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If Tan θ = `A/B` Prove that `(A Sin Theta - B Cos Theta)/(A Sin Theta + B Cos Theta) = (A^2 - B^2)/(A^2 + B^2)` - Mathematics

If tan θ = `a/b` prove that `(a sin theta - b cos theta)/(a sin theta + b cos theta) = (a^2 - b^2)/(a^2 + b^2)`

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Solution

Let  `(a sin  theta - b cos theta)/(a sin theta + b cos theta)`

Divide both Nr and Dr with cos θ of (a)

`((a sin theta - b cos theta)/cos theta)/((a sin theta + b cos theta)/cos theta)`

`= (tan theta - b)/(a tan theta + b)`

`=(a xx (a/b) - b)/(a xx (a/b) + b)`

`= (a^2 - b^2)/(a^2 + b^2)`

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 10 Trigonometric Ratios
Exercise 10.1 | Q 12 | Page 24
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