Question

Show that I-tan²A by 1+ tan square A=
2 Cos² A-1​

Answer 1

To Prove :

$\sf \dfrac{1-tan^{2}A}{1+tan^{2}A} = 2cos^{2}A - 1$

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Proof :

LHS : $\sf \dfrac{1-tan^{2}A}{1+tan^{2}A}$

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We know that :

$\Large \underline{\boxed{\bf{tan = \dfrac{sin}{cos}}}}$

$\sf : \implies \dfrac{1- \dfrac{sin^{2}A}{cos^{2}A}}{1+ \dfrac{sin^{2}A}{cos^{2}A}}$

$\sf : \implies \dfrac{\dfrac{cos^{2}A - sin^{2}A}{cos^{2}A}}{\dfrac{cos^{2}A + sin^{2}A}{cos^{2}A}}$

$\sf : \implies \dfrac{cos^{2}A - sin^{2}A}{\cancel{cos^{2}A}} \times \dfrac{\cancel{cos^{2}A}}{cos^{2}A + sin^{2}A}$

$\sf : \implies \dfrac{cos^{2}A - sin^{2}A}{cos^{2}A + sin^{2}A}$

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Now, we know that :

$\Large \underline{\boxed{\bf{cos^{2}A + sin^{2}A = 1}}}$

$\sf : \implies \dfrac{cos^{2}A - sin^{2}A}{1}$

$\sf : \implies cos^{2}A - sin^{2}A$

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Now, we know that :

$\Large \underline{\boxed{\bf{ sin^{2}A = 1 - cos^{2}A}}}$

$\sf : \implies cos^{2}A - (1 - cos^{2}A)$

$\sf : \implies cos^{2}A - 1 + cos^{2}A$

$\sf : \implies 2 cos^{2}A - 1$

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RHS : $\sf 2cos^{2}A - 1$

As, LHS = RHS

Hence, Proved.

Answer 2

To Prove :

$\bull \: \sf \dfrac{1 - tan {}^{2}A }{1 + tan {}^{2}A } = 2cos {}^{2} A - 1$

Proof :

LHS

$\longrightarrow \: \sf \dfrac{1 - tan {}^{2}A }{1 + tan {}^{2}A } \: \red\bigstar$

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$\longrightarrow \: \sf \dfrac{ 1 - \dfrac{sin {}^{2}A }{cos {}^{2} A}}{ 1 +\dfrac{sin {}^{2}A }{cos {}^{2} A} }$

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$\longrightarrow \: \sf \dfrac{ \dfrac{cos {}^{2}A - sin {}^{2}A }{cos {}^{2}A } }{ \dfrac{sin {}^{2} A + cos {}^{2}A }{cos {}^{2} A} }$

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$\longrightarrow \: \sf \dfrac{ \dfrac{cos {}^{2}A - sin {}^{2}A }{ \cancel{cos {}^{2}A }} }{ \dfrac{sin {}^{2} A + cos {}^{2}A }{ \cancel{cos{}^{2} A}} }$

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$\longrightarrow \: \sf \dfrac{cos {}^{2}A - sin {}^{2}A} {sin {}^{2} A + cos {}^{2}A }$

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$\longrightarrow \: \sf \dfrac{cos {}^{2}A - (1 - cos {}^{2}A )}{1}$

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$\longrightarrow \: \sf 2cos {}^{2} A - 1 \: \green \bigstar$

RHS

• Hence Proved!!