# CBSE Class 10th Math 8 – Introduction to Trigonometry MCQs

#### The value of sec4A − tan4A is ____.

Correct! Wrong!

sec4A − tan4A = (sec2 A)2 – (tan2 A)2 = (sec2 A - tan2 A)( sec2 A + tan2 A) = (1) ( sec2 A + tan2 A) (∵sec2 A - tan2 A = 1) = sec2 A + tan2 A ∵ sec4A − tan4A = sec2 A + tan2 A

#### tan 25° tan 45° tan 65° = ____.

Correct! Wrong!

We have, tan 25° tan 45° tan 65° = tan 25° × 1 × tan (90−25)° = tan 25° cot 25°(∵ tan190° − θ) = cotθ) = tan 25° × 1 tan 25° = 1 ∴ tan 25° tan 45° tan 65° = 1

#### The value of sin4 θ – cos4 θ =____.

Correct! Wrong!

sin2θ – cos2 θ = (sin2θ)2 – (cos2 θ)2 = (sin2θ + cos2 θ)( sin2θ – cos2 θ) = (1)( sin2θ – cos2 θ)( ∵ sin2θ + cos2 θ = 1) = sin2θ – cos2 θ Hence, the value of sin4 θ – cos4 θ is sin2θ – cos2 θ.

#### If sin(θ + 24°) = cosθ, where (θ + 24°) is an acute angle, then θ = ____.

Correct! Wrong!

Given sin(θ + 24°) = cosθ ⇒sin(θ + 24) = sin(90°−θ) ⇒θ + 24° = 90°−θ ⇒2θ = 66° ∴ θ = 33°

#### The minimum value of tan2α + cot2α is ____.

Correct! Wrong!

At α π/4 + tan2 α + cot2 α has minimum value. Tan2 π/4 + cot2 π/4 = tan2 450 + cot2 450 = 1 + 1  = 2 Hence, the minimum value of tan2 α + cot2 α is 2.

#### Choose the following correct option.

Correct! Wrong!

sin(900 – θ) =cos θ

#### If cos(81° + θ) = sin (9° + θ), then θ = ____.

Correct! Wrong!

We have, cos(81° + θ) = sin (9° + θ) ⇒cos(81° + θ) = cos(90° − 9° − θ) ⇒81° + θ = 81° − θ ⇒2θ = 0 ∴ θ = 0

#### If A and B are acute angles and sin A = cos B, then A + B is ____.

Correct! Wrong!

Given, sin A = cos B ⇒sin A = sin(90°−B) ⇒A = 90°−B ∴ A +B = 90°

#### tan 11° tan 31° tan 45° tan 79° tan 59° = ____.

Correct! Wrong!

We have, tan 11° tan 31° tan 45° tan 79° tan 59° = tan 11° tan 31° × 1 × tan(90°−11°) tan(90°−31°) = tan 11° tan 31° × 1 × cot 11° cot 31° (∵ tan(90°−θ) = cotθ) = 1 (∵ tanθ = 1/cotθ) ∴ tan 11° tan 31° tan 45° tan 79° tan 59° = 1.

#### sin 270/cos 630= ____.

Correct! Wrong!

We have, sin 270/cos 630 = sin(900 – 630)/cos 630 = cos 630/cos 630 (∵ sin(90°−θ) = cosθ) = 1 sin 270/cos 630= 1

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