Question :- Prove that
[tex]\sf \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } = 2cosecA \\ \\ \\ [/tex]
Step-by-step explanation:
Consider,
[tex]\sf \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } \\ \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \dfrac{ \sqrt{secA - 1} }{ \sqrt{secA + 1} } + \dfrac{ \sqrt{secA + 1} }{ \sqrt{secA - 1} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ (\sqrt{secA - 1} )^{2} + {( \sqrt{secA + 1} )}^{2} }{ \sqrt{secA + 1} \: \sqrt{secA - 1} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{secA - 1 + secA + 1}{ \sqrt{(secA + 1)(secA - 1)} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {sec}^{2} A - 1} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {tan}^{2} A} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {sec}^{2}A - {tan}^{2}A = 1 \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ tanA } \\ \\ [/tex]
can be further rewritten as
[tex]\sf \: = \: \dfrac{2 \times \dfrac{1}{cosA} }{\dfrac{sinA}{cosA} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2}{sinA} \\ \\ [/tex]
[tex]\sf \: = \: 2cosecA \\ \\ [/tex]
Hence,
[tex]\bf\implies \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } = 2cosecA \\ \\ [/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]
Answer:
secA - 1 secA +1 V secA +1 + √ secA - 1
secA-1 √secA + 1 + secA +1 √secA - 1
= 2 (√secA − 1)² + (√secA +1)² √secA + 1 √secA – 1
secA 1 + secA + 1 √(secA + 1)(secA — 1)
2secA /sec²A - 1(x + y)(x − y) = x² - y²
2secA √tan2A
sec2A - tan2A = 1
2secA tanA
2 x
1
cosA
sinA
sinAsecA-1
secA + 1 + secA +1 secA 1 2
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Verified answer
Question :- Prove that
[tex]\sf \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } = 2cosecA \\ \\ \\ [/tex]
Step-by-step explanation:
Consider,
[tex]\sf \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } \\ \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \dfrac{ \sqrt{secA - 1} }{ \sqrt{secA + 1} } + \dfrac{ \sqrt{secA + 1} }{ \sqrt{secA - 1} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ (\sqrt{secA - 1} )^{2} + {( \sqrt{secA + 1} )}^{2} }{ \sqrt{secA + 1} \: \sqrt{secA - 1} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{secA - 1 + secA + 1}{ \sqrt{(secA + 1)(secA - 1)} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {sec}^{2} A - 1} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ \sqrt{ {tan}^{2} A} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {sec}^{2}A - {tan}^{2}A = 1 \: }} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2secA}{ tanA } \\ \\ [/tex]
can be further rewritten as
[tex]\sf \: = \: \dfrac{2 \times \dfrac{1}{cosA} }{\dfrac{sinA}{cosA} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2}{sinA} \\ \\ [/tex]
[tex]\sf \: = \: 2cosecA \\ \\ [/tex]
Hence,
[tex]\bf\implies \: \sqrt{\dfrac{secA - 1}{secA + 1} } + \sqrt{\dfrac{secA + 1}{secA - 1} } = 2cosecA \\ \\ [/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]
Answer:
secA - 1 secA +1 V secA +1 + √ secA - 1
can be rewritten as
secA-1 √secA + 1 + secA +1 √secA - 1
= 2 (√secA − 1)² + (√secA +1)² √secA + 1 √secA – 1
secA 1 + secA + 1 √(secA + 1)(secA — 1)
2secA /sec²A - 1(x + y)(x − y) = x² - y²
2secA √tan2A
sec2A - tan2A = 1
2secA tanA
can be further rewritten as
2 x
1
cosA
sinA
cosA
sinAsecA-1
secA + 1 + secA +1 secA 1 2