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If tan θ = a/b then 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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Question

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

Theorem
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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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Chapter 10: Trigonometric Ratios - Exercise 10.1 [Page 24]

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R.D. Sharma Mathematics [English] Class 10
Chapter 10 Trigonometric Ratios
Exercise 10.1 | Q 12 | Page 24
R.S. Aggarwal Mathematics [English] Class 10
Chapter 10 Trignometric Ratios
EXERCISE 10 | Q 11. | Page 547
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