2
Question
Show that 2tan−1(−3)=−π2+tan−1(−43).
Show that 2tan−1(−3)=−π2+tan−1(−43).
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Solution
LHS=2tan−1(−3)=−2tan−13 [∵ tan−1(−x)=−tan−1 x, x∈ R]=−[cos−11−321+32] [∵ 2tan−1 x=cos−11−x21+x2,x≥0]=[cos−1(−810)]=−[cos−1(−45)]=[π−cos−1(45)] {∵ cos−1(−x)=π−cos−1 x, x∈[−1, 1]}=−π+cos−1(45)[let cos−1(45)=θ⇒ cos θ=45⇒ tan θ =34⇒ θ =tan−134]=−π+tan−1(34)=−π+[π2−cot−1(34)]=−π2−cot−134=−π2−tan−143=−π2+tan−1(−43) [∵ tan−1(−x)=−tan−1 x]=RHS Hence proved.
LHS=2tan−1(−3)=−2tan−13 [∵ tan−1(−x)=−tan−1 x, x∈ R]=−[cos−11−321+32] [∵ 2tan−1 x=cos−11−x21+x2,x≥0]=[cos−1(−810)]=−[cos−1(−45)]=[π−cos−1(45)] {∵ cos−1(−x)=π−cos−1 x, x∈[−1, 1]}=−π+cos−1(45)[let cos−1(45)=θ⇒ cos θ=45⇒ tan θ =34⇒ θ =tan−134]=−π+tan−1(34)=−π+[π2−cot−1(34)]=−π2−cot−134=−π2−tan−143=−π2+tan−1(−43) [∵ tan−1(−x)=−tan−1 x]=RHS Hence proved.
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