Question

$\frac{ { \tan }^{2} }{( \sec-1) } = \frac{1 + \ \cos }{1 - \cos }$
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Answer 1

Correct Identity:-

• $\sf{\dfrac{tan^2\theta}{Sec\theta - 2} = \dfrac{1+cos\theta}{cos\theta}}$

To prove:-

• LHS = RHS

Solution:-

Taking LHS,

$\sf{\dfrac{tan^2\theta}{sec\theta - 1}}$

We know,

$\sf{tan\theta = \dfrac{sin\theta}{cos\theta}}$

And

$\sf{sec\theta = \dfrac{1}{cos\theta}}$

Hence,

$\sf{\dfrac{\dfrac{sin^2\theta}{cos^2\theta}}{\dfrac{1}{cos\theta} - 1}}$

= $\sf{\dfrac{\dfrac{sin^2\theta}{cos^2\theta}}{\dfrac{1-cos\theta}{cos\theta}}}$

We know,

cos²θ + sin²θ = 1

=> sin²θ = 1 - cos²θ

Therefore,

= $\sf{\dfrac{1-cos^2\theta}{cos^2\theta}\times \dfrac{cos\theta}{1-cos\theta}}$

= $\sf{\dfrac{(1+cos\theta)(1-cos\theta)}{cos^2\theta}\times \dfrac{cos\theta}{1-cos\theta}}$

= $\sf{\dfrac{1+cos\theta}{cos\theta}}$

Hence,

LHS = RHS [Proved]

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Some other identities:-

• Sin²θ + Cos²θ = 1
• => Cos²θ = 1 - Sin²θ
• => Sin²θ = 1 - Cos²θ
• Tan²θ + 1 = Sec²θ
• => Tan²θ = Sec²θ - 1
• => Sec²θ - Tan²θ = 1
• Cot²θ + 1 = Cosec²θ
• => Cot²θ = Cosec²θ - 1
• => Cosec²θ - Cot²θ = 1

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