In triangle A B C, if |1 a b 1 c a 1 b c|=0 then the value of sin 2 A+ cos 2 B+ tan 2 C is equal to

Question

In triangle A B C, if |1 a b 1 c a 1 b c|=0 then the value of sin 2 A+ cos 2 B+ tan 2 C is equal to (A) 2 (B) 4 (C) (9/2) (D) (11/2)

Tardigrade Admin · Accepted Answer

|0 a-c b-c 0 c-b a-c 1 b c|=0 (a-c)2+(b-c)(b-a)=0 a2+c2-a c+b2-a b-b c=0 (a-b)2+(b-c)2+(c-a)2=0 ∴ a=b=c ⇒ text triangle is equilateral. ∴ sin 2 A+ cos 2 B + tan 2 C ⇒ °=4

Q. In triangle $A BC$ , if $∣ ∣ 111 a c b b a c ∣ ∣ = 0$ then the value of $sin^{2} A + cos^{2} B + tan^{2} C$ is equal to

2

4

$\frac{9}{2}$

$\frac{11}{2}$

$∣ ∣ 001 a - c c - b b b - c a - c c ∣ ∣ = 0$
$(a - c)^{2} + (b - c) (b - a) = 0$
$a^{2} + c^{2} - a c + b^{2} - ab - b c = 0$
$(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$
$∴ a = b = c \Rightarrow triangle is equilateral.$
$∴ sin^{2} A + cos^{2} B + tan^{2} C$
$\Rightarrow^{\circ} = 4$

Q. In triangle ABC, if ∣∣​111​acb​bac​∣∣​=0 then the value of sin2A+cos2B+tan2C is equal to

2

4

29​

211​

∣∣​001​a−cc−bb​b−ca−cc​∣∣​=0 (a−c)2+(b−c)(b−a)=0 a2+c2−ac+b2−ab−bc=0 (a−b)2+(b−c)2+(c−a)2=0 ∴a=b=c⇒ triangle is equilateral. ∴sin2A+cos2B+tan2C ⇒∘=4

Q. In triangle $A BC$ , if $∣ ∣ 111 a c b b a c ∣ ∣ = 0$ then the value of $sin^{2} A + cos^{2} B + tan^{2} C$ is equal to

$\frac{9}{2}$

$\frac{11}{2}$

$∣ ∣ 001 a - c c - b b b - c a - c c ∣ ∣ = 0$
$(a - c)^{2} + (b - c) (b - a) = 0$
$a^{2} + c^{2} - a c + b^{2} - ab - b c = 0$
$(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$
$∴ a = b = c \Rightarrow triangle is equilateral.$
$∴ sin^{2} A + cos^{2} B + tan^{2} C$
$\Rightarrow^{\circ} = 4$