Question

plz tell the answer fast​

Answer 1

Question:

There is a misprint in the question. The correct question is :

Prove that :

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

Step-by-step explanation:

L.H.S. = ${\sf{\ \ {\dfrac{1 + sec A}{sec A}} }}$

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$\Rightarrow{\sf{ {\dfrac{1}{sec A}} + {\dfrac{sec A}{sec A}} }}$

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$\Rightarrow{\sf{ {\dfrac{1}{sec A}} + 1}}$

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${\boxed{\sf{\red{Identity \ : \ cos \theta = {\dfrac{1}{sec \theta}}}}}}$

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$\Rightarrow{\boxed{\sf{cos A + 1}}}$

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

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${\boxed{\sf{\red{Identity \ : \ sin^2 A + cos^2 A = 1}}}}$

${\sf{\red{From \ this, \ we \ get \ [ sin^2 A = 1 - cos^2 A] }}}$

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$\Rightarrow{\sf{ {\dfrac{1 - cos^2 A}{1 - cos A}}}}$

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We can write this as :

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

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

${\sf{\red{Here, \ a = 1, \ b = cos A}}}$

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$\Rightarrow{\sf{ {\dfrac{ (1 + cos A)(1 - cos A)}{1 - cos A}} }}$

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$\Rightarrow{\sf{ {\dfrac{ (1 + cos A){\cancel{(1 - cos A)}}}{{\cancel{1 - cos A}}}} }}$

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$\Rightarrow{\boxed{\sf{1 + cos A}}}$

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L.H.S. = R.H.S.

Hence, proved !!

Answer 2

your question has proved

there was mistake in question