Solution: Given,tan (A + B) = 3As we know, tan 60° = 3Thus, we can write;tan (A + B) = tan 60°(A + B) = 60° …… (i)Now again given;tan (A – B) = 1/3Since, tan 30° = 1/3Thus, we can write;tan (A – B) = tan 30°(A – B) = 30° ….. (ii)Adding the equation (i) and (ii), we get;A + B + A – B = 60° + 30°2A = 90°A= 45°Now, put the value of A in eq. (i) to find the value of B;45° + B = 60°B = 60° – 45°B = 15°Therefore A = 45° and B = 15°Question 8: Show that :(i) tan 48° tan 23° tan 42° tan 67° = 1(ii) cos 38° cos 52° – sin 38° sin 52° = 0Solution:(i) tan 48° tan 23° tan 42° tan 67°We can also write the above given tan functions in terms of cot functions, such as;tan 48° = tan (90° – 42°) = cot 42°tan 23° = tan (90° – 67°) = cot 67°Hence, substituting these values, we get= cot 42° cot 67° tan 42° tan 67°= (cot 42° tan 42°) (cot 67° tan 67°)= 1 × 1 [since cot A.tan A = 1]= 1(ii) cos 38° cos 52° – sin 38° sin 52°We can also write the given cos functions in terms of sin functions.cos 38° = cos (90° – 52°) = sin 52°cos 52°= cos (90° – 38°) = sin 38°Hence, putting these values in the given equation, we get;sin 52° sin 38° – sin 38° sin 52° = 0