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Miscellaneous Exercise · Q21

Q.Find the derivative of sin⁡(x+a)cos⁡x\dfrac{\sin(x + a)}{\cos x}.

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Apply the Quotient Rule; the numerator collapses via cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B+\sin A\sin B=\cos(A-B) to cos⁡a\cos a, giving the derivative cos⁡a sec⁡2x\cos a\,\sec^2 x.

We differentiate y=sin⁡(x+a)cos⁡xy=\dfrac{\sin(x+a)}{\cos x}, where aa is a constant.

Step 1 — Set up the Quotient Rule.

Let u=sin⁡(x+a)u=\sin(x+a) and v=cos⁡xv=\cos x. Then u′=cos⁡(x+a)u'=\cos(x+a) and v′=−sin⁡xv'=-\sin x.

Step 2 — Apply (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}:

dydx=cos⁡(x+a)cos⁡x−sin⁡(x+a)(−sin⁡x)cos⁡2x=cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡xcos⁡2x.\frac{dy}{dx}=\frac{\cos(x+a)\cos x-\sin(x+a)(-\sin x)}{\cos^2 x}=\frac{\cos(x+a)\cos x+\sin(x+a)\sin x}{\cos^2 x}.

Step 3 — Simplify the numerator.

Using cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B+\sin A\sin B=\cos(A-B) with A=x+aA=x+a and B=xB=x, …

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