\(\mathrm{LHS}=\frac{\cos 7 A+\cos 5 A}{\sin 7 A-\sin 5 A}\)
\(=\frac{2 \cos \frac{7 A+5 A}{2} \cos \frac{7 A-5 A}{2}}{2 \sin \frac{7 A-5 A}{2} \cos \frac{7 A+5 A}{2}}\)
\(=\frac{\cos A}{\sin A}=\cot A=\mathrm{RHS}\)
Hence proved.

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 = \(\left( \frac { 3 }{ 5 } \right) \left( -\frac { 12 }{ 13 } \right) -\left( \frac { 4 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) \)
= \(\frac { -36 }{ 65 } +\frac { 20 }{ 65 } =\frac { -16 }{ 65 } \)

Let \(\tan^{-1}\left(\frac34\right)=\theta\Rightarrow\tan\theta=\frac34\)
\(\sec ^2 \theta=1+\tan ^2 \theta=1+\frac{9}{16}=\frac{25}{16}\)
\(\sec \theta=\frac{5}{4} \Rightarrow \cos \theta=\frac{4}{5}\)
\(\therefore\cos\left[\tan^{-1}\left(\frac34\right)\right]=\cos \theta=\frac45=RHS\)

380° : I quadrant (360° + 20°)

Let \(y=sec^{-1}(-\sqrt2)\) where \(0 \leq y \leq \pi\)
\(y=\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)
\(\cos y=\frac{-1}{\sqrt{2}}=\cos (\pi-\pi / 4)\)
\(\cos y=\cos \frac{3 \pi}{4}\)
\(y=3 \frac{\pi}{4}\)

cos x sin x - cos x sin x = 0

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

\(\tan 2 A= \frac{2 \tan A}{1-\tan ^2 A}=\frac{2(1 / 2)}{1-1 / 4}=\frac{1}{3 / 4}=4 / 3 \)
\(\tan (2 A+B)= \frac{\tan 2 A+\tan B}{1-\tan 2 A \cdot \tan B}=\frac{\frac{4}{3}+\frac{1}{3}}{1-4 / 9} \)
\(=\frac{5 / 3}{5 / 9}=3 \)

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

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

\(\sec A(-\cos A)=\frac{1}{\cos A}(-\cos A)=-1\)

\(\sin (28+17)=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\)

\(\sin15^o = \sin(45^o - 30^o) = \sin45^o \cos 30^o - \cos 45^o \sin 30^o\)
\(= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}\)

\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=22 \frac{1}{2}^{\circ}=22^{\circ} 30^{\prime}\)

Since \(\cfrac { 3\pi }{ 2 } <\left( A,B \right) <2\pi \), both A and B lie in the fourth quadrant,
 \(\therefore \) sinA and sinB are negative
Given \(cosA=\cfrac { 4 }{ 5 } \ cosB=\cfrac { 12 }{ 13 } \)
Therefore  \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} B =-\sqrt{1-\cos ^2 B} \)
\(=-\sqrt{1-\frac{144}{169}} \)
\(=-\sqrt{\frac{169-144}{169}} \)
\(=-\frac{5}{13}\)
\(sin(A-B)=sinAcosB-cosAsinB\\ =\left( \cfrac { -3 }{ 5 } \right) \left( \cfrac { 12 }{ 13 } \right) -\left( \cfrac { 4 }{ 5 } \right) \left( \cfrac { -5 }{ 13 } \right) \\ =\cfrac { -36 }{ 65 } +\cfrac { 20 }{ 65 } =\cfrac { -6 }{ 65 } \)

\(\mathrm{LHS} =\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} \)
\(=\cos 20^{\circ} \cos 40^{\circ}\left(\frac{1}{2}\right) \cos 80^{\circ}\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos 40^{\circ} \cos 80^{\circ}\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos \left(60^{\circ}-20^{\circ}\right) \cos \left(60^{\circ}+20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos ^2 60^{\circ}-\sin ^2 20\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1}{4}-\left(1-\cos ^2 20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1-4+4 \cos ^2 20^{\circ}}{4}\right)\)
\(=\frac{1}{8}\left(4 \cos ^3 20^{\circ}-3 \cos 20^{\circ}\right)\)
\(=\frac{1}{8} \cos 3\left(20^{\circ}\right)=\frac{1}{8} \cos 60^{\circ}\)
\(=\frac{1}{8}\left(\frac{1}{2}\right)=\frac{1}{16}\)
Hence proved.

