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Q16  If \cos y = x \cos (a + y), with \cos a \neq \pm 1 , prove that  \frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }

Answers (1)

best_answer

Given  function is
\cos y = x \cos (a + y)
Now,  Differentiate w.r.t x
\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\ \\ -\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\ \\ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\ \\ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) \ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\ \\ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\ \\ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b) \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\ \\ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}
Hence proved

Posted by

Gautam harsolia

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