Question

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prove that :-

$\frac{1 + \ \sec 2A }{ \tan2 \: a } = \cot a$
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Answer 1

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Answer 2

$\huge\red{\boxed{\blue{\mathcal{\overbrace{\underbrace{\fcolorbox{blue}{aqua}{\underline{\red{QuestioN}}}}}}}}}$

Prove that :-

$\frac{1+sec2A}{tan2a} =\:cot \:a$

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LHS = $\frac{1+tan^2A}{1+cot^2A}$

                                                 [ ∴sec²A - tan²A = 1, and cosse²A - cot²A=1 ]

       = $\frac{sec^2A}{cossee^2A}$

       = $\frac{sec^2A}{cos^2A}$

       = $(\frac{sinA}{losA} )^2\:\:\:\:\:\:=\:tan^2A$

MHS = $(\frac{1-tanA}{1-cotA} )^2$

        = $[\frac{1-tanA}{1-\frac{1}{tanA} } ]^2$

       = $[\frac{1-tanA}{-(1-tanA)} *tanA]^2\:\:\:\:\:\:=tan^2A$

∴ LHS = MHS = RHS        [ Proved ]

∴Hence, Proved

"See the given attachment".

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