Question

#Quality Question
@Trigonometry

Prove that
$\bf( {1 - tanA)}^{2} (1 - {cotA)}^{2} = (secA - cosecA) {}^{2}$
​

Answer 1

༒ To Prove :-

$\Large\bf( {1 - tanA)}^{2} (1 - {cotA)}^{2} \\\Large= \bf(secA - cosecA) {}^{2}$

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༒ Formulae Used :-

$\bf \maltese \: \: \: 1 + {tan}^{2} A = sec {}^{2} A \\ \\ \bf \maltese \: \: \: 1 + {cot} {}^{2} A = {cosec}^{2} A \\ \\ \bf \maltese \: \: \: sin^2\theta+cos^2\theta=1 \\ \\ \bf \maltese \: \: \: \frac{1}{sin\theta}=cosec\theta \\ \\ \bf \maltese \: \: \: \frac{1}{cos\theta}=sec\theta$

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༒ Proof :-

$\sf {(1 - tanA) }^{2} + {(1 - cot A)}^{2} \\ \\ \sf = (1 + {tan}^{2}A - 2tanA) \\ \sf+ ( 1 + {cot}^{2} A - 2cotA) \\ \\ \sf = {sec}^{2} A - 2tanA + {cosec}^{2} A - 2cotA \\ \{ \bf\because1 + {tan}^{2} A = sec {}^{2} A \\ \bf and1 + {cot} {}^{2} A = {cosec}^{2} A \} \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( tanA + cotA) \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( \frac{sinA }{cosA }+ \frac{cosA}{sinA} ) \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2( \frac{ {sin}^{2} A + {cos}^{2}A)}{cosA \: sin A } \\ \\ = \sf {sec}^{2} + {cosec}^{2} A - 2( \frac{1}{sinA \: cos A } )\\\{\bf\because sin^2\theta+cos^2\theta=1\} \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2secA \: cosecA \\\\ \{\bf\because\frac{1}{sin\theta}=cosec\theta\\ \bf and\:\:\frac{1}{cos\theta}=sec\theta\} \\ \\ \sf = (secA - cosecA) {}^{2}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

$\huge{\mathbb{Solution↓}}$

$\begin{gathered} \sf {(1 - tanA) }^{2} + {(1 - cot A)}^{2} \\ \\ \sf = (1 + {tan}^{2}A - 2tanA) \\ \sf+ ( 1 + {cot}^{2} A - 2cotA) \\ \\ \sf = {sec}^{2} A - 2tanA + {cosec}^{2} A - 2cotA \\ \{ \bf\because1 + {tan}^{2} A = sec {}^{2} A \\ \bf and1 + {cot} {}^{2} A = {cosec}^{2} A \} \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( tanA + cotA) \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( \frac{sinA }{cosA }+ \frac{cosA}{sinA} ) \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2( \frac{ {sin}^{2} A + {cos}^{2}A)}{cosA \: sin A } \\ \\ = \sf {sec}^{2} + {cosec}^{2} A - 2( \frac{1}{sinA \: cos A } )\\\{\bf\because sin^2\theta+cos^2\theta=1\} \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2secA \: cosecA \\\\ \{\bf\because\frac{1}{sin\theta}=cosec\theta\\ \bf and\:\:\frac{1}{cos\theta}=sec\theta\} \\ \\ \sf = (secA - cosecA) {}^{2} \end{gathered}$

Hence proved

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