Question

If $A+B={90}^{}$, $secA=a,cotB=b$ then " $a^2-b^2=0$. "if this is right then true and otherwise false ?

• A. True
• B. False

Answer 1

The correct option is B False

$A+B={90}^{\circ }$

$\Rightarrow A={90}^{\circ }-B$

$\sec A=a$

$\Rightarrow \sec ({90}^{\circ }-B)=a$

$\Rightarrow a=\csc B$

Given $b=\cot B$

Now $a^2-b^2={\csc }^{2}B-{\cot }^{2}B=1\ne 0$

Hence $a^2-b^2=0$ is a false statement for the above values of $a$ and $b$