Question

Prove that: (1/cosecA-cotA) - (1/sinA) = (1/sinA) - (1/cosecA+cotA)​

Answer 1

Prove that:

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$\sf \bigg(\dfrac{1}{cosec\;A - cot\; A}\bigg) - \bigg(\dfrac{1}{sin\;A}\bigg) = \bigg(\dfrac{1}{sin\;A}\bigg) - \bigg(\dfrac{1}{cosec\;A + cot \;A}\bigg)$

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$\underline{\bf{\dag} \:\mathfrak{Taking\; LHS\: :}}$⠀⠀⠀⠀

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$:\implies\sf \dfrac{1}{ \frac{1}{sin\;A} - \frac{cos\;A}{sin\;A}} - \dfrac{1}{sinA} \\\\\\:\implies\sf \dfrac{sin\; A}{1 -\; cosA} - \dfrac{1}{sin\; A} \\\\\\:\implies\sf \dfrac{sin^2\;A - 1 - cos \; A}{sin\; A(1 - cos \; A)}\\\\\\:\implies\sf \dfrac{\cancel{\:1} - cos^2\; A -\:\cancel{\;1} - cos\; A}{sin\; A(1 - cos\; A)}\\\\\\:\implies\sf \dfrac{cos\; A \: \;\cancel{(1 - cos \; A)}}{sin\; A\: \: \cancel{1 - cos\; A}}\\\\\\:\implies\bf\pink{Cot \; A \: \: \: \: \: \: \: \: \: \qquad\bigg\lgroup\sf \because \dfrac{cos\;A}{sin\;A} = Cot \; A \; \bigg\rgroup}$

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$\underline{\bf{\dag} \:\mathfrak{Now,\; taking \; RHS \: :}}$⠀⠀⠀⠀

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$:\implies\sf \dfrac{1}{sin\;A} - \dfrac{1}{cosec\;A + cot\; A}\\\\\\:\implies\sf \dfrac{1}{sin\;A} - \dfrac{1}{\frac{1}{sin\;A} + \frac{cos\; A}{sin\; A}} \\\\\\:\implies\sf \dfrac{1}{sin\; A} - \dfrac{sin\; A}{ 1 + cos\; A}\\\\\\:\implies\sf \dfrac{1 + cos\;A - sin^2\;A}{sin\; A(1 + cos\; A)}\\\\\\:\implies\sf \dfrac{\cancel{\;1} + cos\; A - \;\cancel{\;1} + cos^2 \; A}{sin\; A(1 + cos\;A)}\\\\\\:\implies\sf\dfrac{cos\;A \;\cancel{(1 + cos\;A)}}{sin\;A\:\cancel{1 + cos\; A}}\\\\\\:\implies\sf\dfrac{cos\;A}{sin\;A}\\\\\\:\implies\bf\pink{Cot\; A \: \: \: \: \: \: \: \: \qquad\bigg\lgroup\sf \because \dfrac{cos\;A}{sin\;A} = Cot \; A \; \bigg\rgroup}$

⠀⠀⠀⠀⠀$\therefore$ LHS = RHS; Hence Proved!

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$\bigstar\:\sf Trigonometric\:Values :\\\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}Radians/Angle & 0 & 30 &45 &60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} &1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}& Not D \hat{e} fined \\\cline{1-6}\end{tabular}$⠀⠀

Answer 2

$\huge\cal{\underline{\underline{\bigstar\:Question:-}}}$

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$\sf \bigg(\dfrac{1}{cosec\;A - cot\; A}\bigg) - \bigg(\dfrac{1}{sin\;A}\bigg) = \bigg(\dfrac{1}{sin\;A}\bigg) - \bigg(\dfrac{1}{cosec\;A + cot \;A}\bigg)$

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$\huge\cal{\underline{\underline{\bigstar\:Answer:-}}}$

$\underline{\bf{\dag} \:\mathfrak{Taking\;\sf{LHS}\: :}}$

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$:\implies\sf \dfrac{1}{ \frac{1}{sin\;A} - \frac{cos\;A}{sin\;A}} - \dfrac{1}{sinA} \\\\\\:\implies\sf \dfrac{sin\; A}{1 -\; cosA} - \dfrac{1}{sin\; A} \\\\\\:\implies\sf \dfrac{sin^2\;A - 1 - cos \; A}{sin\; A(1 - cos \; A)}\\\\\\:\implies\sf \dfrac{\cancel{\:1} - cos^2\; A -\:\cancel{\;1} - cos\; A}{sin\; A(1 - cos\; A)}\\\\\\:\implies\sf \dfrac{cos\; A \: \;\cancel{(1 - cos \; A)}}{sin\; A\: \: \cancel{1 - cos\; A}}\\\\\\:\implies\bf\pink{Cot \; A \: \: \: \: \: \: \: \: \: \qquad\bigg\lgroup\sf \because \dfrac{cos\;A}{sin\;A} = Cot \; A \; \bigg\rgroup}$

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$\underline{\bf{\dag} \:\mathfrak{Now,\; taking \; \sf{RHS} \: :}}$

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$:\implies\sf \dfrac{1}{sin\;A} - \dfrac{1}{cosec\;A + cot\; A}\\\\\\:\implies\sf \dfrac{1}{sin\;A} - \dfrac{1}{\frac{1}{sin\;A} + \frac{cos\; A}{sin\; A}} \\\\\\:\implies\sf \dfrac{1}{sin\; A} - \dfrac{sin\; A}{ 1 + cos\; A}\\\\\\:\implies\sf \dfrac{1 + cos\;A - sin^2\;A}{sin\; A(1 + cos\; A)}\\\\\\:\implies\sf \dfrac{\cancel{\;1} + cos\; A - \;\cancel{\;1} + cos^2 \; A}{sin\; A(1 + cos\;A)}\\\\\\:\implies\sf\dfrac{cos\;A \;\cancel{(1 + cos\;A)}}{sin\;A\:\cancel{1 + cos\; A}}\\\\\\:\implies\sf\dfrac{cos\;A}{sin\;A}\\\\\\:\implies\bf\pink{Cot\; A \: \: \: \: \: \: \: \: \qquad\bigg\lgroup\sf \because \dfrac{cos\;A}{sin\;A} = Cot \; A \; \bigg\rgroup}$

⠀⠀⠀⠀⠀$\bf\underline{\therefore Hence\:Proved!!}$

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