Evaluate : `(2sin68^0)/(cos22^0)-(2cot15^0)/(5tan75^0)-(3tan45^0tan20^0tan40^0tan50^0tan70^0)/5`

Explanation

As know,


`cos (90-theta)=sin theta `


`tan (90-theta)=cot theta `


`cot theta=frac{1}{tan theta} `


`cos 22^circ=cos (90^circ-68^circ)=sin 68^circ`


`tan 75^circ=tan (90^circ-15^circ)=cot 15^circ`


`tan 20^circ=tan (90^circ-70^circ)=cot 70^circ `


`tan 40^circ=tan (90^circ-50^circ)=cot 50^circ`


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