Question

Question of Trigonometry

Answer 1

I hope it helps your

Answer 2

Question:

To Prove :

${\sf{ {\sqrt{ {\dfrac{1 + cos A}{1 - cos A}} }} = (cosec A + cot A)}}$

Step-by-step explanation:

L.H.S. = ${\sf{\ \ {\sqrt{ {\dfrac{1 + cos A}{1 - cos A}} }}}}$

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Dividing and multiplying with 1 + cos A, we get

$\implies{\sf{ {\sqrt{ {\dfrac{1 + cos A}{1 - cos A}} \times {\dfrac{1 + cos A}{1 + cos A}} }} }}$

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$\implies{\sf{ {\sqrt{ {\dfrac{ (1 + cos A)^2}{(1 - cos A)(1 + cos A)}} }}}}$

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

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

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$\implies{\sf{ {\sqrt{ {\dfrac{ (1 + cos A)^2}{(1)^2 - (cos A)^2}} }}}}$

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$\implies{\sf{ {\sqrt{ {\dfrac{ (1 + cos A)^2}{1 - cos^2 A}} }}}}$

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${\boxed{\tt{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}$

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

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$\implies{\sf{ {\sqrt{ {\dfrac{ (1 + cos A)^2}{sin^2 A}} }}}}$

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We can write this as :

$\implies{\sf{ {\sqrt{ {\dfrac{ (1 + cos A)^2}{(sin A)^2}} }}}}$

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$\implies{\sf{ {\sqrt{ \left( {\dfrac{1 + cos A}{sin A}} \right) ^2 }}}}$

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

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We can write this as :

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

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${\boxed{\tt{Identity \ : \ {\dfrac{1}{sin A}} = cosec A}}}$

${\boxed{\tt{Identity \ : \ {\dfrac{cos A}{sin A}} = cot A}}}$

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$\implies{\sf{ cosec A + cot A}}$

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= R.H.S.

Hence, proved !!