Composition of Inverse Trigonometric Functions and Trigonometric Functions
then P -2 Q +...
Question
If P=∞∑r=1tan−1(1r+3)Q=∞∑r=1tan−1(1r+1)R=∞∑r=1tan−1(1r) , then P−2Q+R is
A
limx→π−sinx(π−x)2
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B
limx→π2π(x−π)21+cosx
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C
3∑k=1cot−1k
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D
tan−1tan(9)+sin−1sin(9)
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Solution
The correct option is Dtan−1tan(9)+sin−1sin(9) P−2Q+R=∞∑r=1tan−11r+3−2tan−1(1r+1)+tan−11r=tan−114−2tan−112+tan−111+tan−115−2tan−113+tan−112+tan−116−2tan−114+tan−113+tan−117−2tan−115+tan−114⋮⋮⋮⋮⋮⋮ =tan−11−tan−112−tan−113=π4−π4=0 Now a.)L=limx→π−sinx(π−x)2=limx→π−sin(π−x)(π−x)2=limh→0−sinhh2 As h→0−,L→∞ and as h→0+,L→−∞