Question

Evaluate WITHOUT using trignometric table:
Question is in the attached image. ​

Answer 1

Answer:

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Answer 2

Answer:

(Sin 18°/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°) = 2

Step-by-step explanation:

We have,

(Sin 18°/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°)

Now, we know that,

Sin A = Cos (90° - A)

Tan A = Cot (90° - A)

and Cot A = (1/Tan A)

So, solving,

(Sin 18°/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°)

= (Cos (90° - 18°)/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°)

= (Cos 72°/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°)

= 1 + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°)

Similarly,

= 1 + √3(Cot (90° - 10°) × Tan 30° × Cot (90° - 40°) × Tan 50° × Tan 80°)

= 1 + √3(Cot 80° × Tan 30° × Cot 50° × Tan 50° × Tan 80°)

= 1 + √3(Cot 80° × Tan 80° × Cot 50° × Tan 50° × Tan 30°)

= 1 + √3((1/Tan 80°) × Tan 80° × (1/Tan 50°) × Tan 50° × Tan 30°)

= 1 + √3(1 × 1 × Tan 30°)

We know that,

Tan 30° = (1/√3)

So,

= 1 + √3(1 × 1 × (1/√3))

= 1 + √3(1/√3)

= 1 + 1

= 2

Hence,

(Sin 18°/Cos 72°) + √3(Tan 10° × Tan 30° × Tan 40° × Tan 50° × Tan 80°) = 2

Hope it helped and believing you understood it........All the best