If $a+b+c=0$ then $\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}$ is equal to

Q. If three real numbers $a,b,c$ none of which is zero are related by: $a^2=b^2+c^2-2bc\sqrt{1-a^2},b^2=c^2+a^2-2ca\sqrt{1-b^2},c^2=a^2+b^2-2ab\sqrt{1-c^2}$, then prove that $a=c\sqrt{1-b^2}+b\sqrt{1-c^2}$.

Q. If a, b, c $\in R^{+}$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$, $\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ and $-\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ has?

Q. If $a=1$, $b=-1$ and $c=0$, find the value of expression $\left(a^2-b^2\right)+\left(c^2-b^2\right)+\left(a^2-c^2\right)$

Q.

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(c) c² + a² − b² + 2ca = 4s (s − b)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Q.

If a, b, c, d are in G.p., prove that :

(i) $(a^2+b^2),(b^2+c^2),(c^2+d{)}^2$ are in G.P.

(ii) $(a^2-b^2),(b^2-c^2),(c^2-d{)}^2$ are in G.P.

(iii) $\frac{1}{a^2+b^2},\frac{1}{b^2+c^2},\frac{1}{c^2+d^2}$ are in G.P.

(iv) $(a^2+b^2+c^2),(ab+bc+cd),(b^2+c^2+d^2)$