Question

In a triangle ABC, tan C < 0. Then

• A. (A) tan A . tan B < 1
• B. (B) tan A . tan B > 1
• C. (C) tan A + tan B + tan C < 0
• D. (D) tan A + tan B + tan C > 0

Answer 1

Answer: (A), (D)

(A) tan A . tan B < 1

Answer 2

Explanation:
Here, tan⁡C<0⇒C is obtuse.⇒A and B are acute and A+B<π2 tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B>0⇒1−tan⁡A⋅tan⁡B>0⇒tan⁡Atan⁡B<1tan⁡A+tan⁡B+tan⁡C=(+)(+)(−)<0=tan⁡A⋅tan⁡B⋅tan⁡CinΔABC