You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of n Terms

If A, B, ...

Question

If $A$ , $B$ , $C$ be the angles of a triangle and
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + sin A & 1 + sin B & 1 + sin C sin A + {sin}^{2} A & sin B + {sin}^{2} B & sin C + {sin}^{2} C \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
then prove that $Δ$ must be isosceles.

Open in App

Solution

Apply $R_{2} - R_{1}$ and then $R_{3} - R_{2}$ (new).
$Δ = (sin A - sin B) (sin B - sin C) (sin C - sin A) = 0$
Hence $Δ$ is isosceles as in part $(b)$ .

Suggest Corrections

Similar questions

A, B

C

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + sin A & 1 + sin B & 1 + sin C sin A + {sin}^{2} A & sin B + {sin}^{2} B & sin C + {sin}^{2} C \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + sin A & 1 + sin B & 1 + sin C sin A + {sin}^{2} A & sin B + {sin}^{2} B & sin C + {sin}^{2} C \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + s i n A & 1 + s i n B & 1 + s i n C s i n A + s i n^{2} A & s i n B + s i n^{2} B & s i n C + s i n^{2} C \end{matrix} ∣ ∣ ∣ ∣ = 0

A, B a n d C

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + sin A & 1 + sin B & 1 + sin C sin A + {sin}^{2} A & sin B + {sin}^{2} B & sin C + {sin}^{2} C \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + s i n A & 1 + s i n B & 1 + s i n C s i n A + s i n^{2} A & s i n B + s i n^{2} B & s i n C + s i n^{2} C \end{matrix} ∣ ∣ ∣ ∣

a = 1 + 2 + 4 + - - -

n

b = 1 + 3 + 9 + - - -

n

c = 1 + 5 + 25 + - - - -

n

Δ ∣ ∣ ∣ ∣ \begin{matrix} a & 2 b & 4 c 2 & 2 & 2 2^{n} & 3^{n} & 5^{n} \end{matrix} ∣ ∣ ∣ ∣

Join BYJU'S Learning Program