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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
$\text{If}\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\ \text{with}\cos\text{a}\neq\pm1,\ \text{prove that}\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
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Answer

It is given that, $\cos\text{y}=\text{x}\cos(\text{a}+\text{y})$ $\therefore\ \frac{\text{d}}{\text{dx}}[\cos\text{y}]=\frac{\text{d}}{\text{dx}}[\text{x}\cos(\text{a}+\text{y})]$ $\Rightarrow\ -\sin\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y}).\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}[\cos(\text{a}+\text{y})]$ $\Rightarrow\ -\sin\text{y}\frac{\text{dy}}{\text{dt}}=\cos(\text{a}+\text{y})+\text{x}.[-\sin(\text{a}+\text{y})]\frac{\text{dy}}{\text{dx}}$ $\Rightarrow\ [\text{x}\sin(\text{a}+\text{y})-\sin\text{y}]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})\ \dots(1)$Since $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{x}=\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}$
Then, equation (1) reduces to $\Big[\frac{\cos\text{y}}{\cos(\text{a}+\text{y})}.\sin(\text{a}+\text{y})-\sin\text{y}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$ $\Rightarrow\ \ [\cos\text{y}.\sin(\text{a}+\text{y})-\sin\text{y}.\cos(\text{a}+\text{y})].\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{y})$ $\Rightarrow\ \sin(\text{a}+\text{y}-\text{y})\frac{\text{dy}}{\text{dx}}=\cos^2(\text{a}+\text{b})$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{b})}{\sin\text{a}}$ Hence, proved.

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