Question:
If in ΔABC, ∠C = 90°. Then, the value of cos(A + B) is
(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) $\frac{3}{5}$
Solution:
In ∆ABC,
∠A + ∠B + ∠C = 180º (Angle sum property of triangle)
⇒ ∠A + ∠B + 90º = 180º (∠C = 90º)
⇒ ∠A + ∠B = 180º − 90º = 90º
∴ cos(A + B) = cos90º = 0
Thus, the value of cos(A + B) is 0.
Hence, the correct answer is option (a).
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