Question

सिद्ध कीजिए कि:
19. (1- tan A)2 + (1 - cot A)2 = (sec A - cosec A)2.​

Answer 1

LHS :

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

= RHS

Answer 2

∆ Given :-

It is given to prove that,

$\sf(1- tan A {)}^{2} + (1 - cot A {)}^{2} = (sec A - cosec A {)}^{2}$

∆ Solution :-

$\sf \: (1- tan A {)}^{2} + (1 - cot A {)}^{2}$

$\sf \to \: (1 +tan^{2}A - 2tanA) + (1+cot^{2}A-2cotA)$

$\sf \to \: sec^{2}A - 2tanA + cosec^{2}A - 2cotA$

$\sf \to \: {sec}^{2} A + cose {c}^{2} A - 2 (\frac{si {n}^{2} A+ co {s}^{2} A}{cosA \: \: sinA})$

$\sf \to \: se {c}^{2} + cose {c}^{2} A - 2( \frac{1}{sinA \: \: cosA})$

$\sf \to \: se {c}^{2} A + cose {c}^{2} A - 2secA \: \: cosecA$

$\bf \red { = (secA \: - \: cosecA {)}^{2} }$

Hence,

$\sf \green{(1- tan A {)}^{2} + (1 - cot A {)}^{2} = (sec A - cosec A {)}^{2} }$

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