Question

please answer the question​

Answer 1

Question :-

$\sf{if \: CosecA + CotA = m \: then \: , show \: that \: \dfrac{m^{2} - 1}{m^{2} + 1} = CosA}$

Given :-

$\sf{(CosecA + CotA) = m}$

To prove :-

$\sf{\dfrac{m^{2} - 1}{m^{2} + 1} = CosA}$

Using Identity :-

$\sf{(Cosec^{2} A - Cot^{2} A) = 1}$

Solution :-

$\sf{(CosecA + CotA = m}$ ---(1)

$\sf{From \: using \: the \: above \: identity-}$

$\sf{(CosecA + CotA)(CosecA - CotA) = 1}$

$\sf{m(CosecA - CotA) = 1}$

$\sf{(CosecA - CotA) =\dfrac{1}{m}}$ --(2)

$\sf{From \: eq.(1) + eq.(2)-}$

$\sf{2CosecA =(m + \dfrac{1}{m})}$

$\sf{2CosecA =\dfrac{(m^{2} + 1)}{m}}$

$\sf{CosecA =\dfrac{(m^{2} + 1)}{2m}}$

$\sf{SinA =\dfrac{1}{CosecA}=\dfrac{2m}{(m^{2}+1)}}$

$\sf{CosA =\sqrt{(1 - Sin^{2} A)}}$

$\sf{CosA =\sqrt{1 - \dfrac{4m^{2}}{(m^{2}+1)^{2}}}}$

$\sf{CosA =\sqrt{\dfrac{(m^{2}+1)^{2} - 4m^{2}}{(m^{2}+1)^{2}}}}$

$\sf{(a + b)^{2} - 4ab = (a - b)^{2}}$

$\sf{So \: ,}$

$\sf{CosA =\sqrt{\dfrac{(m^{2}-1)^{2}}{(m^{2}+1)^{2}}}}$

$\sf{CosA =\sqrt{[\dfrac{(m^{2} - 1)}{(m^{2}+1)}]^{2}}}$

$\sf{CosA =\dfrac{(m^{2} -1)}{(m^{2}+1)}}$

$\sf{HENCE \: PROOF}$