Question

Prove the Identity:

Answer 1

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• $\sf{\dfrac{cos( - \theta)}{1 - sin \theta} - \dfrac{cos( \theta + \pi)}{1 + sin \theta} }$

[$\tt{cos(-\theta)=cos\theta }$ ]

• $\sf{\dfrac{cos \theta}{1 - sin \theta} - \dfrac{cos( \theta + \pi)}{1 + sin \theta}}$

[$\tt{cos(\theta+\pi)=-cos\theta }$ ]

• $\sf{\dfrac{cos \theta}{1 - sin \theta} - \dfrac{-cos\theta }{1 + sin \theta}}$

• $\sf{\dfrac{cos \theta}{1 - sin \theta} + \dfrac{cos\theta }{1 + sin \theta}}$

• $\sf{cos\theta(\dfrac{1}{1-sin\theta}+\dfrac{1}{1+sin\theta} ) }$

• $\sf{cos\theta(\dfrac{1+sin\theta+1-sin\theta}{(1)^2-(sin\theta)^2 })}$

• $\sf{cos\theta(\dfrac{1\cancel{+sin\theta}+1\cancel{-sin\theta}}{(1)^2-(sin\theta)^2 })}$

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

[$\tt{1-sin^2\theta=cos^2\theta }$ ]

• $\sf{cos\theta(\dfrac{2}{cos^2\theta })}$

• $\sf{cos\theta(\dfrac{2}{cos\theta×cos\theta })}$

• $\sf{\cancel{cos\theta}×(\dfrac{2}{cos\theta×\cancel{cos\theta} })}$

• $\sf{\dfrac{2}{cos\theta} }$

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$\therefore{ \frac{cos( - \theta)}{1 - sin \theta} - \frac{cos( \theta + \pi)}{1 + sin \theta} = \frac{2}{cos \theta} }$(proved)

Answer 2

Answer:

Use properties:

• cos(-θ) = cosθ
• cos(θ + π) = - cosθ
• sin²θ + cos²θ = 1

Solution:

• cosθ/(1 - sinθ) - (-cosθ)/(1 + sinθ) =
• cosθ([1/(a- sinθ) + 1/(1 + sinθ)] =
• cosθ[(1 + sinθ + 1 - sinθ)/(1 - sin²θ)] =
• cosθ(2/cos²θ) =
• 2/cosθ

Proved