Question

tan-1(√3)-sec-1(-2)=?

Answer 1

Let tan−13–√=xtan−1⁡3=x

⇒tanx=3–√=tanπ3⇒tan⁡x=3=tanπ3

\( \therefore\), x=π3ϵ[−π2,π2]x=π3ϵ[−π2,π2]

Let sec−1(−2)=ysec−1⁡(−2)=y

⇒secy=−2⇒sec⁡y=−2

\( \therefore\), y=−secπ2y=−secπ2

⇒secy=sec(π−π2)=sec(2π3)⇒secy=sec(π−π2)=sec(2π3)

\(\therefore\), y=2π3ϵ[0,π]−(π2)y=2π3ϵ[0,π]−(π2)

tan−13–√−sec−1(−2)=x−y=π3−2π3=−π3tan−1⁡3−sec−1⁡(−2)=x−y=π3−2π3=−π3

The correct answer is −π3