Question

Evaluate:

`(4  cos 57^circ)/(sin 33^circ) - (sqrt(3) · cos 38^circ "cosec"  52^circ)/(tan 20^circ tan 60^circ tan 70^circ)`

Answer 1

Given: `(4  cos 57^circ)/(sin 33^circ) - (sqrt(3) · cos 38^circ "cosec"  52^circ)/(tan 20^circ tan 60^circ tan 70^circ)`.

Step-wise calculation:

1. sin 33° = cos 57°

Since 33° + 57° = 90°.

So, `(4  cos 57^circ)/(sin 33^circ)`

= `(4  cos 57^circ)/(cos 57^circ)`

= 4

2. cosec 52° = `1/(sin 52^circ)` and sin 52° = cos 38°

Since 52° + 38° = 90°. 

Hence, `cos 38^circ ·  "cosec"  52^circ`

= `cos 38^circ · 1/(cos 38^circ)`

= 1 

So, the numerator of the second term is `sqrt(3)`.

3. tan 70° = cot 20°

= `1/(tan 20^circ)`

70° + 20° = 90° 

So, tan 20° · tan 70° = 1. 

Therefore, tan 20° · tan 60° · tan 70°

= 1 · tan 60° 

= tan 60° 

= `sqrt(3)`

4. Thus, the second term = `sqrt(3)/sqrt(3)` = 1.

5. Whole expression = 4 – 1 = 3.

Value = 3.