Question : $\frac{2\sin68°}{\cos22°}-\frac{2\cot15°}{5\tan75°} -\frac{3\tan45°\cdot\ \tan20°\cdot\ \tan40°\cdot\ \tan50°\cdot\ \tan70°}{5}$ is equal to:
Option 1: –1
Option 2: 0
Option 3: 1
Option 4: 2
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Correct Answer: 1
Solution : $\frac{2\sin68°}{\cos22°}-\frac{2\cot15°}{5\tan75°} - \: \frac{3\tan45°.\tan20°.\tan40°.\tan50°.\tan70°}{5}$ = $\frac{2\sin68°}{\cos(90°-68°)}-\frac{2\cot(90°-75°)}{5\tan75°} - \frac{3\cdot 1\cdot\tan20°\cdot\tan40°\cdot\tan(90°-40°)\cdot\tan(90°-20°)}{5}$ = $\frac{2\sin68°}{\sin68°}-\frac{2\tan75°}{5\tan75°} -\frac{3\cdot 1\cdot\tan20°\cdot\tan40°\cdot\cot40°\cdot\cot20°}{5}$ = $2×1 - \frac{2}{5}× 1 - \frac{3}{5}$ = $2-\frac{2}{5}-\frac{3}{5}$ = $\frac{10-2-3}{5}$ = $1$ Hence, the correct answer is 1.
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