Question

Prove the following identity.
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Answer 1

Given

$\small\tt\longmapsto{\dfrac{1}{cosecA - cotA} - \dfrac{1}{sinA} = \dfrac{1}{sinA} - \dfrac{1}{cosecA - cotA}}$

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Formula used

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

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Solution

• Taking L.H.S

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$\tt:\implies{L.H.S = \dfrac{1}{cosecA - cotA} - \dfrac{1}{sinA}}$

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★ Rationalising

$\tiny\tt:\implies{L.H.S = \dfrac{1}{cosecA - cotA} \times \dfrac{cosecA + cotA}{cosecA + cotA} - \dfrac{1}{sinA}}$

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$\small\tt:\implies{L.H.S = \dfrac{cosecA + cotA}{cosec^2A - cot^2A} - cosecA}$

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$\: \: \: \: \: \: \boxed{\orange{cosec^2A - cot^2 = 1}}$

$\tt:\implies{L.H.S = \cancel{cosecA} + cotA - \cancel{cosecA}}$

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$\tt:\implies{L.H.S = cotA}$⠀⠀.....[1]

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• Now, taking R.H.S

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$\tt:\implies{R.H.S = \dfrac{1}{sinA} - \dfrac{1}{cosecA + cotA}}$

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★ Rationalising

$\tiny\tt:\implies{R.H.S = \dfrac{1}{sinA} - \dfrac{1}{cosecA + cotA} \times \dfrac{cosecA - cotA}{cosecA - cotA}}$

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$\small\tt:\implies{R.H.S = \dfrac{1}{sinA} - \dfrac{cosecA - cotA}{cosec^2A - cot^2A}}$

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$\: \: \: \: \: \: \boxed{\orange{cosec^2A - cot^2 = 1}}$

$\tt:\implies{R.H.S = cosecA - (cosecA - cotA)}$

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$\tt:\implies{R.H.S = \cancel{cosecA} - \cancel{cosecA} + cotA}$

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$\tt:\implies{R.H.S = cotA}$⠀⠀....[2]

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★ From [1] and [2] we find that

$\large{\underline{\sf{\red{L.H.S = R.H.S}}}}$

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