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  If sintheta=n"sin"(theta+2alp...
  
  Updated On: 27-06-2022
  
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  Solution : Given <br> sintheta=nsin(theta+2alpha)<br> =>(sintheta)/(sin(theta+2alpha))=n <br> apply componendo dividendo rule,<br> =>(sintheta+sin(theta+2alpha))/(sintheta−sin(theta+2alpha))=(n+1)/(n-1)<br> applying sinC+sinD=2sin((C+D)/2)cos((C-D)/2) and sinC−sinD=2sin((C-D)/2)cos((C+D)/2) formula<br> =>(2sin(theta+alpha)cosalpha)/(2sinalphacos(theta+alpha))=(n+1)/(n-1)<br> =>(tan(theta+alpha))/(tanalpha)=(n+1)/(n-1)<br> =>tan(theta+alpha)=(n+1)/(n-1)tanalpha