Question

Answer the question ​

Answer 1

Answer:

Proved!

Step-by-step explanation:

Given :

$\sf \left(\;\dfrac{1 + tan^2}{1 + cot^2A}\right)=\left(\dfrac{(1 - tanA}{1 - cotA)^2}\right)^2 = tan^2A$

To Prove :

$\sf \left(\;\dfrac{1 + tan^2}{1 + cot^2A}\right)=\left(\dfrac{(1 - tanA}{1 - cotA)^2}\right)^2 = tan^2A.$

Solution :

L.H.S  :

1 + tan² A/1+cot²A

⇒ (1 + Sin²A /Cos²A)/(1 + Cos²A/Sin²A)

⇒ [(Cos²A + Sin^2A)/cos²A)/(Sin²A + Cos²A)/ Sin²A

⇒ (1 / Cos²A) / (1 / sin²A)

⇒ 1 / Cos²A × sin²A / 1

⇒ sin²A / cos^2A

⇒ tan²A

R.H.S :

⇒ (1 - tanA / 1-cotA)²

⇒ (1 + tan²A - 2tanA) / (1 + cot² - 2cotA)

⇒ (sec²A  - 2tanA) / (Cosec^2A - 2cosA / sinA)

⇒ (sec^2A-2 × sinA / cos A) / (cosec^2A - 2cosA/sinA)

⇒ (1/cos²A-2sinA/cosA)/(1/sin²A-2cosA/sinA)

⇒ [(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin²A]

⇒ (1-2sinAcosA)/cos^2A × sin²A / (1-2sinAcosA)

⇒ sin²A/cos²A

⇒ tan²A

R.H.S :

⇒ tan²A

Hence, Proved,

$\sf \left(\;\dfrac{1 + tan^2}{1 + cot^2A}\right)=\left(\dfrac{(1 - tanA}{1 - cotA)^2}\right)^2 = tan^2A$

Answer 2

SOLUTION:-

CASE 1:

Take L.H.S

$\frac{1 + {tan}^{2}A }{1 + {cot}^{2} A} \\ \\ = > \frac{1 + \frac{ {sin}^{2} A}{ {cos}^{2}A} }{1 + \frac{ {cos}^{2} A}{ {sin}^{2} A} } \\ \\ = > \frac{ \frac{ \frac{ {cos}^{2} A + {sin}^{2} A}{ {cos}^{2} A} }{ {sin}^{2}A + {cos}^{2}A } }{ {sin}^{2}A } \\ \\ = > \frac{ \frac{ \frac{1}{ {cos}^{2} A} }{1} }{ {sin}^{2} A} \\ \\ = > \frac{1}{ {cos}^{2} A} \times \frac{ {sin}^{2} A}{1} = \frac{ {sin}^{2} A}{ {cos}^{2}A } \\ \\ = > {tan}^{2} A \: \: \: \: [R.H.S]$

CASE 2:

Take L.H.S

$( \frac{1 + tan \: A}{1 - cot \: A} ) {}^{2} \\ \\ = > ( \frac{1 + \frac{sin \: A}{cos \: A} }{1 - \frac{cos \: A}{sin \: A} } ) {}^{2} \\ \\ = > (\frac{ \frac{ \frac{cos \: A +sin \: A}{cos \: A} }{sin \: A - cos \: A} }{sin \: A} ) {}^{2} \\ \\ = > ( \frac{ \frac{ \frac{ (sin \: A +cos \: A)}{cos \: A} }{-sin \: A - cos \: A} }{sin \: A} ) {}^{2} \\ \\ = > ( \frac{ (sin \: a + cos \: A)}{-sin \: A - cos \: A} \times \frac{sin \: A}{cos \: A} ) {}^{2} \\ \\ = > (- \frac{sin \: A}{cos \: A} ) {}^{2} \\ \\ = > \frac{ {sin}^{2} A}{ { cos}^{2} A} = {tan}^{2} A \: \: \: \: \: [R.H.S]$

Hence proved.

Hope it helps ☺️