Expression : $$sec12^{0}sin12^{0}tan38^{0}tan78^{0}tan52^{0}$$
= $$\frac{1}{cos(12^\circ)}.sin(12^\circ).tan(38^\circ).tan(78^\circ).tan(52^\circ)$$
= $$[tan(12^\circ).tan(78^\circ)].[tan(38^\circ).tan(52^\circ)]$$
Using, $$tan(90^\circ-\theta)=cot(\theta)$$
= $$[tan(12^\circ).cot(12^\circ)].[tan(38^\circ).cot(38^\circ)]$$
Also, $$tan(\theta)cot(\theta)=1$$
= $$1\times1=1$$
=> Ans - (A)
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