Question

if cot theta +tan theta=2 then prove:cot²theta +tan²theta=2​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt if\: cot\:\theta +tan\:\theta=2 \:then \:prove:\\\tt {cot}^{2}\:\theta +{tan}^{2}\:\theta=2$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf \implies cot\:\theta +tan\:\theta=2$

${\underbrace {\overbrace {\color {orange} {\sf \bf Square \:the\:both\:sides}}}}$

$\sf \bf \implies {\Big(cot\:\theta +tan\:\theta\Big)}^{2}={\Big(2\Big)}^{2}$

$\sf \bf \implies {cot}^{2}\:\theta +{tan}^{2} \:\theta+2 \times cot\:\theta \times tan\:\theta=4$

$\sf \bf \implies {cot}^{2}\:\theta +{tan}^{2} \:\theta+2 \times \dfrac{1}{tan\:\theta} \times tan\:\theta=4$

$\sf \bf \implies {cot}^{2}\:\theta +{tan}^{2} \:\theta+2 \times \dfrac{1}{\cancel {tan\:\theta}} \times \cancel {tan\:\theta}=4$

$\sf \bf \implies {cot}^{2}\:\theta +{tan}^{2} \:\theta+2 =4$

$\sf \bf \implies {cot}^{2}\:\theta +{tan}^{2} \:\theta=4-2$

$\implies {\boxed {\boxed {\color{blue} {\sf \bf {cot}^{2}\:\theta +{tan}^{2} \:\theta=2}}}}$

$\tt Used \:identity$

$\sf \bf {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\color {green} {\tt Identities}}$

$\sf \bf {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$

$\sf \bf {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$

$\sf \bf (a+b) (a-b) ={a}^{2}-{b}^{2}$

Answer 2

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