Coseca +cota+1 /1 +coseca - cota=coseca +cota =1 +cosa /sinaa

Answers 1

[tex]\large\underline{\sf{Solution-}}[/tex]

Consider,

[tex]\sf \:\dfrac{coseca + cota + 1}{1 + coseca - cota} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: \dfrac{coseca + cota + ( {cosec}^{2}a - {cot}^{2}a)}{1 + coseca - cota} \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: {cosec}^{2}a - {cot}^{2}a = 1 \: }} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{coseca + cota + ( coseca + cota)(coseca - cota)}{1 + coseca - cota} \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{(coseca + cota)(1 + coseca - cota)}{1 + coseca - cota} \\ \\ [/tex]

[tex]\bf \: = \: coseca + cota\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{1}{sina} + \dfrac{cosa}{sina} \\ \\ [/tex]

[tex]\bf \: = \: \dfrac{1 + cosa}{sina} \\ \\ [/tex]

Hence,

[tex]\boxed{ \bf{\dfrac{coseca + cota + 1}{1 + coseca - cota} = coseca + cota = \dfrac{1 + cosa}{sina} \: }} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

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