Question

(1+cot78)(1+cot57) is equal to​

Answer 1

Step-by-step explanation:

(1+cot78)(1+cot57)

Now,cot (90−θ)=tanθ

∴cot(78)=tan12

cot(57)=tan33

(1+ tan 12) (1+ tan 33)

=1+tan12+tan33+tan12.tan33 ... (1)

Now,

45 = 12 + 33

tan45=tan(12+33)

1=

1−tan12.tan33

tan12+tan33

∴ 1-tan12.tan33=tan12+tan33

∴1 = tan12.tan33+tan12+tan33

Thus, substituting above result in eq (1)

= 1 + 1

= 2