Question

tan10°×tan20°×tan30°••••×tan70°×tan80°

Answer 1

= tan10.tan20.tan30.tan40.tan(90–40).tan(90–30).tan(90–20). tan(90–10) ( 90=90°)

= tan10.tan20.tan30.tan40.cot40.cot30.cot20.cot10. { as tan(90-x)=cotx}

= tan10.cot10.tan20.cot20.tan30.cot30.tan40.cot40

= 1*1*1*1=1 (we know tanx.cotx=1)

Answer 2

Answer:

$\tan(10°) \times \tan(20°) \times \tan(30°) \times \\ \tan(40°) \times \tan(50°) \times \tan(60°) \times \\ \tan(70°) \times \tan(80°) =1$

Step-by-step explanation:

We know that from complementary formula

$tan(90° - x) = \cot(x) \\ \\ and \: cot(90° - x) = tan \: (x)$

we also know that

$tan \: x \times cot \: x = 1$

here in the given question

$\tan(10°) \times \tan(20°) \times \tan(30°) \times \\ \tan(40°) \times \tan(50°) \times \tan(60°) \times \\ \tan(70°) \times \tan(80°) \\ \\ tan(90° - 80°) \times \tan(90° - 70°) \times \\ \tan(90° - 60°) \times \tan(90° - 50°) \\ \times \tan(50°) \times \tan(60°) \times \tan(70°) \\ \times \tan(80°) \\ \\ \cot(80°) \times \cot(70°) \times \cot(60°) \\ \times \cot(50°) \times \tan(50°) \times \tan(60°) \times \\ \tan(70°) \times \tan(80°) \\ \\ \cot(80°) \times \tan(80°) \\ \times \cot(70°) \times \tan(70°) \times \\ \cot(60°) \times \tan(60°) \times \\ \cot(50°) \times \tan(50°) \\ \\ = 1 \times 1 \times 1 \times 1 \\ \\ = 1$

Hope it helps you.