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applied mathematics
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Asked by pardeepsinghkarn412 | 17 Jan, 2023, 05:23: AM
answered-by-expert Expert Answer
Do not post multiple number of questions in single posting. Ask only one question per posting.
 
Only few questions are answered below
 
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begin mathsize 14px style fraction numerator tan 70 space plus space tan 65 over denominator 1 space minus space tan 70 space tan 65 end fraction space equals space ? end style
 
Let us apply the formula  begin mathsize 14px style fraction numerator tan space A space plus space tan space B over denominator 1 space minus space tan A space tan B end fraction space equals space tan left parenthesis A plus B right parenthesis end style
 
begin mathsize 14px style fraction numerator tan 70 space plus space tan 65 over denominator 1 space minus space tan 70 space tan 65 end fraction space equals space tan left parenthesis 70 plus 65 right parenthesis space equals space tan 135 end style
 
tan135 = tan(180-45) = -tan45 = -1
 
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independent term in the expansion  begin mathsize 14px style open parentheses 3 over 2 x squared minus space fraction numerator 1 over denominator 3 x end fraction close parentheses to the power of 9 end style
 
Term Independent of x in above expansion is   begin mathsize 14px style C presuperscript 9 subscript 6 end style begin mathsize 14px style open parentheses 3 over 2 space x squared close parentheses cubed space open parentheses fraction numerator 1 over denominator 3 x end fraction close parentheses to the power of 6 end style = begin mathsize 14px style 7 over 18 end style
 
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Partial fraction of begin mathsize 14px style fraction numerator 2 x plus 1 over denominator x squared minus 3 x plus 2 end fraction end style
 
 
begin mathsize 14px style fraction numerator 2 x plus 1 over denominator x squared minus 3 x plus 2 end fraction space equals space fraction numerator 2 x plus 1 over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis end fraction space equals space fraction numerator A over denominator left parenthesis x minus 1 right parenthesis end fraction plus fraction numerator B over denominator left parenthesis x minus 2 right parenthesis end fraction end style
 
 
begin mathsize 14px style fraction numerator 2 x plus 1 over denominator x squared minus 3 x plus 2 end fraction space equals space fraction numerator A left parenthesis x minus 2 right parenthesis plus B left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis end fraction space end style
 
begin mathsize 14px style A left parenthesis space x space minus 2 space right parenthesis space plus space B space left parenthesis x minus 1 right parenthesis space equals space 2 space x space plus 1 end style
 
if we substitute x =2 in above expression, we get B = 5
 
if we substitute x =1 in above expression, we get A = -3
 
begin mathsize 14px style fraction numerator 2 x plus 1 over denominator x squared minus 3 x plus 2 end fraction space equals space space fraction numerator 5 over denominator left parenthesis x minus 2 right parenthesis end fraction minus fraction numerator 3 over denominator left parenthesis x minus 1 right parenthesis end fraction end style
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S = 0.9 + 0.09 + 0.009 ................................to n-th term
 
S = 0.9 [ 1 + 0.1 + (0.1)2 + .........................(0.1)n-1 ]
 
begin mathsize 14px style S space equals space 0.9 space cross times fraction numerator 1 space minus space left parenthesis 0.1 right parenthesis to the power of n over denominator 1 space minus space left parenthesis 0.1 right parenthesis end fraction space equals space 1 space minus space left parenthesis 0.1 right parenthesis to the power of n end style
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Answered by Thiyagarajan K | 17 Jan, 2023, 09:32: AM
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