Question

Evaluate:

`(sin 40^circ)/(cos 50^circ) + (2  "cosec"  50^circ)/(sec 40^circ) - 3  cos 50^circ  "cosec"  40^circ + 1`

Answer 1

Given: `(sin 40^circ)/(cos 50^circ) + (2  "cosec"  50^circ)/(sec 40^circ) - 3  cos 50^circ  "cosec"  40^circ + 1`.

Step-wise calculation:

1. Note cos 50° = sin 40° and sin 50° = cos 40° since cos (90° – θ) = sin θ.

2. `(sin 40^circ)/(cos 50^circ)`

= `(sin 40^circ)/(sin 40^circ)` 

= 1

3. `(2  "cosec"  50^circ)/(sec 40^circ)`

= `2 xx (1/sin 50^circ)/(1/cos 40^circ)`

= `2 xx ((cos 40^circ)/(sin 50^circ))`

= `2 xx ((cos 40^circ)/(cos 40^circ))`

= 2

4. 3 cos 50° cosec 40°

= `3 xx (cos 50^circ)(1/(sin 40^circ))` 

= `3((sin 40^circ)/(sin 40^circ))` 

= 3

So, –3 cos 50° cosec 40° = –3.

5. Sum:

1 + 2 – 3 + 1 = 1