. Prove that: (sec theta + tan theta)² = cosec theta+1 ÷ cosec theta-1
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Prove that: (sec theta + tan theta)² = cosec theta+1 ÷ cosec theta-1
. Prove that: (sec theta + tan theta)² = cosec theta+1 ÷ cosec theta-1
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Step-by-step explanation:
LHS =tanθ+
tanθ
1
=
tanθ
tan
2
θ+1
=
tanθ
sec
2
θ
=
cosθ
sinθ
cos
2
θ
1
=secθcosecθ
=RHS
Hence proved
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider,
[tex]\sf \: {(sec\theta + tan\theta )}^{2} \\ \\ [/tex]
We know,
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: sec\theta = \dfrac{1}{cos\theta } \qquad \: \\ \\& \qquad \:\sf \: tan\theta = \dfrac{sin\theta }{cos\theta } \end{aligned}} \qquad \: \\ \\ [/tex]
So, using these results, we get
[tex]\sf \: = \: {\left(\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \right)}^{2} \\ \\ [/tex]
[tex]\sf \: = \: {\left(\dfrac{1 + sin\theta }{cos\theta } \right)}^{2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ {(1 + sin\theta )}^{2} }{ {cos}^{2}\theta } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ {(1 + sin\theta )}^{2} }{ 1 - {sin}^{2}\theta } \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: {sin}^{2}\theta + {cos}^{2} \theta = 1 \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ {(1 + sin\theta )}^{2} }{ (1 - {sin}\theta)(1 + sin\theta ) } \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1 + sin\theta }{1 - sin\theta } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1 + \dfrac{1}{cosec\theta } }{1 - \dfrac{1}{cosec\theta } } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{\dfrac{cosec\theta + 1}{cosec\theta } }{\dfrac{cosec\theta - 1}{cosec\theta } } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{cosec\theta + 1}{cosec\theta - 1} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: {(sec\theta + tan\theta )}^{2} = \: \dfrac{cosec\theta + 1}{cosec\theta - 1} \\ \\ [/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]