Math, asked by novakharsha, 11 months ago

(cos+sin/cos-sin) - (cos-sin/cos+sin) = 4tan/1-tan2

Answers

Answered by Neerajpprajapati
1

Step-by-step explanation:

Taking LHS :-

  • ( \frac{ \cos(a) +  \sin(a)  }{ \cos(a) -  \sin(a)  } ) - ( \frac{ \cos(a) -  \sin(a)  }{ \cos(a) +  \sin(a)  } )
  •  \frac{( \cos(a) +  \sin(a) ) {}^{2} - ( \cos(a)  -  \sin(a)  ) {}^{2}  }{( \cos(a)  -  \sin(a))( \cos(a)  +  \sin(a) ) }
  •  because \:  \:  \:  {(a + b)}^{2}  =  {a}^{2}  +  {b}^{2}  + 2ab \:  \: and \:  {(a - b)}^{2}  =  {a}^{2}  +  {b}^{2}  - 2ab \:  \:  \:  \:  \: and \:  \:   (a - b)(a + b) =  {a}^{2}  -  {b}^{2}
  • Then, after solve this we get,
  •  \frac{4 \sin(a)  \cos(a) }{ \cos {}^{2} (a)  \:  -  \sin {}^{2} (a) }
  • Now, because
  • 2 \sin(a)  \cos(a)  =  \sin(2a)  \:  \: and \:  \cos {}^{2} (a)  -  \sin {}^{2} (a)  =  \cos(2a)
  • Then, we get
  •  \frac{2 \sin(2a) }{ \cos(2a) }
  •  =  > 2 \tan(2a)
  • But,
  •  \tan(2a \:  ) =  \frac{2 \tan(a) }{1 -  \tan {}^{2} (a) }
  • Then, we get : -
  •  \frac{</strong><strong>4</strong><strong> \tan(a) }{1 -  \tan {}^{2} (a) }  \:  = rhs
  • Hence proved.
  • I hope it will help you. Thank you
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