Question

sec2 A . cosec2 A = tan2 A + cot2 A + 2 prove ​

Answer 1

Question :-

$sec^{2}A \times cosec^{2}A = tan^{2}A+cot^{2}A + 2$

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Answer :-

We know that :-

${ \sec}^{2} \theta = 1 + { \tan}^{2} \theta$

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${ \cosec}^{2} \theta = 1 + { \cot}^{2} \theta$

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$\tan \theta \times \cot \theta = 1$

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Now, we have :-

$sec^{2}A \times cosec^{2}A = (1+tan^{2}A)(1+cot^{2}A)\\ \\ \\sec^{2}A \times cosec^{2}A =1 + cot^{2}A + tan^{2}A \\ + tan^{2}A.cot^{2}A\\ \\ \\sec^{2}A \times cosec^{2}A =2+cot^{2}A + tan^{2}A(RHS)$

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LHS = RHS

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Answer 2

Given

$\sec {}^{2} \theta \times cosec {}^{2} \theta = { \tan}^{2} \theta + { \cot }^{2} \theta + 2$

Taking

LHS

${sec}^{2} \theta \times {cosec}^{2} \theta \\ \\ {sec}^{2} \theta - {tan}^{2} \theta = 1 \\ \\ sec {}^{2} \theta = 1 + tan {}^{2} \theta \\ \\ {cosec}^{2} \theta - {cot}^{2} \theta = 1 \\ \\ cosec {}^{2} \theta = 1 + {cot}^{2} \theta$

Now LHS can Written as

$(1 + {tan}^{2} \theta)(1 + {cot}^{2} \theta) \\ \\ 1(1 + cot {}^{2} \theta) + {tan}^{2} \theta(1 + {cot}^{2} \theta) \\ \\ 1 + {cot}^{2} \theta + {tan}^{2} \theta + (tan {}^{2} \theta \times {cot}^{2} \theta)$

Now we know

${cot} \theta = \frac{1}{ {tan} \theta} \\ \\ cot \theta \times tan \theta = 1$

Now

$1 + {cot}^{2} \theta + {tan}^{2} \theta + (tan {}^{2} \theta \times {cot}^{2} \theta) \\ \\ = 1 + {cot}^{2} \theta + tan {}^{2} \theta + (tan \theta \times cot\theta) {}^{2} \\ \\ 1 + cot {}^{2} \theta + {tan}^{2} \theta + (1) {}^{2} \\ \\ 1 + {cot}^{2} \theta + tan {}^{2} \theta + 1 \\ \\ {tan}^{2} \theta + {cot}^{2} \theta + 2 \\ \\ lhs = {tan}^{2} \theta + {cos}^{2} \theta + 2$

Hence It's Proved

LHS=RHS