Question

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

Answer 1

Answer:

A+∠B+∠C=180

∘

....1

(angle sum property)

then it is given that \angle C = 90^\circ∠C=90

∘

putting the value of C in equation 1

we get

\begin{gathered}\angle A + \angle B + 90^\circ= 180^\circ\\\angle A + \angle B= 180^\circ -90^\circ\\\angle A + \angle B= 90^\circ\\\end{gathered}

∠A+∠B+90

∘

=180

∘

∠A+∠B=180

∘

−90

∘

∠A+∠B=90

∘

now ,

sec (A+B) = sec 90°

sec 90° = \frac{1}{cos 90}

cos90

1

= \frac{1}{0}

0

1

= not defined

hence ,

The value of sec(A+B) is not defined

hope it helps you.