text: Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
  
  # Find the value
  
  Question:
  
  If $y=\frac{\cos x-\sin x}{\cos x+\sin x}$, show that $\frac{d y}{d x}+y^{2}+1=0$
  
  Solution:
  
  $=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  
  Formula:
  
  $\frac{d(\sin x)}{d x}=\cos x$ and $\frac{d(\cos x)}{d x}=-\sin x$
  
  According to the quotient rule of differentiation
  
  If $y=u / v$
  
  $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$
  
  $=\frac{(\cos x+\sin x) \times(-\sin x-\cos x)-(\cos x-\sin x) \times(-\sin x+\cos x)}{(\cos x+\sin x)^{2}}$
  
  $=\frac{-(\cos x+\sin x)^{2}-(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$
  
  $=-\frac{(\cos x+\sin x)^{2}}{(\cos x+\sin x)^{2}}-\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$
  
  $=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  
  $\frac{d y}{d x}+y^{2}+1=0$
  
  HENCE PROVED.
  
  #### Leave a comment
  
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  Free Study Material
explanation: The document contains a math problem and its solution. The problem is to show that \(\frac{d y}{d x} + y^{2} + 1 = 0\) given \(y = \frac{\cos x - \sin x}{\cos x + \sin x}\). The solution is provided, and it concludes with "HENCE PROVED."
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
  
  # Find the value
  
  Question:
  
  If $y=\frac{\cos x-\sin x}{\cos x+\sin x}$, show that $\frac{d y}{d x}+y^{2}+1=0$
  
  Solution:
  
  $=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  
  Formula:
  
  $\frac{d(\sin x)}{d x}=\cos x$ and $\frac{d(\cos x)}{d x}=-\sin x$
  
  According to the quotient rule of differentiation
  
  If $y=u / v$
  
  $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$
  
  $=\frac{(\cos x+\sin x) \times(-\sin x-\cos x)-(\cos x-\sin x) \times(-\sin x+\cos x)}{(\cos x+\sin x)^{2}}$
  
  $=\frac{-(\cos x+\sin x)^{2}-(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$
  
  $=-\frac{(\cos x+\sin x)^{2}}{(\cos x+\sin x)^{2}}-\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$
  
  $=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  
  $\frac{d y}{d x}+y^{2}+1=0$
  
  HENCE PROVED.
  
  #### Leave a comment
  
  None
  Free Study Material
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem and its solution. The problem is to show that \(\frac{d y}{d x} + y^{2} + 1 = 0\) given \(y = \frac{\cos x - \sin x}{\cos x + \sin x}\). The solution is provided, and it concludes with "HENCE PROVED."
  
  Label: question-and-answer