ABCD is a rectangle of which AC is a diagonal. The value of (tan2 ∠CAD + 1)sin2 ∠BAC is?

= (tan2α + 1 )sin2β = (tan245° + 1 )sin245° = (1 + 1) ( 1sqrt 2 )^2 = 2 12 = 1

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This question belongs to Arithmetic Ability Trigonometry

Trigonometry

?

ABCD is a rectangle of which AC is a diagonal. The value of (tan² ∠CAD + 1)sin² ∠BAC is?

A) 2 B) $$\frac{1}{4}$$ C) 1 D) 0

Answer & Solution

Correct Answer: Option C

= (tan²α + 1 )sin²β
= (tan²45° + 1 )sin²45°
$$ = (1 + 1) {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$$
$$ = 2 \times \frac{1}{2} = 1$$

Question posted by Examveda

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Habibur Rahaman 2 years ago

Solution is wrong , because two angles not equal

Mohit Kumar 6 years ago

WHY ALPHA AND BEETA EQUAL TO 45

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