If $\cos \theta-\sin \theta=\sqrt{3} \cos \left(90^{\circ}-\theta\right), 0^{\circ}<\theta<90^{\circ}$ then find the value of $\tan \theta-\cot \theta$.
Answer & explanation
Correct answer: option 2
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 )
= \(\frac{1 - (√3 + 1 )² }{ (√3 + 1)}\)
= \(\frac{1 - 3 - 1 - 2√3 }{ (√3 + 1)}\)
= \(\frac{-3 - 2√3 }{ (√3 + 1)}\)
= - \(\frac{3 + 2√3 }{ (√3 + 1)}\)