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| 1. |
Tan a=1÷3, tan b=1÷2 prove that sin2(a+b)=1 |
| Answer» sin 2(a + b) = 12 tan(a + b)/{1 - tan2(a + b)} = 12[(tan a + tan b)/(1 - tana.tanb)]/[1 + {(tan a + tan b)/(1 - tana.tanb)}2] = 12{(1/3 + 1/2)/1 - 1/3 x 1/2)]/[1 +{(1/3 + 1/2)/(1 - tan a .tan b)}2] = 12[(5/6)/(5/6)]/[1 + {(5/6)/(5/6)}2] = 12/[1 + 1] = 12/2 = 11 = 1Hence proved.\xa0\xa0 | |