Question

Let $a,b,c$ be the length of the sides of a triangle ABC such that $a\ne 1,b+c\ne 1,c-b\ne 1$ if ${\log }_{b+c}a+{\log }_{c-b}a=2{\log }_{c+b}a{\log }_{c-b}a$ then

• A. ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$
• B. $\tan A+\tan B=1$
• C. $A+B=C$
• D. ${\cos }^{2}A+{\cos }^{2}B=1$

Answer 1

Answer: (D)

${\cos }^{2}A+{\cos }^{2}B=1$

$\log b=c^a+\log c-b^a=2\log c+b^a\log c-b^a$

$\frac{{\log }^a}{\log b+c}+\frac{{\log }^a}{\log c-b}=\frac{2{\log }^a}{\log c+b}.\frac{{\log }^a}{\log c-b}$

$\frac{\log a\log c-b+\log a\log b+c}{\log \left(b+c\right)\log \left(c-b\right)}=\frac{2{\left(\log a\right)}^2}{\log \left(c+b\right)\log \left(c-b\right)}$

$\log a\left(\log \left(c-b\right)+\log \left(b+c\right)-\log a\right)=0$

$a\ne 1so\log a\ne 0$

$\log \frac{\left(c-b\right)\times \left(b+c\right)}{a^2}=0$

$\frac{c^2-b^2}{a^2}=1$

$c^2-b^2=a^2$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{2b^2}{2bc}=\frac{b}{c}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{2a^2}{2ac}=\frac{a}{c}$

${\cos }^{2}A+{\cos }^{2}B=\frac{b^2}{c^2}+\frac{a^2}{c^2}$

$\frac{b^2+a^2}{c^2}$

$\frac{c^2}{c^2}=1$

$\left(D\right)Option$