Question

Evaluate:

`(tan 70^circ)/(cot 20^circ) + (sin 38^circ)/(cos 52^circ) - 4 sin^2 30^circ`

Answer 1

Given: `(tan 70^circ)/(cot 20^circ) + (sin 38^circ)/(cos 52^circ) - 4 sin^2 30^circ`.

Step-wise calculation:

1. `(tan 70^circ)/(cot 20^circ) = tan 70^circ ·  tan 20^circ` 

Since `cot 20^circ = 1/(tan 20^circ)`. 

But tan 70° = tan (90° – 20°)

= cot 20°

= `1/(tan 20^circ)`

So, tan 70° · tan 20° = 1.

2. `(sin 38^circ)/(cos 52^circ) = (sin 38^circ)/(cos 52^circ)`. 

Since cos 52° = sin (90° – 52°)

= sin 38°

This ratio = 1.

3. `4 sin^2 30^circ = 4 xx (1/2)^2`

= `4 xx 1/4` 

= 1

Combine:

1 + 1 – 1 = 1