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CBSE Class 12-science Answered

Kindly explain your answer.  
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Asked by Varsneya Srinivas | 09 Feb, 2018, 09:07: AM
answered-by-expert Expert Answer
begin mathsize 16px style tan to the power of negative 1 end exponent straight a plus tan to the power of negative 1 end exponent straight b plus tan to the power of negative 1 end exponent straight c equals tan to the power of negative 1 end exponent space open parentheses fraction numerator straight a plus straight b over denominator 1 minus ab end fraction close parentheses plus tan to the power of negative 1 end exponent straight c space space space space space space space space space space space space space tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y equals tan to the power of negative 1 space end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses equals tan to the power of negative 1 end exponent space open parentheses fraction numerator fraction numerator straight a plus straight b over denominator 1 minus ab end fraction plus straight c over denominator 1 minus fraction numerator straight a plus straight b over denominator 1 minus ab end fraction cross times straight c end fraction close parentheses After space simplifying space we space get equals tan to the power of negative 1 end exponent space open parentheses fraction numerator straight a plus straight b plus straight c minus abc over denominator 1 minus ac minus bc end fraction close parentheses According space to space the space question space straight a plus straight b plus straight c minus abc equals 0 equals tan to the power of negative 1 end exponent left parenthesis 0 right parenthesis equals straight pi end style
Answered by Sneha shidid | 09 Feb, 2018, 09:35: AM

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CBSE 12-science - Maths
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