Test: Trigonometric Ratios - SAT MCQ

Cos A sec A = 1
Similarly sin A cosec A = 1 and tan A cot A = 1
∴ cos A sec A + sin A cosec A – tan A cot A = 1 + 1 – 1 = 1

Hence, BC = 3k, AC = 4k
Using Pythagoras theorem
AB² = AC² + BC²
AB² = 4k² + 3k²
AB = 5k

tan α = √3 and tan 60° = √3
∴ α = 60°
Cosec β = 1 and cosec 90° = 1
∴ β = 90°
α – β = 60 – 90 = -30°

sin A = 8/17
cos A sec A can be written as
∴ cos A sec A = 1

The value of sin (A + B) = √3/2 and sin 60° = √3/2
∴ A + B = 60     (1)
The value of tan (A – B) = 1 and tan 45° = 1
∴ A – B = 45     (2)
Adding equation (1) and (2)
A + B = 60
+ A – B = 45
– – – – – – – – – – – – –
2 A = 105
A = 52.5
∴ B = 7.5

In this ∆ ABC,

sin30°cos15° + cos30°sin15° = sin⁡45° = 1/√2

Since the triangle is right angles at C,
The sum of the remaining two angles will be 90
∴ cosec(A + B) = Cosec 90° = 1

Consider, two triangles ABC and DEF
In ∆ABC,

From these examples it is evident that the value of the trigonometric ratios depends on their angle and not on their lengths.    

cos 2θ = 2cos θ² – 1
cos θ = 3/4
cos 2θ = 2(3/4)² – 1
= 1/8