Question

If in a $\Delta$ ABC, $\angle C={90}^0$ then value of tan (A-B) is

• A. $\frac{a^2+b^2}{2ab}$
• B. $\frac{|a^2-b^2|}{2ab}$
• C. $\frac{c^4}{2ab}$
• D. $\frac{ca}{2b}$

Answer 1

Answer: (B)

$\frac{|a^2-b^2|}{2ab}$

$\sec (A-B)=\frac{1}{cos(A-B)}$

$\frac{1}{cosAcosB+sinAsinB}=\frac{1}{\frac{b}{a}.\frac{a}{c}+\frac{a}{c},\frac{b}{c}}=\frac{c^2}{2ab}$

Now $\tan (A-B)=\sqrt{se{c}^2(A-B)-1}$

$\sqrt{\frac{c^4}{4a^2{b}^2}-1}=\sqrt{\frac{c^4-4a^2{b}^2}{4a^2{b}^2}}=\frac{|a^2-b^2|}{2ab}$

(using $c^2=a^2+b^2$)