Let \({ \cos }^{ -1 }\left( \cfrac { 4 }{ 5 } \right) =\theta \) .Then \(cos\theta =\cfrac { 4 }{ 5 } \) 
\(\Rightarrow \sin\theta =\cfrac { 3 }{ 5 } \) 
\(\therefore \tan\theta =\cfrac { 3 }{ 4 } \Rightarrow \theta ={ \tan }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) \)
\(\therefore {\cos }^{ -1 }\left( \cfrac { 4 }{ 5 } \right) ={ \tan }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) \) 
Now, \({ \cos }^{ -1 }\left( \cfrac { 4 }{ 5 } \right) +{\tan }^{ -1 }\left( \cfrac { 3 }{ 5 } \right) \) 
\(={ \tan }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) +{ \tan }^{ -1 }\left( \cfrac { 3 }{ 5 } \right) \)
\(={ \tan }^{ -1 }\left( \cfrac { \frac { 3 }{ 4 } +\frac { 3 }{ 5 } }{ 1-\frac { 3 }{ 4 } \times \frac { 3 }{ 5 } } \right) \)
\(={ \tan }^{ -1 }\left( \cfrac { 27 }{ 11 } \right) \)

2cos13Asin15A = sin(13A + 15A) - sin(13A - 15A)
= sin28A + sin2A

\(\)\(\cos \frac{3 A}{2} \cos \frac{5 A}{2}= \frac{1}{2} \cos \left(\frac{3 A}{2}+\frac{5 A}{2}\right) +\cos \left(\frac{3 A}{2}-\frac{5 A}{2}\right) \)
\(=\frac{1}{2}\left[\cos \frac{8 A}{2}+\cos \left(\frac{-2 A}{2}\right)\right]\)
\(=\frac{1}{2}[\cos 4 A+\cos (-A)]\)
\(=\frac{1}{2}[\cos 4 A+\cos A]\)

\(2 \sin 4 \theta \sin 2 \theta= \cos (4 \theta-2 \theta)- \cos (4 \theta+2 \theta)\)
\(=\cos 2 \theta-\cos 6 \theta\)

\(2 \cos 3 \theta \cos \theta= \cos (3 \theta+\theta)+ \cos (3 \theta-\theta)\)
\(=\cos 4 \theta+\cos 2 \theta\)

The terminal side of -70^o lies in IV quadrant.

\(\cfrac { 1 }{ 4 } radian=\cfrac { 1 }{ 4 } \cfrac { 180 }{ \pi } \)
\( =\cfrac { 1 }{ 4 } \times 180\times \cfrac { 7 }{ 22 } ={ 14 }^{ o }19'5''\)

We have \(\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}\)
\(\therefore LHS=\cos2x+\sin^2\phi=\frac{1-\tan^2x}{1+\tan^2x}+\sin^2\phi\)
\(=\frac{1-\left(2\tan^\phi+1\right)}{1+(2\tan^2\phi)+1}+\sin^2\phi\)      \([\because tan^{ 2 }x=2tan^{ 2 }\phi +1]\)
\(=\frac{-2\tan^2\phi}{2(1+\tan^2\phi)}+\sin^2\phi=\frac{-\tan^2\phi}{\sec^2\phi}+\sin^2\phi\)
\(=\frac{-\sin^2\phi}{\cos^2\phi.\frac{1}{\cos^2\phi}}+\sin^2\phi=-\sin^2\phi+\sin^2\phi=0=RHS\)
Hence Proved.

LHS=cos18^o-sin 18^o
=cos18^o-cos72^o[sin18^o=sin(90-72^o)=cos72^o]
\(=2\sin\left(\frac{18^o+72'}{2}\right).\sin\left(\frac{72^o-18'}{2}\right)\left[\because\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\right]\)
\(=2\sin45^o\sin27^o=2\times\frac{1}{\sqrt2}\sin27^o=\sqrt{2}\sin27^o=RHS\)
Hence proved.

Since A, B, C, D are angles of a quadrilateral, A + B + C + D =\(2\pi\)
A + B + C + D =\(2\pi\)
\(\Rightarrow A+B=2\pi-(C+D)\)
\(\Rightarrow\sin(A+B)=\sin[2\pi-(C+D)]\)
=-sin(C+D)[\(\therefore2\pi-(C+D)\)is in the IV quadrant]
\(\Rightarrow\sin(A+B)+\sin(C+D)=0\)

\(\frac{15\pi}{4}=15\times45^o=675^o\)
\(\cot\left(\frac{-15\pi}{4}\right)=\)cot (-675°)= - cot 675° = - cot (720 -45°) = -cot (2 x 360° - 45°)
= -(-cot 45°)(\(\therefore\) 675° is in the IV quadrant) =-(-1) = 1.

 \(\frac { 11\pi }{ 18 } \) Radians = \(\frac { 11\pi }{ 18 } \times \frac { 180 }{ \pi } \)\(=11\times 10={ 110 }^{ o }\)

-3 Radians = \(-3\times \frac { { 180 }^{ o } }{ \pi } =-3\times \frac { 180 }{ 22 } \times 7\)
\(=-\frac { { 3780 }^{ o } }{ 22 } ={ -171.81 }^{ o }\)

\({ -320 }^{ o }=-320\times \frac { \pi }{ 180 } \)= \(\frac { -16\pi }{ 9 } \)

\({ 240 }^{ o }=240\times \frac { \pi }{ 180 } \)= \(\frac { 4\pi }{ 3 } \)