Question

If cosec (A + B) = 1 and cosec (A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of:

(i) sin A cos B + cos A sin B

(ii) `(tan A - tan B)/(1 + tan A tan B)`

Answer 1

Given:

cosec (A + B) = 1

cosec (A – B) = 2

0° < (A + B) ≤ 90° and A > B

Step-wise calculation:

1. cosec (A + B) = 1 

⇒ sin (A + B) = 1

Since 0° < (A + B) ≤ 90°, the only possibility is A + B = 90°.

2. cosec (A – B) = 2

⇒ `sin (A - B) = 1/2`

With A > B and A + B = 90°, we have 0° < A – B < 90°, so A – B = 30°.

3. Solve for A and B:

`A = ((A + B) + (A - B))/2`

= `(90^circ + 30^circ)/2`

= 60°

`B = ((A + B) - (A - B))/2`

= `(90^circ - 30^circ)/2`

= 30°

(i) sin A cos B + cos A sin B = sin (A + B) 

= sin 90°

= 1

Check with values: sin 60 · cos 30 + cos 60 · sin 30 

= `(sqrt(3)/2)(sqrt(3)/2) + (1/2)(1/2)`

= `3/4 + 1/4`

= 1

(ii) `(tan A - tan B)/(1 + tan A tan B) = tan (A - B)`

= tan 30°

= `1/sqrt(3)` 

= `sqrt(3)/3`