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)`
Concept: Trigonometric Ratios
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