prove that sinθ/1+cosθ + 1+cosθ/sinθ = 2 cosecθ
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Subject:
Math -
Author:
cheyanne9pwa -
Created:
1 year ago
Answers 1
Question :- Prove that
[tex]\qquad\sf \:\dfrac{sin\theta }{1 + cos\theta } + \dfrac{1 + cos\theta }{sin\theta } = 2 \: cosec\theta \\ \\ [/tex]
Step-by-step explanation:
Consider,
[tex]\qquad\sf \:\dfrac{sin\theta }{1 + cos\theta } + \dfrac{1 + cos\theta }{sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ {sin}^{2}\theta + {(1 + cos\theta )}^{2} }{(1 + cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ {sin}^{2}\theta + 1 + {cos}^{2} \theta + 2cos\theta }{(1 + cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ 1 + {sin}^{2} \theta + {cos}^{2}\theta + 2cos\theta }{(1 + cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ 1 + 1 + 2cos\theta }{(1 + cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ 2 + 2cos\theta }{(1 +cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{ 2(1 + cos\theta) }{(1 +cos\theta )sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \:\dfrac{2}{sin\theta } \\ \\ [/tex]
[tex]\qquad\sf \: = \:2 \: cosec\theta \\ \\ [/tex]
Hence,
[tex]\qquad\sf\implies \bf \:\dfrac{sin\theta }{1 + cos\theta } + \dfrac{1 + cos\theta }{sin\theta } = 2 \: cosec\theta \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used :-
[tex]\sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \\ \\ [/tex]
[tex]\sf \: {sin}^{2}\theta + {cos}^{2}\theta = 1 \\ \\ [/tex]
[tex]\sf \:cosec\theta \: = \: \dfrac{1}{sin\theta } \\ \\ [/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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Author:
mcmahon
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