Question

Evaluate:

`((3tan41^circ)/(cot49^circ))^2 - ((sin35^circ sec55^circ)/(tan10^circ tan20^circ tan 60^circ tan 70^circ tan 80^circ))^2`

Answer 1

Given: `((3tan41^circ)/(cot49^circ))^2 - ((sin35^circ sec55^circ)/(tan10^circ tan20^circ tan 60^circ tan 70^circ tan 80^circ))^2`.

Step-wise calculation:

1. cot 49° = tan(90° – 49°)

= tan 41°

so `(3 tan 41^circ)/(cot 49^circ) = 3`.

Squaring: (3)² = 9.

2. `sec 55^circ = 1/cos 55^circ` and cos 55° = sin(90° – 55°) = sin 35° 

So `sin 35^circ·sec 55^circ = sin 35^circ · (1/sin 35^circ) = 1`. 

Thus the second fraction becomes `1/(tan 10^circ tan 20^circ tan 60^circ tan 70^circ tan 80^circ)`.

3. Use complementary-angle cotangent identities:

`tan 70^circ = cot 20^circ = 1/tan 20^circ`

`tan 80^circ = cot 10^circ = 1/tan 10^circ`

Therefore the product tan 10°·tan 20°·tan 60°·tan 70°·tan 80°

= `tan 10^circ·tan 20^circ·tan 60^circ·(1/tan 20^circ)·(1/tan 10^circ)` 

= tan 60°

= `sqrt(3)`

4. So the second term is `(1/sqrt(3))^2 = 1/3`.

5. Finally: `9 - 1/3 = (27/3 - 1/3)`

= `26/3`