Question

1. In Brainly, why I'm getting that type of answers from all.
See that attachment.

2. Show that
$\dfrac{1 + {tan}^{2} A}{1 + {cot}^{2} A} = { \dfrac{1 -tanA }{1 - cotA} }^{2}$
!!!!!PLEASE DON'T SPAM!!!!!

Answer my both questions!!​

Answer 1

Step-by-step explanation:

2. LHS= (1+tan²A)/(1+cot²A)

= sec²A/cosec²A

= (1/cos²A) / (1/sin²A)

=sin²A/cos²A

=tan²A

RHS= {(1-tanA)/(1-cotA)}²

= {(1-tanA)/(1-1/tanA)}²

= {(1-tanA).tanA/{-(1-tanA)}²

= tan²A

LHS = RHS, proved

Answer 2

1. You can use Brainly website (i.e) Browser, or contact BRAINLY LEADERS .

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

2. Show that :

$\dfrac{1 + {tan}^{2} A}{1 + {cot}^{2} A} = \large ( { \dfrac{1 -tanA }{1 - cotA} \large) }^{2}$

$\bf \: L.H.S :$

$\implies\dfrac{1 + {tan}^{2} A}{1 + {cot}^{2} A}$

$\implies \dfrac{1 + { \tan }^{2} }{1 + \dfrac{1}{ { \tan}^{2}A } }$

$\implies \dfrac{ 1 + { \tan }^{2}A }{ \dfrac{ { \tan}^{2}A + 1 }{ { \tan}^{2} A} }$

$\implies \dfrac{1}{ \dfrac{1}{ { \tan}^{2} A} }$

$\implies \underline{ \: \bold{{ tan}^{2} A} \: } \: --->(1)$

$\bf \: R.H.S :$

$\implies \large ( { \dfrac{1 -tanA }{1 - cotA} \large) }^{2}$

$\implies { \large[ \frac{1 - \: \tan A }{1 - \frac{1}{ \tan A} } \large ]}^{2}$

$\implies { \large[ \frac{1 - \tan A }{ \frac{\tan A - 1}{ \tan A} } \large ]}^{2}$

$\implies { \large( - \tan A \large )}^{2}$

$\implies \underline{ \: \bold{{ tan}^{2} A} \: } \: --->(2)$

$From \: (1) \: and \: (2),$

$\boxed { \bf L.H.S = R.H.S = {tan A}^{2} }$

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$