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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ with $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
✓

Answer

We have, $\cos\text{y}=\text{x}\cos(\text{a}+\text{y})$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\cos\text{y})=\frac{\text{d}}{\text{dx}}\big\{\text{x}\cos(\text{a}+\text{y})\big\}$
$\Rightarrow -\sin\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})$
$\Rightarrow -\sin\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})+\text{x}\big[-\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\sin(\text{a}+\text{y})-\sin\text{y}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y}) \\ \Big[\because \cos\text{y}=\text{x}\cos(\text{a}+\text{y})\Rightarrow\text{x}=\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}\Big] $
$\Rightarrow\big[\cos\text{y}\sin(\text{a}+\text{y})-\sin\text{y}\cos(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \sin(\text{a}+\text{y}-\text{y})\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$

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