Q108 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

If $x \cos (k+y)=\cos y$ then $\frac{d y}{d x}$ at $y=\frac{\pi}{2}$ is

  • Correct answer
    A
    $\sin k$
  • B
    $\cos k$
  • C
    1
  • D
    0

Solution

Given $x \cos (k+y)=\cos y$ $$ x=\frac{\cos y}{\cos (k+y)} $$ Differentiate $x$ w.r.t ' $y$ ' $$ \begin{aligned} & \frac{d y}{d x}=\frac{\sin y \cos (k+y)+\cos y \sin (k+y)}{\cos ^2(k+y)} \\ & \frac{d x}{d y}=\frac{\sin (k+y-y)}{\cos ^2(k+y)}=\frac{\sin k}{\cos ^2(k+y)} \\ & \frac{d y}{d x}=\frac{\cos ^2(k+y)}{\sin k} \end{aligned} $$ Put $y=\frac{\pi}{2}$ in the above expression. $$ \frac{d y}{d x}=\frac{\left[\cos \left(\frac{\pi}{2}+k\right)\right]^2}{\sin k}=\frac{\sin ^2 k}{\sin k}=\sin k $$ So, option (a) is correct.

Hence, the correct answer is option: 'A'.

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