tan (-(23 π/3))- cot (θ-(13 π/3))=

Question

tan (-(23 π/3))- cot (θ-(13 π/3))= (A) √3+ cot θ (B) √3- tan ((π/6)+θ) (C) √3+ tan θ (D) √3+ cot ((π/3)-θ)

Tardigrade Admin · Accepted Answer

tan (-(23 π/3))- cot (θ-(13 π/3)) =- tan (7 π+(2 π/3))+ cot (4 π+((π/3)-θ)) =- tan (2 π/3)+ cot ((π/3)-θ) =- tan (π-(π/3))+ cot ((π/3)-θ) = tan (π/3)+ cot ((π/3)-θ)=√3+ cot ((π/3)-θ)

Q. $tan (- \frac{23 π}{3}) - cot (θ - \frac{13 π}{3}) =$

$3 + cot θ$

$3 - tan (\frac{π}{6} + θ)$

$3 + tan θ$

$3 + cot (\frac{π}{3} - θ)$

Q. tan(−323π​)−cot(θ−313π​)=

3​+cotθ

3​−tan(6π​+θ)

3​+tanθ

3​+cot(3π​−θ)

tan(−323π​)−cot(θ−313π​) =−tan(7π+32π​)+cot(4π+(3π​−θ)) =−tan32π​+cot(3π​−θ) =−tan(π−3π​)+cot(3π​−θ) =tan3π​+cot(3π​−θ)=3​+cot(3π​−θ)

Q. $tan (- \frac{23 π}{3}) - cot (θ - \frac{13 π}{3}) =$

$3 + cot θ$

$3 - tan (\frac{π}{6} + θ)$

$3 + tan θ$

$3 + cot (\frac{π}{3} - θ)$