Evaluate : `(sin50^0)/(cos40^0)+(cos e c40^0)/(sec50^0)-4cos50^0cos e c40^0`

Explanation

`cos (90-theta)=sin theta `


`sec (90-theta)={cosec} theta `


`cosectheta=frac{1}{sin theta}`


So,


`cos 40^circ=cos (90^circ-50^circ)=sin 50^circ`


`sec 50^circ=tan (90^circ-40^circ)=cosec 40^circ`


`cos 50^circ=sin (90^circ-40^circ)=sin 40^circ`


`frac{sin 50^circ}{cos 40^circ}+frac{csc 40^circ}{sec 50^circ}-4 cos 50^circ cosec 40^circ`


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