If triangle $ABC$ is right angled at $C$, then find the value of $sec ( A+B)$.

Given: Triangle $ABC$ is right angled at $C$, 

To do: To find the value of $sec ( A+B)$.

Solution:

As given, triangle $ABC$ is right angled at $C$.

$\because \vartriangle ABC$ is right angled at $C$.

$\therefore \angle A+\angle B+\angle C=180^o$

$\Rightarrow \angle A+\angle B+90^o=180^o$

$\Rightarrow \angle A+\angle B=180^o-90^o$

$\Rightarrow \angle A+\angle B=90^o$

$\therefore cos( A+B)=cos90^o=0$

$\Rightarrow sec( A+B)=\frac{1}{cos( A+B)}=\frac{1}{0}=$ Not defined

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Updated on: 10-Oct-2022

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