If $\cos \theta-\sin \theta=\sqrt{3} \cos \left(90^{\circ}-\theta\right), 0^{\circ}<\theta<90^{\circ}$ then find the value of $\tan \theta-\cot \theta$.

$-\frac{3+2 \sqrt{3}}{(1+\sqrt{3})}$

We are given that :-

cosθ - sinθ = √3 cos (90º - θ)

{ using , sinθ = cos (90º - θ)  }

cosθ - sinθ = √3sin θ

cosθ  = ( √3 + 1 ) sin θ

cot θ  = ( √3 + 1 )

& tanθ = \(\frac{1}{ √3 + 1}\)

Now,

tanθ - cotθ

= \(\frac{1}{ (√3 + 1)}\) - ( √3 + 1 )

= \(\frac{1 - (√3 + 1 )² }{ (√3 + 1)}\)

= \(\frac{1 - 3 - 1 - 2√3 }{ (√3 + 1)}\)

= \(\frac{-3 - 2√3 }{ (√3 + 1)}\)

=  - \(\frac{3 + 2√3 }{ (√3 + 1)}\)