Question

Prove it
$\boxed{\sf \dfrac{1-tan^2\theta}{cot^2\theta-1}=tan^2\theta}$

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Answer 1

Prove it

$\\\: \: \: \: \: \sf \dfrac{1-tan^2\theta}{cot^2\theta-1}=tan^2\theta$

Solution

$\\~~~~~\implies\sf \dfrac{1-tan^2\theta}{cot^2\theta-1}=tan^2\theta$

• We know cot²θ = 1/tan²θ

$~~~~~\implies\sf \dfrac{1-tan^2\theta}{ \dfrac{1} {tan^2\theta} - 1}=tan^2\theta$

• Taking LCM

$~~~~~\implies\sf \dfrac{1-tan^2\theta}{ \dfrac{1 - tan^2\theta} {tan^2\theta}}=tan^2\theta$

$~~~~~\implies\sf \dfrac{ \cancel{1 -tan^2\theta}}{ \cancel{1 - t an^2\theta}} \times {tan^2\theta}=tan^2\theta$

$~~~~~\implies\bf { {tan^2\theta}=tan^2\theta} \: \: _{proved}$

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Relations among trigonometrical ratios:

• sinθ×cosecθ=1
• sinθ=1∕cosecθ
• cosecθ=1∕sinθ
• cosθ×secθ=1
• cosθ=1∕secθ
• secθ=1∕cosθ
• tanθ×cotθ=1
• tanθ=1∕cotθ
• cotθ=1∕tanθ
• sin²θ+cos²θ=1
• sin²θ=1-cos²θ
• cos²θ=1-sin²θ
• tanθ=sinθ∕cosθ
• cotθ=cosθ∕sinθ
• sec²θ-tan²θ=1
• sec²θ=1+tan²θ=1
• tan²θ=sec²θ-1
• cosec²θ-cot²θ=1
• cosec²θ=1+cot²θ
• cot²θ=cosec²θ-1

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