Question

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

Answer 1

Answer:

Not defined

Step-by-step explanation:

We know that angle c is 90 degree

Using angle sum property we will get

Angle a +angle b =90

Therefore sec (A+B)=sec (90)

Amd the value of sec 90 is not defined

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Answer 2

The value of  sec(A+B) is not defined

Step-by-step explanation:

in triangle ABC

$\angle A + \angle B + \angle C = 180^\circ$....1

(angle sum property)

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

putting the value of C in equation 1

we get

$\angle A + \angle B + 90^\circ= 180^\circ\\\angle A + \angle B= 180^\circ -90^\circ\\\angle A + \angle B= 90^\circ$

now ,

sec (A+B) = sec 90°

sec 90° = $\frac{1}{cos 90}$ = $\frac{1}{0}$ = not defined

hence ,

The value of  sec(A+B) is not defined

#Learn more:

If a+b+c = 5 , ab+bc+ca=10 and abc=3 then find the vaue of a^3+b^3+c^3

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