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

#### The minimum value of tan^{2}α + cot^{2}α is ____.

At α π/4 + tan^{2} α + cot^{2} α has minimum value. Tan^{2} π/4 + cot^{2} π/4 = tan^{2} 45^{0} + cot^{2} 45^{0} = 1 + 1 = 2 Hence, the minimum value of tan^{2} α + cot^{2} α is 2.

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

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.

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

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

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

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

#### Choose the following correct option.

sin(90^{0} – θ) =cos θ

#### The value of sec^{4}A − tan^{4}A is ____.

sec^{4}A − tan^{4}A = (sec^{2} A)^{2} – (tan^{2} A)^{2} = (sec^{2} A - tan^{2} A)( sec^{2} A + tan^{2} A) = (1) ( sec^{2} A + tan^{2} A) (∵sec^{2} A - tan^{2} A = 1) = sec^{2} A + tan^{2} A ∵ sec^{4}A − tan^{4}A = sec^{2} A + tan^{2} A

#### The value of sin^{4} θ – cos^{4} θ =____.

sin^{2}θ – cos^{2} θ = (sin^{2}θ)^{2} – (cos^{2} θ)^{2} = (sin^{2}θ + cos^{2} θ)( sin^{2}θ – cos^{2} θ) = (1)( sin^{2}θ – cos^{2} θ)( ∵ sin^{2}θ + cos^{2} θ = 1) = sin^{2}θ – cos^{2} θ Hence, the value of sin^{4} θ – cos^{4} θ is sin^{2}θ – cos^{2} θ.

#### tan 25° tan 45° tan 65° = ____.

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

#### sin 27^{0}/cos 63^{0}= ____.

We have, sin 27^{0}/cos 63^{0} = sin(90^{0} – 63^{0})/cos 63^{0} = cos 63^{0}/cos 63^{0} (∵ sin(90°−θ) = cosθ) = 1 sin 27^{0}/cos 63^{0}= 1