Question

Root (1-cosA/1+cosA)+root( 1+cosA/ 1-cosA)=2cosecA

Answer 1

√[(1-cosA)²/(1-cos²A)] + √[(1+cosA)²/(1-cos²A)],

√[(1-cosA)²/sin²A] + √[(1+cosA)²/(sin²A)],

(1-cosA)/sinA + (1+cosA)/sinA,

(1-cosA+1+cosA)/sinA,

2/sinA,

2cosecA

Answer 2

$\huge \mathfrak \red {ANSWER}$

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$\sqrt{ \frac{1 + \cos(a) }{1 - \cos(a) } } + \sqrt{ \frac{1 - \cos(a) }{1 + \cos(a) } } \\ \\ = \sqrt{ \frac{1 + \cos(a) (1 + \cos(a) )}{(1 - \cos( a) )(1 + c \cos(a) )} } + \sqrt{ \frac{1 - \cos(a) (1 - \cos(a) )}{(1 + \cos(a) )(1 - \cos(a) )} } \\ \\ = \sqrt{ \frac{ {(1 + \cos(a) )}^{2} }{ {1}^{2} + {( \cos(a) )}^{2} } } + \sqrt{ \frac{ {(1 - \cos(a) }^{2} }{ {1}^{2} + { (\cos(a)) }^{2} } } \\ \\ = \sqrt{ \frac{ {(1 + \cos(a)) }^{2} + {(1 - \cos(a) )}^{2} }{ { \sin(a) }^{2} } } \\ \\ = \frac{1 + \cos(a) + 1 - \cos(a) }{ \sin(a) } \\ \\ = \frac{2}{ \sin(a) } \\ \\ = 2 \times \frac{1}{ \sin(a) } \\ \\ = 2 \times cosec(a) \\ \\ = 2 \: cosec(a)$

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@REVOLVER_RANI