Question

Solve question no 8 at the earliest

Answer 1

Answer:

K=-2

Step-by-step explanation:

$\dfrac{cotA}{1+cosecA}-\dfrac{cotA}{1-cosecA}=\dfrac{K}{cosA}$

Taking L.H.S

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

$=\dfrac{(sinA-1)cosA-(sinA+1)cosA}{sin^2A-1}$

$=\dfrac{cosA(sinA-1-sinA-1)}{-cos^2A}$

Here sinA and cosA got cancelled, so we are left with,

$=\dfrac{-2}{-cosA}$

Comparing it with R.H.S, we get

$\bold{K=2}$

Answer 2

Answer:

K = 2

Step-by-step explanation:

Given :

${\sf{\ \ {\dfrac{cot A}{1 + cosec A}} - {\dfrac{cot A}{1 - cosec A}} = {\dfrac{K}{cos A}} }}$

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

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

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$\Rightarrow{\sf{ {\dfrac{ {\dfrac{cos A}{sin A}} }{ 1 + {\dfrac{1}{sin A}} }} - {\dfrac{ {\dfrac{cos A}{sin A}} }{ 1 - {\dfrac{ 1}{sin A}} }} = {\dfrac{K}{cos A}}}}$

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$\Rightarrow{\sf{{\dfrac{{\dfrac{cos A}{sin A}}}{{\dfrac{sin A + 1}{sin A}}}} - {\dfrac{{\dfrac{cos A}{sin A}}}{{\dfrac{sin A - 1}{sin A}}}} = {\dfrac{K}{cos A}}}}$

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$\Rightarrow{\sf{ {\dfrac{cos A}{sin A}} \times {\dfrac{sin A}{sin A + 1}} - {\dfrac{cos A}{sin A}} \times {\dfrac{sin A}{sin A - 1}} = {\dfrac{K}{cos A}} }}$

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$\Rightarrow{\sf{ {\dfrac{cos A}{{\cancel{sin A}}}} \times {\dfrac{{\cancel{sin A}}}{sin A + 1}} - {\dfrac{cos A}{{\cancel{sin A}}}} \times {\dfrac{{\cancel{sin A}}}{sin A - 1}} = {\dfrac{K}{cos A}} }}$

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$\Rightarrow{\sf{ {\dfrac{cos A}{sin A + 1}} - {\dfrac{cos A}{sin A - 1}} = {\dfrac{K}{cos A}} }}$

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

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

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

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Taking cos A common from the numerator.

$\Rightarrow{\sf{ {\dfrac{cos A [ (sin A - 1) - (sin A + 1) ] }{ (sin A)^2 - (1)^2}} = {\dfrac{K}{cos A}}}}$

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

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$\Rightarrow{\sf{ {\dfrac{cos A ({\cancel{sin A}} - 1 - {\cancel{sin A}} - 1)}{sin^2 A - 1}} = {\dfrac{K}{cos A}} }}$

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

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

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

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

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$\Rightarrow{\sf{ {\dfrac{2}{cos A}} = {\dfrac{K}{cos A}} }}$

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$\Rightarrow{\sf{ {\dfrac{2}{{\cancel{cos A}}}} = {\dfrac{K}{{\cancel{cos A}}}} }}$

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$\Rightarrow{\boxed{\sf{\green{K = 2}}}}$