Question

# The value of $${ \left( \dfrac { \cos { A } +\cos { B } }{ \sin { A } -\sin { B } } \right) }^{ n }+{ \left( \dfrac { \sin { A } +\sin { B } }{ \cos { A } -\cos { B } } \right) }^{ n }$$ is

A
2tannAB2
B
2cotnAB2:n is even
C
0:n is odd
D
none

Solution

## The correct option is C $$0:n\ is\ odd$$$$\begin{array}{l} { \left( { \frac { { \cos A+\cos B } }{ { \sin A-\sin B } } } \right) ^{ n } }+{ \left( { \frac { { \sin A+\sin B } }{ { \cos A-\cos B } } } \right) ^{ n } } \\ { \left( { \frac { { 2\cos \frac { { A+B } }{ 2 } \cdot \cos \frac { { A+B } }{ 2 } } }{ { 2\cos \frac { { A+B } }{ 2 } \cdot \sin \frac { { A+B } }{ 2 } } } } \right) ^{ n } }+{ \left( { \frac { { 2\sin \left( { \frac { { A+B } }{ 2 } } \right) \cdot \cos \left( { \frac { { A-B } }{ 2 } } \right) } }{ { -2\sin \frac { { A+B } }{ 2 } \cdot \sin \frac { { A-B } }{ 2 } } } } \right) ^{ n } } \\ { \left[ { \cot \frac { { A-B } }{ 2 } } \right] ^{ n } }+{ \left[ { -\cot \frac { { A-B } }{ 2 } } \right] ^{ n } } \end{array}$$If his odd its value is (0)Mathematics

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