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CBSE Class 12-science Answered

Question written in the image .
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Asked by ritikabehera01 | 18 Feb, 2019, 04:13: PM
answered-by-expert Expert Answer
Formula to be used in this question :  begin mathsize 12px style tan to the power of negative 1 end exponent x space minus space tan to the power of negative 1 end exponent y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator x minus y over denominator 1 plus space x space y end fraction close parentheses end style
begin mathsize 12px style tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus 1.2 end fraction close parentheses space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 minus 1 over denominator 1 plus 1.2 end fraction close parentheses space equals space tan to the power of negative 1 end exponent 2 space minus space tan to the power of negative 1 end exponent 1 end style
begin mathsize 12px style tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus 2.3 end fraction close parentheses space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 minus 2 over denominator 1 plus 2.3 end fraction close parentheses space equals space tan to the power of negative 1 end exponent 3 space minus space tan to the power of negative 1 end exponent 2 end style
...............etc
 
begin mathsize 12px style S space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus 1.2 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus 2.3 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus 3.4 end fraction close parentheses plus............................ plus tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 plus n left parenthesis n plus 1 right parenthesis end fraction close parentheses end style
begin mathsize 12px style S space equals space tan to the power of negative 1 end exponent 2 space minus tan to the power of negative 1 end exponent 1 plus tan to the power of negative 1 end exponent 3 minus tan to the power of negative 1 end exponent 2 space....................... plus tan to the power of negative 1 end exponent open parentheses n plus 1 right parenthesis close parentheses minus space tan to the power of negative 1 end exponent n S space equals space tan to the power of negative 1 end exponent open parentheses n plus 1 close parentheses space minus space tan to the power of negative 1 end exponent 1 space space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator n plus 1 minus 1 over denominator 1 plus left parenthesis n plus 1 right parenthesis end fraction close parentheses space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator n over denominator n plus 2 end fraction close parentheses space equals space tan to the power of negative 1 end exponent theta H e n c e space theta space equals space fraction numerator n over denominator n plus 2 end fraction end style
Answered by Thiyagarajan K | 19 Feb, 2019, 08:09: AM

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