# Solve: (1 + tan^{2}A) / (1 + cot^{2}A) = [(1 - tan A) /(1 - cotA)]^{2} = tan^{2}A.

Trigonometric Identities are the equations that are true for right-angled triangles. We can use different trigonometric identities to prove the above expression.

## Answer: (1 + tan^{2}A) / (1 + cot^{2}A) = [(1 - tan A) /(1 - cotA)]^{2} = tan^{2}A.

Let us proceed from LHS. We will apply trigonometric identities to prove the result.

**Explanation:**

LHS: (1+ tan² A) / (1+ cot² A)

From the trigonometric identities, we know that

1+ tan² A = sec² A and 1 + cot²A = cosec² A

Therefore, LHS = sec² A / cosec² A

After taking the reciprocal of sec² A and cosec² A we get

⇒ sin² A / cos² A = tan² A

RHS: (1 - tan A)² / (1 - cot A)²

On substituting the reciprocal of tan A and cot A we get,

= (1- sin A / cos A)² / (1- cos A / sin A)²

= [(cos A - sin A) / cos A]² / [(sin A - cosA) / sin A)²]

= [(cos A - sin A)² × sin² A] / [cos²A × (sin A - cos A)²]

= (sin² A) / (cos² A)

= tan² A

Therefore, LHS is equal to RHS.

### Hence we proved (1 + tan^{2}A) / (1 + cot^{2}A) = [(1 - tan A) /(1 - cotA)]^{2} = tan^{2}A.

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