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prove that i. (1 + sin2A - cos2A)/(1 + sin2A + cos2A) = tanA ii. (sinA + sin2A)/(1 + cosA + cos2A) = tanA iii. 2cosA = [2(2 + 2cos4A) 1/2 ] 1/2 iv. 1 + cos56 + cos 58 - cos66 = 4cos28.cos29.sin33

prove that 


i.  (1 + sin2A - cos2A)/(1 + sin2A + cos2A) = tanA


ii.  (sinA + sin2A)/(1 + cosA + cos2A) = tanA


iii.  2cosA = [2(2 + 2cos4A)1/2]1/2


iv. 1 + cos56 + cos 58 - cos66 = 4cos28.cos29.sin33

Grade:11

1 Answers

vikas askiitian expert
509 Points
11 years ago

1) (1 + SIN2A - COS2A)/(1+SIN2A + COS2A) = TANA

 

SIN2A =2TANA/1+TAN2A      &  COS2A =1-TAN2A/1+TAN2A

PUTTING THESE

 LHS =(2TAN2A + 2TANA)/2TANA+2

       =TANA = RHS

2)   (SINA + SIN2A)/(1+COSA+COS2A)    =   TANA

 COS2A = 2COS2A - 1       &   SIN2A =2SINACOSA

AFTER PUTTING THESE

 

LHS = (SINA+2SINACOSA)/COSA+2COS2A

      =SINA(1+2COSA) / COSA(1+2COSA)

      =SINA/COSA=TANA =RHS

HENCE PROVED

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