Prove the following trigonometric identities:
1+tan²θ/1+cot²θ=(1-tanθ/1-cotθ)²=tan²θ
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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.
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