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EXERCISE 9.1 · Q6

Q.tan⁡(2x+3)\tan(2x+3)

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Let f(x)=tan⁡(2x+3)f(x)=\tan(2x+3), so f(x+h)=tan⁡(2x+2h+3)f(x+h)=\tan(2x+2h+3).

f(x+h)−f(x)=sin⁡(2x+2h+3)cos⁡(2x+2h+3)−sin⁡(2x+3)cos⁡(2x+3)=sin⁡(2h)cos⁡(2x+2h+3)cos⁡(2x+3)f(x+h)-f(x)=\dfrac{\sin(2x+2h+3)}{\cos(2x+2h+3)}-\dfrac{\sin(2x+3)}{\cos(2x+3)}=\dfrac{\sin(2h)}{\cos(2x+2h+3)\cos(2x+3)}

(numerator collapses via sin⁡Acos⁡B−cos⁡Asin⁡B=sin⁡(A−B)\sin A\cos B-\cos A\sin B=\sin(A-B) with A=2x+2h+3, B=2x+3A=2x+2h+3,\ B=2x+3, so A−B=2hA-B=2h). …

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