Question

Answer the question pls quickly

Answer 1

Step-by-step explanation:

LHS:

= 1+tan²A / 1+cot²A

Using the trignometric identities we know that 1+tan²A= Sec²A and 1+cot²A= Cosec²A

= Sec²A/ Cosec²A

On taking the reciprocals we get

= Sin²A/Cos²A

= tan²A

RHS:

=(1-tanA)²/(1-cotA)²

Substituting the reciprocal value of tan A and cot A we get,

=(1-sinA/cosA)²/(1-cosA/sinA)²

=[(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²

=(cosA-sinA)²×sin²A /Cos²A. /(sinA-cosA)²

=1×sin²A/Cos²A×1.

=tan

The values of LHS and RHS are same.

Hence proved

Answer 2

$\huge\green {\underline {\mathfrak {Answer}}}$

$\underline {\bold{\pink {To \ prove :}}}$

➣$\dfrac{1+tan^2A}{1+cot^2A} = \dfrac{sin^2A}{cos^2A} = tan^2A$

$\underline {\bold {\pink{Solution:}}}$

✦Solving LHS :

➫$\dfrac{1 + tan^2A}{1 + cot^2A}$

➫$\dfrac{sec^2A}{cosec^2A}$($Identity \ 1, \ 2$⬇️)

➫$\dfrac{1}{cos^2A} \times \dfrac{sin^2A}{1}$($Identity \ 3 ,\ 4$⬇️)

➫$\dfrac{sin^2A}{cos^2A}$

➫$\boxed{\implies{\boxed{tan^2A}}}$

($Identity \ 5$⬇️)

✮$\red{\underline{Trigonometric \; Identities \; used}}$

1️⃣ $\boxed{1 + tan^2A = sec^2A}$

2️⃣ $\boxed{1 + cot^2A = cosec^2A}$

3️⃣ $\boxed{sec^2A = \dfrac{1}{cos^2A}}$

4️⃣ $\boxed{cosec^2A = \dfrac{1}{sin^2A}}$

5️⃣ $\boxed{\dfrac{sin^2A}{cos^2A} = {tan^2A}}$