Question

Sec2 A . cosec2 A = tan2 A + cot2 A + 2 prove ​

Answer 1

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = tan^2 A + cot^2A + 2 }}$

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$\sf{\red{\boxed{\bold{TanA = \dfrac{SinA}{CosA}}}}}$

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$\sf{\red{\boxed{\bold{CotA = \dfrac{CosA}{SinA}}}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = \dfrac{Sin^2A}{Cos^2A} + \dfrac{Cos^2A}{Sin^2A} + 2 }}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = \dfrac{ (Sin^2A)^2 + ( Cos^2A)^2 + 2 Sin^2A Cos^2A }{ Sin^2A Cos^2 A}}}$

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$\sf{\red{\boxed{\bold{a^2+ b^2+ 2ab = ( a + b ) ^2 }}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = \dfrac{ ( sin^2A + Cos^2A) ^2}{ Sin^2A Cos^2A} }}$

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$\sf{\red{\boxed{\bold{Sin^2A + Cos^2A = 1 }}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = \dfrac{1^2}{ Sin^2A. Cos^2 A}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = \dfrac{1}{Sin^2A} . \dfrac{1}{Cos^2A} }}$

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$\sf{\red{\boxed{\bold{Cosec A = \dfrac{1 }{SinA}}}}}$

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$\sf{\red{\boxed{\bold{sec A = \dfrac{1 }{CosA}}}}}$

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$\sf : \implies\:{\bold{ Sec^2A.cosec^2A = sec^2A.cosec^2A}}$

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LHS = RHS

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀..... Hence Proved

Answer 2

Answer:

See the attachment.

Step-by-step explanation: