Question

1+tan^2A/1+cot^2A = _ _ _ _ _
A) sec^2 A
B) tan^2 A
C) cot^2 A
D) sin^2 A ​

Answer 1

$\large{\underline{\underline{\mathfrak{\red{\sf{Answer-}}}}}}$

Option B) Tan²A

$\large{\underline{\underline{\mathfrak{\red{\sf{Explanation-}}}}}}$

★ To find :

• Value of $\sf{\dfrac{1+Tan^2\:A}{1+Cot^2\:A}}$

★ Identities used :

• Sec²A - Tan²A = 1

• Cosec²A - Cot²A = 1

★ Ratios used :

• Sec²A = $\sf{\dfrac{1}{Cos^2\:A}}$

• Cosec²A = $\sf{\dfrac{1}{Sin^2\:A}}$

• Tan²A = $\sf{\dfrac{Sin^2\:A}{Cos^2\:A}}$

★ Solution :

$\sf{\dfrac{1+Tan^2\:A}{1+Cot^2\:A}}$

Put 1 + Tan²A = Sec²A and 1 + Cot²A = Cosec²A

$\implies$ $\sf{\dfrac{Sec^2\:A}{Cosec^2\:A}}$

Now, change it in the ratios sin and cos.

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

$\implies$ $\sf{\dfrac{1}{Cos^2\:A}×\dfrac{Sin^2\:A}{1}}$

$\implies$ $\sf{\dfrac{Sin^2\:A}{Cos^2\:A}}$

$\implies$ Tan²A

$\large{\underline{\boxed{\mathfrak{\sf{\blue{\therefore\:\dfrac{1+Tan^2\:A}{1+Cot^2\:A}=Tan^2\:A}}}}}}$

Answer 2

$\huge\bold\pink{ANSWER}$

Option[B] is the answer ✔️

$\huge\bold\pink{THANKS}$