# If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, then find A and B

Trigonometry is basically the study of the relationship between the angles and the sides of a triangle. It is one of the widely used topics of mathematics that is used in daily life. It involves operations on a right-angled triangle i.e. a triangle having one of the angles equal to 90°. There are some terms that we should know before going further. These terms are,

- Hypotenuse – It is the side opposite to the right angle in a right-angled triangle. It is the longest side of a right-angled triangle. In figure1 side, AC is the hypotenuse.
- Perpendicular – The perpendicular of a triangle, corresponding to a particularly acute angle θ is the side opposite to the angle θ. In figure1 side, AB is the perpendicular corresponding to angle θ.
- Base – It is the side adjacent to a particularly acute angle θ. In figure 1 side BC is the base corresponding to angle θ.

### Trigonometric Functions

As earlier said, trigonometry depicts the relationship between the angles and sides of a right-angled triangle. This relationship is represented by standard ratios and is given as follows:

**Sine (sin)**The sine of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the hypotenuse of the triangle.

sin θ = perpendicular/hypotenuse =p/h

**Cosine (cos)**The cosine of an angle θ is the ratio of the length of the base, corresponding to the angle θ, to the length of the hypotenuse of the triangle.

cos θ = base/hypotenuse=b/h

**Tangent (tan)**The tangent of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the base for the particular angle of the triangle.

tan θ = perpendicular/base=p/b

**Cotangent (cot)**It is the reciprocal of tangent.

cot θ = 1/tan θ = base/perpendicular=b/p

**Secant (sec)**It is the reciprocal of cosine.

sec θ = 1/cos θ = hypotenuse/base=h/b

**Cosecant (cosec)**It is the reciprocal of sine.

cosec θ =1/sin θ = hypotenuse/perpendicular=h/p

Some of the trigonometric ratios along with some of the standard angles are given in the table below,

0^{°} | 30° | 45° | 60° | 90° | |

sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |

cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |

tan | 0 | 1/√3 | 1 | √3 | ∞ |

cot | ∞ | √3 | 1 | 1/√3 | 0 |

sec | 1 | 2/√3 | √2 | 2 | ∞ |

cosec | ∞ | 2 | √2 | 2/√3 | 1 |

### If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, then find A and B

**Solution:**

Given,

tan(A + B) = √3

Since, A + B ≤ 90°,

Therefore,

A ≤ 90° and B ≤ 90°

According to the table, the angle at which tan acquires the value √3, is at 60°.

Therefore,

tan(A + B) = √3 = tan 60°

tan(A + B) = tan 60°

A + B = 60° ⇢ (i)

tan(A – B) = 1/√3

According to the table, the angle at which tan acquires the value 1/√3, is at 30°. Therefore,

tan(A – B) = 1/√3 = tan 30°

tan(A – B) = tan 30°

A – B = 30° ⇢ (ii)

Adding equation (i) and (ii),

2A = 90°

A = 45°

Putting the value of A in equation (i),

45° + B = 60°

B = 60°- 45°

B = 15°

Therefore, the value of A and B that satisfy the given equation is 45° and 15°, respectively.

### Similar Problems

**Question 1: If 2 sin ^{2} θ – 1= 0, and 0°< θ< 90° then find the values of the following**

**a. cos θ + cos 2θ**

**b. sin θ × sin 2θ**

**Solution:**

2 sin

^{2}θ – 1 = 02sin

^{2}θ = 1sin

^{2 }θ = 1/2sin θ = 1/√2

The acute angle for which the value of sin is 1/√2 = 45°. Therefore,

sin θ = 1/√2 = sin 45°

θ = 45°

Therefore,

a. cos θ + cos 2θcos 45° + cos 2.45°

cos 45° + cos 90° (putting the values from the table)

1/√2 + 0 = 1/√2

b. sin θ × sin 2θsin 45° × sin 2.45°

sin 45° × sin 90°

1/√2 × 1

1/√2

**Question 2: Find A and B if sin(A – B) = 1/2 = cos(A + B) and A, B are acute angles.**

**Solution:**

sin(A – B) = 1/2

Since A, B are acute angles,

Therefore A – B should also be acute angle

The acute angle for which the value of sin is 1/2 = 30°. Therefore,

sin(A – B) = sin 30° = 1/2

A – B = 30° ⇢ (i)

The acute angle for which the value of cos is 1/2 = 60°, therefore,

cos(A + B) = cos 60° = 1/2

A + B = 60° ⇢ (ii)

Adding equations (i) & (ii),

2A = 90°

A = 45°

Putting the value of A in equation (i), we get

45° – B = 30°

B = 45° – 30°

B = 15°

**Question 3: Find 2 tan ^{2} θ + cos^{2 }θ – 1, if sin θ = cos θ, where 0°< θ< 90°.**

**Solution:**

The acute angle for which the value of cos and sin are equal =45°, therefore,

cos θ = sin θ = cos 45°

θ = 45°

Therefore,

2 tan

^{2}θ + cos^{2}θ -1= 2 tan 45° × tan 45° + cos 45° × cos 45° -1

= 2 × 1 × 1 + 1/√2 × 1/√2 -1

= 2 + 1/2 -1

= 3/2

**Question 4: Find θ for 0°<θ <90°, where (2sin θ – 1)(sin θ – 2)=0.**

**Solution:**

(2sin θ – 1)(sin θ – 2) = 0

By looking at the equation, to satisfy the equation, either

(2sin θ – 1) = 0

or, (sin θ – 2) = 0

sin θ = 1/2

sin θ = 2 (since sin cannot exceed 1, therefore sin θ – 2 cannot be 0.)

sin θ = 1/2 = sin 30°

θ =30°