LHS \(=\cos^{-1}\left(\frac{12}{13}\right)+\sin^{-1}\left(\frac35\right)\)
\(=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{3}{5}\)
\(=\sin^{-1}\left[\frac{5}{13}\sqrt{1-\left(\frac35\right)^2}+\frac35\sqrt{1-\left(\frac{5}{13}\right)^2}\right]\)
\(=\sin^{-1}\left[\frac{5}{13}\times\frac{4}{5}+\frac35\times\frac{1}{13}\right]\)
\(=\sin^{-1}\left(\frac{56}{65}\right)\)
= RHS. hence proved.

Given \(\sin { \theta = \frac { 3 }{ 5 } } \) and \(\frac { \pi }{ 2 } <\theta <\pi \)
\(\therefore\) \(\theta\) is in II quadrant, only sin and its reciprocal is positive
\(\cos { \theta } =\frac { adj }{ hyp } =-\frac { 4 }{ 5 } ,\tan { \theta } =\frac { -3 }{ 4 } \quad .....(1)\)

Also \(\tan { \varphi } =\frac { 1 }{ 2 } \) and \(\pi <\varphi <\frac { 3\pi }{ 2 } \)
\(\therefore\) \(\phi \) is in III quadrants, tan \(\varphi \) and its reciprocal alone are positive.
\(\therefore \quad \sec { \phi } =\frac { hyp }{ adj } =-\frac { \sqrt { 5 } }{ 2 } \)...(2)

\(8\tan { \theta } -\sqrt { 5 } \sec { \phi } \)
\(=8\left( \frac { -3 }{ 4 } \right) -\sqrt { 5 } \left( -\frac { \sqrt { 5 } }{ 2 } \right) \)   [using (1) and (2)]
\(=+2(-3)+\frac { 5 }{ 2 } =-6+\frac { 5 }{ 2 } \)
\(=\frac { -12+5 }{ 2 } =\frac { -7 }{ 2 } \) 
Hence Proved.

\({ \tan }^{ -1 }\left[ \cfrac { \left( x+2 \right) +\left( 2-x \right) }{ 1-\left( x+2 \right) \left( 2-x \right) } \right] ={ \tan }^{ -1 }\left( \cfrac { 2 }{ 3 } \right)\)
\( \Rightarrow { \tan }^{ -1 }\left( \cfrac { 4 }{ 1-\left( 4-{ x }^{ 2 } \right) } \right) ={ \tan }^{ -1 }\left( \cfrac { 2 }{ 3 } \right) \)
\( \Rightarrow { 2x }^{ 2 }-6=12\)
\(\Rightarrow { x }^{ 2 }=9\)
\( \therefore x=\pm 3\)

\(\cos 75^{\circ}+\cos 45^{\circ}=2 \cos \left(\frac{75^{\circ}+45^{\circ}}{2}\right) \cos \left(\frac{75^{\circ}-45^{\circ}}{2}\right)\)
\(=2 \cos \left(\frac{120^{\circ}}{2}\right) \cos \left(\frac{30^{\circ}}{2}\right)\)
\(=2 \cos 60^{\circ} \cos 15^{\circ}\)
\(=2 \times \frac{1}{2} \cos 15^{\circ}=\cos 15^{\circ}\)

\(\mathrm{LHS}=\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}\)
\(=\sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)\)
\(=\sin 20^{\circ}\left[\sin ^2 60^{\circ}-\sin ^2 20^{\circ}\right]\)
\(=\sin 20^{\circ}\left[\frac{3}{4}-\sin ^2 20^{\circ}\right]\)
\(=\sin 20^{\circ}\left[\frac{3-4 \sin ^2 20^{\circ}}{4}\right]\)
\(=\frac{3 \sin 20^{\circ}-4 \sin ^3 20^{\circ}}{4}=\frac{\sin 3\left(20^{\circ}\right)}{4}\)
\(=\frac{\sin 60^{\circ}}{4}=\frac{\sqrt{3} / 2}{4}=\frac{\sqrt{3}}{8} .\)

  \(cos2A=\cfrac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A } =\cfrac { 1-\frac { 1 }{ 49 } }{ 1+\frac { 1 }{ 49 } } =\cfrac { 48 }{ 49 } \times \cfrac { 49 }{ 50 } \) = \(\cfrac { 24 }{ 25 } \)   ..(1)
Now, sin4B = 2sin2B cos2B
\(\\ \\ \\ =2\cfrac { 2tanB }{ 1+{ tan }^{ 2 }B } \times \cfrac { 1-{ tan }^{ 2 }B }{ 1+{ tan }^{ 2 }B } \)   
\( =\cfrac { 4\times \frac { 1 }{ 3 } }{ 1+\frac { 1 }{ 9 } } \times \cfrac { 1-\frac { 1 }{ 9 } }{ 1+\frac { 1 }{ 9 } } =\cfrac { 24 }{ 25 } \) ..(2)
From(1) and (2) we get, cos2A = sin4B.

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