Question

cos²θ - sin²θ =(1-tan²θ)/(1+tan²θ) prove it​

Answer 1

Answer

1+tan²θ/1+cot²θ = (1-tanθ/1-cotθ)² = tan²θ

Step-by-step explanation:

To prove: 1+tan²θ/1+cot²θ=(1-tanθ/1-cotθ)²=tan²θ

Proof:  

1+tan²θ/1+cot²θ=(1-tanθ/1-cotθ)²=tan²θ

1+tan²θ/1+cot²θ = 1 + Sin²θ/cos²θ / 1 +cos²θ/sin²θ

= (cos²θ + sin²θ)/ cos²θ / (sin²θ + cos²θ)/sin²θ

= 1/cos²θ / 1/sin²θ

= sin²θ/cos²θ

1+tan²θ/1+cot²θ = tan²θ -----------(1)

(1-tanθ/1-cotθ)² = (1 - tanθ)² / (1-cotθ)²

= (1 + tan²θ + 2tanθ) /  (1 + cot²θ + 2cotθ)

= (1 + sin²θ/cos²θ + 2sinθ/cosθ) / (1 + cos²θ/sin²θ + 2cosθ/sinθ)

=  (cos²θ + sin²θ + 2sinθ.cosθ)/cos²θ / (sin²θ + cos²θ + 2cosθ/sinθ)/ sin²θ

= sin²θ/ cos²θ

(1-tanθ/1-cotθ)² = tan²θ ----------(2)

From (1) and (2), we get:

1+tan²θ/1+cot²θ = (1-tanθ/1-cotθ)² = tan²θ

Hence proved.

Answer 2

Answer:

1+tan²θ/1+cot²θ = (1-tanθ/1-cotθ)² = tan²θ

Step-by-step explanation:

To prove: 1+tan²θ/1+cot²θ=(1-tanθ/1-cotθ)²=tan²θ

Proof:  

1+tan²θ/1+cot²θ=(1-tanθ/1-cotθ)²=tan²θ

1+tan²θ/1+cot²θ = 1 + Sin²θ/cos²θ / 1 +cos²θ/sin²θ

= (cos²θ + sin²θ)/ cos²θ / (sin²θ + cos²θ)/sin²θ

= 1/cos²θ / 1/sin²θ

= sin²θ/cos²θ

1+tan²θ/1+cot²θ = tan²θ -----------(1)

(1-tanθ/1-cotθ)² = (1 - tanθ)² / (1-cotθ)²

= (1 + tan²θ + 2tanθ) /  (1 + cot²θ + 2cotθ)

= (1 + sin²θ/cos²θ + 2sinθ/cosθ) / (1 + cos²θ/sin²θ + 2cosθ/sinθ)

=  (cos²θ + sin²θ + 2sinθ.cosθ)/cos²θ / (sin²θ + cos²θ + 2cosθ/sinθ)/ sin²θ

= sin²θ/ cos²θ

(1-tanθ/1-cotθ)² = tan²θ ----------(2)

From (1) and (2), we get:

1+tan²θ/1+cot²θ = (1-tanθ/1-cotθ)² = tan²θ

Hence proved.

Step-by-step explanation: