Question

If : a^2 sec^2 - b^2 tan^2 = c^2
Prove that : sin ^2 = c^2 - a^2 / c^2 - b^2

Is there any trigonometry expert out there?? plz help

Answer 1

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Answer 2

Proof:

We have,

a² sec²θ - b² tan²θ = c²

⇒ a²/cos²θ - b² sin²θ/cos²θ = c²

⇒ a²- b² sin²θ/cos²θ = c²

⇒ a² - b² sin²θ = c²cos²θ

⇒ a² - b² sin²θ = c² (1 - sin²θ)            [ ∵ sin²θ + cos²θ = 1]

⇒ a²- b² sin²θ = c² - c² sin²θ

⇒ sin²θ (c² - b²) = c² - a²

⇒ sin²θ = c²- a²/c² - b²

Hence Proved.

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Regards,

Soham Patil.