Question

9. If ΔABC is right angled at C, then the value of cos (A + B) is (a) 0 (6) 1 (c)1/2 (d) √3/2 ​

Answer 1

Answer:

a) 0

Step-by-step explanation:

Given that a ∆ABC is right angled at C. We need to find out the value of cos (A + B).

Since, in ∆ABC right angles at C means the value of C is 90°. And we are aware about the fact that the sum of all sides of the triangle is 180°.

So, we can write like-

∠A + ∠B + ∠C = 180°

∠A + ∠B + 90° = 180°

∠A + ∠B = 180° - 90°

∠A + ∠B = 90°

As per given question we need to find the value of cos (A + B). So, multiply by cos on both sides,

→ cos (∠A + ∠B) = cos 90°

→ cos (∠A + ∠B) = 0

Hence, the value of cos (A + B) is 0.

TRIGONOMETRY FORMULAS:

secØ = 1/cosØ,                    cosØ = 1/secØ

cosecØ = 1/sinØ,                 sinØ = 1/cosecØ

tanØ = sinØ/cosØ,              cotØ = cosØ/sinØ

cotØ = 1/tanØ,                      tanØ = 1/cotØ    

sin²Ø + cos²Ø = 1

tan²Ø + 1 = sec²Ø

1 + cot²Ø = cosec²Ø

sinØ = cos(90° - Ø)

cosØ = sin(90° - Ø)

tanØ = cot(90° - Ø)

Answer 2

Step-by-step explanation:

$\sf \: given \: \sf \angle \: C \: = 90 \degree \\ \\ \sf \: now \: by \: angle \: sum \: property \: in \: \triangle \: abc \\ \sf \angle \: A + B + C = 180 \degree \\ \sf\angle \: A + B + 90 \degree \: = 180 \degree \\ \sf \: \angle \: A + B = 180 \degree - 90 \degree \\ \angle \: A + \angle \: B = 90 \degree$

$\sf \: cos \: ( A+ B) = cos \: 90 \degree \\ \sf \: = 0 \\ \\ \sf \red{so \: the \: correct \: answer \: is \: (a)}$