LHS = \(\frac { sin(-\theta )tan({ 90 }^{ o }-\theta )sec\left( { 180 }^{ o }-\theta \right) }{ sin(180+\theta )cot(360-\theta )cosec({ 90 }^{ o }-\theta ) } =1\)
= \(\cfrac { \left( -sin\theta \right) cot\theta \left( -sec\theta \right) }{ \left( -sin\theta \right) \left( -cot\theta \right) sec\theta } =1\)

\(\sin \frac{A}{8} \sin \frac{3 A}{8}=\frac{1}{2}\left[2 \sin \frac{A}{8} \sin \frac{3 A}{8}\right]\)
\(=\frac{1}{2}\left[\cos \left(\frac{A}{8}-\frac{3 A}{8}\right)-\cos \left(\frac{A}{8}+\frac{3 A}{8}\right)\right]\)
\(=\frac{1}{2}\left[\cos \frac{A}{4}-\cos \frac{A}{2}\right]\)

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)
\(=\frac{\sin (A-B)}{\cos (A-B)} \)
\(=\frac{\frac{-16}{\frac{35}{63}}}{\frac{-63}{63}}=\frac{16}{63} \)

 cot 75°
tan 75° = tan (45° + 30°)
\(=\frac { 1+\frac { 1 }{ \sqrt { 3 } } }{ 1-1\left( \frac { 1 }{ \sqrt { 3 } } \right) } =\frac { \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } }{ \sqrt { 3 } -\frac { 1 }{ \sqrt { 3 } } } =\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \left[ \tan { { 45 }^{ o } } =1,\ \tan { 30^{ o }=\frac { 1 }{ \sqrt { 3 } } } \right] \)
\(\cot { { 75 }^{ o } } =\frac { 1 }{ \tan { { 75 }^{ o } } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \)

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

\({ 60 }^{ o }=60\times \frac { \pi }{ 180 } = \frac { \pi }{ 3 } \)

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

cos²45^o - sin²45^o = cos 2(45^o)
= cos 90^o = 0

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

\(\sin \left(-420^{\circ}\right) =-\sin 420^{\circ}=-\sin (360+60) =-\sin 60^{\circ}=\frac{-\sqrt{3}}{2} \)

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

\(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right) +\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1 \)
\(\operatorname{LHS}= \sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right) +\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right) \)
\(= \sec \left(270^{\circ}-\theta\right) \sec \left(450^{\circ}-\theta\right) +\tan \left(450^{\circ}+\theta\right)\left(-\tan \left(450^{\circ}-\theta\right)\right) \)
\(=-\operatorname{cosec} \theta \sec \left(360^{\circ}+90^{\circ}-\theta\right) -\tan \left(360^{\circ}+90^{\circ}+\theta\right) \tan \left(360^{\circ}+90^{\circ}-\theta\right) \)
\(=-\operatorname{cosec} \theta \sec \left(90^{\circ}-\theta\right) -\tan \left(90^{\circ}+\theta\right) \tan \left(90^{\circ}-\theta\right) \)
\(=-\operatorname{cosec} \theta \operatorname{cosec} \theta+\cot \theta \cot \theta \)
\(=-\left(\operatorname{cosec}{ }^2 \theta-\cot { }^2 \theta\right)=-1 = \text { RHS } \)
Hence proved.

In order to prove that \(\alpha+2 \beta=\pi / 4,\) it is sufficient to prove that
\(\tan (\alpha+2 \beta)=\tan \pi / 4=1\)
\(\sin \beta=\frac{1}{\sqrt{10}}, \ 0<\beta<\pi / 2\)
\(\cos \beta =\sqrt{1-\sin ^2 \beta} \)
\(=\sqrt{1-\frac{1}{10}}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}}\)
\(\tan \beta=\frac{\sin \beta}{\cos \beta}=\frac{1 / \sqrt{10}}{3 / \sqrt{10}}=1 / 3 \)
\(\tan 2 \beta=\frac{2 \tan \beta}{1-\tan ^2 \beta}=\frac{2 / 3}{1-1 / 9} \)
\(=\frac{2 / 3}{8 / 9}=\frac{3}{4} \)
\(\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}\)
\(=\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}=\frac{4+21}{28-3}=1\)
\(\therefore \alpha+2 \beta=\pi / 4\)