Prove that $c o t^{2} A c o s e c^{2} B - c o t^{2} B c o s e c^{2} A = c o s e c^{2} A - c o s e c^{2} B$ .

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Solution

LHS $= {cot}^{2} A {c o s e c}^{2} B - {cot}^{2} B {c o s e c}^{2} A$

$= \frac{{cos}^{2} A}{{sin}^{2} A {sin}^{2} B} - \frac{{cos}^{2} B}{{sin}^{2} A {sin}^{2} B}$

$= \frac{(1 - {sin}^{2} A) - (1 - {sin}^{2} B)}{{sin}^{2} A {sin}^{2} B}$

$= \frac{{sin}^{2} B - {sin}^{2} A}{{sin}^{2} A {sin}^{2} B}$

$= {c o s e c}^{2} A - {c o s e c}^{2} B$

$= R H S$

Hence proved

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Prove that

Cot²A × cosec²A - cot²B × cosec² B = cot²A - cot² B

{sec}^{2} A + cosec^{2} A = {sec}^{2} A \times cosec^{2} A

\frac{{tan}^{2} A}{{tan}^{2} A + 1} + \frac{c o s e c^{2} A}{{sec}^{2} A + c o s e c^{2} A} = 1

∠ O A B = ∠ O B C = ∠ O C A = ω

c o t ω = c o t A + c o t B + c o t C .

c o s e c^{2} ω = c o s e c^{2} A + c o s e c^{2} B + c o s e c^{2} C .

\hat{i} + \hat{j} + \hat{k}, \hat{i} + (\sec^{2} B) \hat{j} + \hat{k}, \hat{i} + \hat{j} + (\sec^{2} C) \hat{k}

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