Question

Prove that+ sec A - cos A ) .(cotA+tanA )=tan A.secA

Answer 1

$\sf{\orange{\boxed{\bold{ Solution }}}}$

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$: \implies\: \sf{\bold{ \bigg( sec A - cos A \bigg)\bigg( cotA + tan A \bigg) = tanA . Sec A }}$

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$\sf{\red{\boxed{\bold{Sec A = \dfrac{1}{cosA}}}}}$

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$\sf{\red{\boxed{\bold{Cot A =\dfrac{1}{tanA}}}}}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{1}{CosA} - CosA \bigg) \bigg( \dfrac{1}{tanA} + tanA \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{1 - cos^2A}{CosA} \bigg) \bigg( \dfrac{1 + tan^2A}{tanA} \bigg) = tanA . Sec A }}$

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$\sf{\red{\boxed{\bold{1 - cos^2A = Sin^2A}}}}$

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$\sf{\red{\boxed{\bold{1 + tan^2A = sec^2A}}}}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A}{CosA} \bigg) \bigg( \dfrac{sec^2A}{tanA}\bigg) = tanA . Sec A }}$

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

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A}{CosA} \bigg) \bigg( \dfrac{sec^2A}{\dfrac{SinA}{CosA}} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A}{CosA} \bigg) \bigg( {sec^2A} \div\dfrac{sinA} {cosA} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A}{CosA} \bigg) \bigg( \dfrac{sec^2A\times SinA}{cosA} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A \times sec^2A \times CosA}{CosA\times SinA} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sin^2A \times sec^2A }{ SinA} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{ \bigg( \dfrac{sinA\times SinA \times secA\times SecA }{SinA} \bigg) = tanA . Sec A }}$

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$: \implies\: \sf{\bold{{sinA \times secA \times secA} = tanA . Sec A }}$

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

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$: \implies\: \sf{\bold{ sinA\times \dfrac{1}{CosA}\times SecA = tanA . Sec A }}$

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

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$: \implies\: \sf{\bold{ tanA . SecA = tanA . Sec A }}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.....HENCE PROVED

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