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

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-1

not defined

answer is C.

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Detailed Solution

Given that, triangle ABC is right-angled at C i.e., ∠C = 90°.
Let us assume below is the given triangle ABC.
Question Image

We know that, according to the Angle sum property of the triangle, the sum of the interior angles of a triangle is 180°.
So,

∠ A + ∠ B + ∠ C = 180 ° ∠ A + ∠ B + 90 ° = 180 ° ∠ A + ∠ B = 180 ° − 90 ° ∠ A + ∠ B = 90 ° ............ (1)

Apply cosine on both sides of the equation (1). We get,

cos (A + B) = cos 90 ° cos (A + B) = 0 … … (2) cos 90 ° = 0

We know that the reciprocal of cos is sec. So,

sec (A + B) = 1 cos (A + B) … … (3) sec A = 1 cos A

By substituting the value of eq 2 in eq 3. We get,

sec (A + B) = 10 sec (A + B) = ∞ [not defined]

Hence, sec (A + B) is not defined.
So, the correct option is (3).

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