Question

Prove that
$\frac{ \tan^{2}e}{1 + \tan^{2}e } + \frac{ { \cot }^{2}e }{1 + { \cot }^{2} e} = 1$
​

Answer 1

$\mathfrak{\huge{Answer:-}} \\ \frac{ {tan}^{2} θ}{1 + {tan}^{2} θ} + \frac{ {cot}^{2} θ}{1 + {cot}^{2}θ } = 1 \\ \frac{ \frac{ {sin}^{2}θ }{ {cos}^{2}θ } }{1 + \frac{ {sin}^{2}θ }{ {cos}^{2}θ } } + \frac{ \frac{ {cos}^{2} θ}{ {sin}^{2}θ } }{ 1 + \frac{{cos}^{2}θ }{ {sin}^{2}θ } } = 1 \\ \frac{ \frac{ {sin}^{2} θ}{ {cos}^{2}θ } }{ \frac{ {cos}^{2} θ + {sin}^{2} θ}{ {cos}^{2}θ } } + \frac{ \frac{ {cos}^{2}θ }{ {sin}^{2} θ} }{ \frac{ {sin}^{2} θ + {cos}^{2}θ }{ {sin}^{2}θ } } = 1 \\ \frac{ \frac{ {sin}^{2} θ}{ {cos}^{2} θ} }{ \frac{1}{ {cos}^{2}θ } } + \frac{ \frac{ {cos}^{2} θ}{ {sin}^{2}θ } }{ \frac{1}{ {sin}^{2}θ } } = 1 \\ \frac{ {sin}^{2}θ }{ {cos}^{2}θ } \times {cos}^{2} θ + \frac{ {cos}^{2} θ}{ {sin}^{2} θ} \times {sin}^{2} θ = 1 \\ {sin}^{2} θ + {cos}^{2} θ = 1 \\ 1 = 1\\LHS=RHS\\Hence\:proved$

Answer 2

Answer:

PROVED

Step-by-step explanation:

$\frac{ { \tan }^{2} e}{1 + { \tan }^{2}e } + \frac{ { \cot}^{2} e}{1 + { \cot}^{2}e } \\ = \frac{ { \tan}^{2}e }{ { \sec}^{2} e} + \frac{ { \cot}^{2}e }{ { \csc }^{2} e } \\ = \frac{ { \sin }^{2} e}{ { \cos }^{2} e} \times { { \cos}^{2} e} + \frac{ { \cos }^{2}e }{ { \sin}^{2}e } \times { \sin }^{2} e \\ = { \sin }^{2} e + { \cos }^{2} e \\ = 1$

IDENTITY USED:

tan²x + 1 = sec²x

cot²x+ 1 = csc²x

sin²x + cos²x = 1

HOPE IT HELPS :D