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Q.Find the derivative of sin⁡(x+a)cos⁡x\dfrac{\sin(x + a)}{\cos x}. OR Evaluate : lim⁡x→01+x−1x\displaystyle \lim_{x \to 0} \dfrac{\sqrt{1 + x} - 1}{x}

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2019Subjective· 4mImportance★★★★★
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Applying the quotient rule to y=sin⁡(x+a)cos⁡xy=\dfrac{\sin(x+a)}{\cos x} and simplifying using cos⁡(A−B)\cos(A-B) gives y′=cos⁡asec⁡2xy' = \cos a \sec^2 x.

Let y=sin⁡(x+a)cos⁡xy = \dfrac{\sin(x+a)}{\cos x}. Using the quotient rule (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2} with u=sin⁡(x+a)u=\sin(x+a), v=cos⁡xv=\cos x:

u′=cos⁡(x+a),v′=−sin⁡xu' = \cos(x+a), \qquad v' = -\sin x

y′=cos⁡(x+a)cos⁡x−sin⁡(x+a)(−sin⁡x)cos⁡2x=cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡xcos⁡2xy' = \dfrac{\cos(x+a)\cos x - \sin(x+a)(-\sin x)}{\cos^2 x} = \dfrac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}

The numerator matches the cosine-difference identity cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B) = \cos A\cos B + \sin A \sin B with A=x+aA = x+a, B=xB = x:

cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡x=cos⁡((x+a)−x)=cos⁡a\cos(x+a)\cos x + \sin(x+a)\sin x = \cos((x+a)-x) = \cos a

So:

y′=cos⁡acos⁡2x=cos⁡a⋅sec⁡2xy' = \dfrac{\cos a}{\cos^2 x} = \cos a \cdot \sec^2 x

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