Question

(1+ Cot A ) ² + (1 - CotA)² = 2 coseca​

Answer 1

Step-by-step explanation:

Correct Question:-

$\sf Prove \: that :-$

$\sf {(1 + cotA)}^{2} + {(1 - cotA)}^{2} = 2 {cosec}^{2} A$

$\bf{\underline{\underline\red{To\:Prove:-}}}$

• $\sf {(1 + cotA)}^{2} + {(1 - cotA)}^{2} = 2cosecA$

$\bf{\underline{\underline\green{Proof:-}}}$

$\sf Identities\: used:-$

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

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

Let us prove by simplifying LHS and RHS seperately

$\sf LHS = {(1 + cotA)}^{2} + {(1 - cotA)}^{2}$

$\sf = \big \{ {(1)}^{2} + {(cotA) }^{2} + 2(1)( cotA) \big\}+ \big \{{(1 )}^{2} + (cotA) ^{2} - 2(1)(cotA)\big \}$

$\sf = \big \{ 1+ cot ^{2} A+ 2cotA \big\}+ \big \{1+ cot ^{2} A - 2cotA\big \}$

$\sf = 1+ cot ^{2} A+ 2cotA + 1+ cot ^{2} A - 2cotA$

$\sf = 1+ cot ^{2} A + 1+ cot ^{2} A + 2cotA- 2cotA$

$\sf = 1+ cot ^{2} A + 1+ cot ^{2} A \: \cancel{ + 2cotA} \: \: \cancel{- 2cotA }$

$\sf = 2+ 2cot ^{2} A$

$\sf = 2(1+ cot ^{2} A )$

$\sf = 2 \bigg(1+ \dfrac{{cos}^{2} A}{ {sin}^{2} A} \bigg ) \: \: \: \:\:\:\:\:\: \: \bigg ( \because {cot} \theta = \dfrac{cos \theta}{ {sin}\theta } \bigg)$

$\sf = 2 \bigg( \dfrac{{{sin}^{2} A + cos}^{2} A}{ {sin}^{2} A} \bigg )$

$\sf = 2 \bigg( \dfrac{1}{ {sin}^{2} A} \bigg ) \: \: \: \: \: \: \:\:\:\: \: \bigg( \because{sin}^{2} A + {cos}^{2} A = 1 \bigg)$

$\sf = 2 cosec^{2}A \: \: \:\:\:\:\:\:\:\:\: \: \: \: \bigg( \because \dfrac{1}{ {sin}^{2} A } = {cosec}^{2} A \bigg)$

$\sf = 2 cosec^{2}A = RHS$

LHS = RHS

Hence Proved