NTSE Test: Introduction To Trigonometry - Grade 10 MCQ

sin 30° = 1/2,
cos 30°=√3/2,
sin 60°=√3/2,
cos 60°=1/2,
By putting the value of sin 30°, cos 30°, sin 60° and cos 60° in equation

We get=
(sin30°+cos30°)-(sin60°+cos60°)=(1/2+√3/2)-(√3/2+1/2)
=0

The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle.

sin 45° + cos 45°

Hence, the answer is = √2

Let,angle= θ
(asinθ + bcosθ)/(asinθ - bcosθ)
Dividing both numerator and denominator from cosθ
We get,
atanθ +b/atanθ - b
= ( a.a/b + b) /(a.a/b - b) =(a²/b +b)/(a²/b - b)
=(a² + b²/a²- b²) 

6cot+2cosec=cot+5cosec
6cot-cot=5cosec-2cosec
5cot=3cosec
5cos/sin=3/sin
cos=3/5

Correct Answer :- b

Explanation : If θ is one of the acute angles in a triangle, then the sine of theta is the ratio of the opposite side to the hypotenuse, the cosine is the ratio of the adjacent side to the hypotenuse, and the tangent is the ratio of the opposite side to the adjacent side.

9 sec² A - 9 tan² A
= 9( sec² A - tan² A)
= 9 × 1
= 9

Solution :- sin^2 30° − cos^2 30°

= (1/2)² −((3)^1/2/2)²

= 1/4 - 3/4

= -1/2

Tanθ = Perpendicular / Base
We are given that TanA = 3/2
On comparing
Perpendicular = 3
Base = 2
To fing hypotenuse
Hypotenuse² = Perpendicular² + Base²
Hypotenuse² = 3² + 2²
Hypotenuse = 
Hypotenuse = 3.6

Cosθ = Base / Hypotenuse
CosA = 2 / 3.6
Hence the value of Cos A is 2/3.6=2/√13

3 cot θ = 2 ⇒ cot θ = 2/3 ⇒ tan θ = 3/2