slove that: tana /1+cota+cota/1-tana=1+seca×coseca
Answers
[tex]\mathfrak{\huge{Answer:-}} \\ \frac{tanA}{1 - cotA} + \frac{cotA}{1 - tanA} = 1 + secAcosecA \\ \frac{ \frac{sinA}{cosA} }{1 - \frac{cosA}{sinA} } + \frac{ \frac{cosA}{sinA} }{ 1 - \frac{sinA}{cosA} } =1 + secAcosecA \\ \frac{ \frac{sinA}{cosA} }{ \frac{sinA -cosA }{sinA} } + \frac{ \frac{cosA}{sinA} }{ \frac{cosA -sinA }{cosA} } = 1 + secAcosecA \\ \frac{ \frac{sinA}{cosA} }{ \frac{sinA - cosA}{sinA} } - \frac{ \frac{cosA}{sinA} }{ \frac{sinA - cosA}{cosA} } =1 + secAcosecA \\ \frac{sinA}{cosA} \times \frac{sinA}{sinA - cosA} - \frac{cosA}{sinA} \times \frac{cosA}{ sinA-cosA } =1 + secAcosecA \\ \frac{ {sin}^{2}A }{cosA(sinA - cosA)} - \frac{ {cos}^{2}A }{sinA(sin A-cos A)} = 1 + secAcosecA\\ \frac{1}{sinA - cosA} ( \frac{ {sin}^{2}A }{cosA} - \frac{ {cos}^{2}A }{sinA}) = 1 + secAcosecA \\ \frac{1}{ sinA-cos A}( \frac{ {sin}^{3}A - {cos}^{3} A }{cosAsinA} = 1 + secAcosecA \\ \frac{1}{sinA - cosA} \times \frac{(sinA - cosA)( {sin}^{2}A + {cos}^{2}A +sinAcosA )}{cosAsinA} =1 + secAcosecA \\ \frac{( sinA-cos A)( {sin}^{2}A + {cos}^{2} A+ sin AcosA }{( sinA- cosA)(cosAsinA)} = 1 + secAcosecA \\ \frac{ {sin}^{2}A + {cos}^{2}A +sinAcos A}{cosAsinA} = 1 + secAcosecA\\ \frac{1 + sinAcosA}{cosAsinA} = 1 + secAcosecA \\ \frac{1}{cosAsinA} + \frac{sinAcosA}{sinAcosA} = 1 + secAcosecA \\ secAcosecA+ 1 = 1 + secAcosecA \\ 1 + secAcosecA =1 + secAcosecA \\ LHS=RHS \\ Hence \: proved[/tex]