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CBSE Class 11-science Answered

The question is attached above. Please send step by step complete solution. I also request you please do not send any web link to this question because I am unable to  open the link every time  .
Asked by sudhanshubhushanroy | 14 Dec, 2017, 06:52: AM
answered-by-expert Expert Answer
begin mathsize 16px style straight A plus straight B equals straight pi over 3
tan open parentheses straight A plus straight B close parentheses equals fraction numerator tanA plus tanB over denominator 1 minus tanAtanB end fraction
tan straight pi over 3 equals fraction numerator tanA plus tanB over denominator 1 minus tanAtanB end fraction
square root of 3 open parentheses 1 minus tanAtanB close parentheses equals tanA plus tanB......... left parenthesis straight i right parenthesis
AM greater or equal than GM
fraction numerator tanA plus tanB over denominator 2 end fraction greater or equal than square root of tanAtanB
tanA plus tanB greater or equal than 2 square root of tanAtanB...... left parenthesis ii right parenthesis
Using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
3 open parentheses 1 minus tanAtanB close parentheses squared greater or equal than 4 tanatanB
Put space straight x equals tanatanB
and space simplify space it space further space you space will space get space straight y less or equal than 1 third space or space straight y greater or equal than 3
According space to space the space question space 0 less than tanA less than square root of 3 space and space 0 less than tanB less than square root of 3
tanAtanB less or equal than 3
straight y less than 3 rightwards double arrow straight y less or equal than 1 third

end style
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