Question

help me.................

Answer 1

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$\underline\bold{\huge{ANSWER \: :}}$
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(1+tan²A)/(1+cot²A)

= {1+(sin²A/cos²A)}/{1+(cos²A/sin²A)}

= {(cos²A+sin²A)/cos²A}/{(sin²A+cos²A)/sin²A}

= (1/cos²A) × (sin²A/1)

= sin²A/cos²A

= tan²A [OPTION (D)]



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$\bold{ALWAYS\: \: BE \: \: BRAINLY}$

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Answer 2

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\dfrac{1+tan^{2}\: A}{1+cot^{2}\: A}$ ———[1]



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\huge{\textsf{Step-1:}}}$
Using the Identities
$\bigstar\: \mathbf{sec^{2}\: \theta-tan^{2}\: \theta=1}$
$\bigstar\: \mathbf{cosec^{2}\: \theta-cot^{2}\: \theta=1}$

$\implies 1+tan^{2}\: A= sec^{2}\: A$ ———[2]
$\implies 1+cot^{2}\: A= cosec^{2}\: A$———[3]



$\underline{\huge{\textsf{Step-2:}}}$
Substitute [2]& [3] in [1]

$\implies \dfrac{sec^{2}\: A}{cosec^{2}\: A}$ ———[4]



$\underline{\huge{\textsf{Step-3:}}}$
Express [4] in terms of sine and cosine and Simplify

$\implies \dfrac{\frac{1}{cos^{2}\: A}}{\frac{1}{sin^{2}\: A}}$

$\implies \dfrac{1}{cos^{2}\: A} \times \dfrac{sin^{2}\: A}{1}$

$\implies \dfrac{cos^{2}\: A}{sin^{2}\: A}$

Since,
$\bigstar\: \mathbf{tan\: \theta= \dfrac{sin\: \theta}{cos\: \theta}}$

$\implies tan^{2}\: A$

This Answer exists in Option-D

$\therefore$
$\blacksquare\: \mathbf{\dfrac{1+tan^{2}\: A}{1+cot^{2}\: A}= \underline{\large{tan^{2}\: A}}}$