Question

please solve my maths problem​

Answer 1

Step-by-step explanation:

We Have :-

$\csc( a ) + \cot(a) = m$

To Prove :-

$\dfrac{m^{2} - 1 }{m^{2} + 1} = \cos(a)$

Formulas Used :-

$(a + b)^{2} = a^{2} + b^{2} + 2ab \\ \\ \\ \csc^{2} (a) - \cot^{2} (a) = 1$

Solution :-

$\csc(a) + \cot(a) = m \\ \\ \\ squaring \: both \: sides \\ \\ \\ (\csc(a) + \cot(a) )^{2} = m^{2} \\ \\ \\ \csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) = m^{2} \\ \\ \\ now \: we \: have \\ \\ \\ \dfrac{m^{2} - 1}{m^{2} + 1} \\ \\ \\ putting \: the \: value \: of \: m^{2} \\ \\ \\ \dfrac{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) - 1}{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) + 1} \\ \\ \\ \dfrac{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) - (\csc^{2} (a) - \cot^{2} (a)) }{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) + (\csc^{2} (a) - \cot^{2} (a))} \\ \\ \\ \dfrac{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) - \csc^{2} (a) + \cot^{2} (a) }{\csc^{2} (a) + \cot^{2} (a) + 2 \csc(a) \cot(a) + \csc^{2} (a) - \cot^{2} (a)} \\ \\ \\ \dfrac{ \cot^{2} (a) + 2 \csc(a) \cot(a) + \cot^{2} (a) }{\csc^{2} (a) + 2 \csc(a) \cot(a) + \csc^{2} (a) } \\ \\ \\ \dfrac{ 2\cot^{2} (a) + 2 \csc(a) \cot(a) }{2\csc^{2} (a) + 2 \csc(a) \cot(a) } \\ \\ \\ \dfrac{ 2\cot(a)( \csc(a) + \cot(a) ) }{2\csc (a) ( \csc(a) + \cot(a)) } \\ \\ \\ \dfrac{ \cot(a) }{ \csc(a) } \\ \\ \\ \dfrac{ \dfrac{ \cos(a) }{ \sin(a) } }{ \dfrac{1}{ \sin(a) } } \\ \\ \\ \dfrac{ \cos(a) }{ \sin(a) } \times \dfrac{ \sin(a) }{1} \\ \\ \\ \cos(a)$