# tan60°-tan30°---------------------=tan30°1+tan60°. tan30°

CHAKRAVARTHI
52 Points
6 years ago
Answer is” tan-1[1+ square root of 3] /2
Given ,
= tan600-[tan300][X]=tan300+tan600.tan300
=square root of 3-1/ square root of 3[X]=1/ square root of 3+[ square root of 3.1/ square root of 3]  …. [CANCEL]
=[3-1] / square root of 3[X]=1/ square root of 3 +1
=finally we get,
X =[1+ square root of 3] /2
CHAKRAVARTHI
52 Points
6 years ago

Answer is” tan-1[1+ square root of 3] /2
Given ,
= tan600-[tan300][X]=tan300+tan600.tan300
=square root of 3-1/ square root of 3[X]=1/ square root of 3+[ square root of 3.1/ square root of 3]  …. [CANCEL]
=[3-1] / square root of 3[X]=1/ square root of 3 +1
=finally we get,
X =[1+ square root of 3] /2
CHAKRAVARTHI
52 Points
6 years ago

Answer is” tan-1[1+ square root of 3] /2
Given ,
= tan600-[tan300][X]=tan300+tan600.tan300
=square root of 3-1/ square root of 3[X]=1/ square root of 3+[ square root of 3.1/ square root of 3]  …. [CANCEL]
=[3-1] / square root of 3[X]=1/ square root of 3 +1
=finally we get,
X =[1+ square root of 3] /2

CHAKRAVARTHI
52 Points
6 years ago

Answer is” tan-1[1+ square root of 3] /2
Given ,
= tan600-[tan300][X]=tan300+tan600.tan300
=square root of 3-1/ square root of 3[X]=1/ square root of 3+[ square root of 3.1/ square root of 3]  …. [CANCEL]
=[3-1] / square root of 3[X]=1/ square root of 3 +1
=finally we get,
X =[1+ square root of 3] /2