Question

tan38°-cot22°=?
give solution with proper steps

Answer 1

LHS

=tan38°-cot22°

=(sin38°/cos38°)-(cos22°/sin22°)

=(sin38°sin22°-cos38°cos22°)/(cos38°)(sin22°)

=-cos(38°+22°)(sec38°)(cosec22°)

=-cos60°cosec22°sec38°

=-½cosec22°sec38°

Answer 2

Answer:

Step-by-step explanation:

This is simple, you just have to change tan and cot terms into sin and cos terms,

tan(380)−cot(220)=sin(380)cos(380)−cos(220)sin(220)

=> tan(380)−cot(220)=sin(380)sin(220)−cos(380)cos(220)sin(220)cos(380) —————(1)

We know that,

cos(A+B)=cosAcosB−sinAsinB

Using above trigonometric formula, in (1), we get-

=> tan(380)−cot(220)=−cos((38+22)0)sin(220)cos(380)

=> tan(380)−cot(220)=−cos(600)sin(220)cos(380)

=> tan(380)−cot(220)=−12sin(22deg)cos(38deg)

=> tan(380)−cot(220)=−12(cosec(220))(sec(380))