Question

TRIGONOMETRY
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Answer 1

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Answer 2

Step-by-step explanation:

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

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R.H.S. = ${\sf{\ \ {\dfrac{tan^2 A}{sec A - 1}}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ 1 + tan^2 \theta = sec^2 \theta }}}$

${\tt{From \ this, \ we \ get \ [ tan^2 \theta = sec^2 \theta - 1 ]}}$

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$\implies{\sf{ {\dfrac{sec^2 A - 1}{sec A - 1}}}}$

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• We can write this as :

$\implies{\sf{ {\dfrac{ (sec A)^2 - (1)^2 }{sec A - 1}}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ a^2 - b^2 = (a + b)(a - b)}}}$

${\tt{Here, \ a = sec A, \ b = 1}}$

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$\implies{\sf{ {\dfrac{ (sec A + 1)(sec A - 1)}{ sec A - 1}}}}$

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• Cancelling the common terms.

$\implies{\sf{sec A + 1}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ sec \theta = {\dfrac{1}{cos \theta}}}}}$

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$\implies{\sf{ {\dfrac{1}{cos A}} + 1}}$

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= R.H.S.

Hence, proved !!