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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```
6 years ago

510 Points
```										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
```
6 years ago
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