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Exercise 3.1 · Q6

Q.Prove that: tan⁡(π4+θ)=1+tan⁡θ1−tan⁡θ\tan\left(\dfrac{\pi}{4}+\theta\right)=\dfrac{1+\tan\theta}{1-\tan\theta}

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Step 1: Use tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} with A=π/4A=\pi/4, B=θB=\theta. …

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