^{ o } \right) } =\sec { { 30 }^{ o } } =\frac { 2 }{ \sqrt { 3 } } \)

\(\sin { { 300 }^{ o } } \)
\(=\sin { 300 }^{ o } =\sin { \left( { 360 }^{ o }-{ 60 }^{ o } \right) } \) (IV quad)
\(= \sin 60^o=-\frac { \sqrt { 3 } }{ 2 } \)

1195° = 3 x 360° + 115°
II quadrant.

cos x sin x - cos x sin x = 0

\(p =\frac{\tan 50^{\circ}}{\sec 50^{\circ}}=\frac{\sin 50^{\circ} / \cos 50^{\circ}}{1 / \cos 50^{\circ}} =\sin 50^{\circ}\)

\(\frac{-1}{\operatorname{cosec} 45^{\circ}}=\frac{-1}{\sqrt{2}}\)

\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=4 / 5\)

\(\frac{3 \tan 10^{\circ}-\tan ^3 10^{\circ}}{1-3 \tan ^2 10^{\circ}} =\tan 3(10) =\tan 30^{\circ}=\frac{1}{\sqrt{3}} \)

\(\sin A=\frac{1}{2} \Rightarrow A=30^{\circ} \)
\(4 \cos ^3 A-3 \cos \mathrm{A} =\cos 3 \mathrm{~A} \)
\(=\cos 3(30) \)
\(=\cos 90^{\circ} =0\)

\(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}} =\sin 2\left(30^{\circ}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \)

4cos³40^o - 3cos40^o = cos 3(40^o)
= cos 120^o
=  cos (180^o - 60^o)
= -cos 60^o
\(= \frac{-1}{2}\)

1 - 2sin²45^o = cos 2(45^o)
= cos 90^o = 0

\(\sec \theta =\sqrt{1+\tan ^2 \theta}=\sqrt{1+\frac{1}{5}}=\sqrt{\frac{6}{5}} \)

LHS = cos²2x - cos²6x
= (cos 2x + cos 6x)(cos 2x - cos 6x) [∴ cos²A - cos²B = (cos A + cos B)(cos A - cos B)
\(=\left[ 2cos\left( \frac { 2x+6x }{ 2 } \right) .cos\left( \frac { 2x-6x }{ 2 } \right) \right] \left[ -2sin\left( \frac { 2x+6x }{ 2 } \right) .sin\left( \frac { 2x-6x }{ 2 } \right) \right] \)
\(=\left[ \because cosC+cosD=2\quad cos\left( \frac { C+D }{ 2 } \right) cos\left( \frac { C-D }{ 2 } \right) and\quad cosC-cosD=-2sin\left( \frac { C+D }{ 2 } \right) .sin\left( \frac { C-D }{ 2 } \right) \right] \)
= [2 cos 4x . cos(-2x)][-2 sin 4x . sin(-2x)]
= (2 cos 4x . cos 2.x) (2 sin4x . sin 2.x) [:. cos (-\(\theta\)) = cos \(\theta\) and sin (-\(\theta\)) = -sin\(\theta\)]
= (2 sin 2.x cos 2x) (2 sin 4x cos 4x)
= sin 4x sin 8x [sin 2A = 2 sin A cos A] = RHS.
Hence proved

LHS = sin(n + 1)x sin(n + 2) x + cos(n + 1)xcos(n + 2)x
Let A = (n + 1)x and B = (n + 2)x
= sinAsinB + cosAcosB = cos(A - B)
= cos[(n + 1)x - (n + 2)x] = cos[nx + x - nx - 2x]
= cos[x - 2x] = cos(-x) = cosx[since cos x is an even function]
= RHS Hence proved.