If θ ∈(-(π/2), 0), then the value of sec ( tan -1( cot θ)- cot -1( tan θ)) equals

Question

If θ ∈(-(π/2), 0), then the value of sec ( tan -1( cot θ)- cot -1( tan θ)) equals (A) 2 (B) √2 (C) -1 (D) 1

Tardigrade Admin · Accepted Answer

We have tan -1( cot θ)- cot -1( tan θ)=((π/2)- cot -1( cot θ))-((π/2)- tan -1( tan θ))= tan -1( tan θ)- cot -1( cot θ) =θ-(π+θ)=-π ( operatornameas θ ∈((-π/2), 0)) text Hence sec ( tan -1( cot θ)- cot -1( tan θ))= sec (-π)= sec π=-1

Q. If $θ \in (- \frac{π}{2}, 0)$ , then the value of $sec (tan^{- 1} (cot θ) - cot^{- 1} (tan θ))$ equals

2

$2$

-1

1

Q. If θ∈(−2π​,0), then the value of sec(tan−1(cotθ)−cot−1(tanθ)) equals

2

2​

-1

1

We have tan−1(cotθ)−cot−1(tanθ)=(2π​−cot−1(cotθ))−(2π​−tan−1(tanθ))=tan−1(tanθ)−cot−1(cotθ) =θ−(π+θ)=−π(asθ∈(2−π​,0)) Hence sec(tan−1(cotθ)−cot−1(tanθ))=sec(−π)=secπ=−1

Q. If $θ \in (- \frac{π}{2}, 0)$ , then the value of $sec (tan^{- 1} (cot θ) - cot^{- 1} (tan θ))$ equals

$2$