Solve it $$ {sin \ θ\over 1+cos\ θ}+{1+cosθ\over sin\ θ}=?$$

Simplify $$ ({Sinθ\over{1+cosθ}}+{1+cosθ\over sinθ})({1\over (tanθ+cotθ)})$$

If $$ {sin\ {θ}}={1\over {\sqrt{2}}}$$ then (tanθ+cosθ)=?

If 2sinθ+cosθ= $${7\over3}$$ then the value of (tan²θ-sec²θ) is

$${sin^2θ\over cosθ(1+cosθ)}+{1+cosθ\over cosθ} =?$$

If secθ + tanθ = 3, then the value of secθ is:

If $${cos^2θ\over cot^2θ-cos^2θ}=3,$$ where $${0^0<θ<90^0}$$ then the value of θ is:

The numerical value of $$ {1\over 1+cot2}+{3\over 1+tan^2θ}+2sin^2θ$$ will be