**Frequently Asked Questions on Chapter 8 – Introduction to Trigonometry**

Q.1(#Introduction To Trigonometry)

**If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B?**

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/AC = BD/BC

Let take a constant value

AD/AC = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = K BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD^{2}=BC^{2}-BD^{2}…(3)

CD^{2}=AC^{2}-AD^{2}..(4)

From the equations (3) and (4) we get,

AC^{2}-AD^{2}=BC^{2}-BD^{2}

Now substitute the equations (1) and (2) in (3) and (4)

k^{2}(BC-BD^{2})=(BC^{2}-BD^{2})K^{2}=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

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

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

Q.2(#Introduction To Trigonometry)

**Evaluate: sin 18°/cos 72°?**

**(i) sin 18°/cos 72°**

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

Q.3(#Introduction To Trigonometry)

**Show that : tan 48° tan 23° tan 42° tan 67° = 1?**

tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

Q.4(#Introduction To Trigonometry)

**If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A?**

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

Q.5(#Introduction To Trigonometry)

**State whether the statement is true or false. Justify your answer. The value of tan A is always less than 1**

**The given statement, value of tan A is always less than 1 is False.**

Proof:

△ABC, in which ∠B = 90,

AB = 3, BC = 4 and AC = 5

Value of tan A = ¾ which is greater than 1.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5, as it follows the pythagoras theorem.

AC^2 = AB^2 + BC^2

5^2 = 3^2 + 4^2

25 = 9 + 16

25 = 25

Q.6(#Introduction To Trigonometry)

**Find the value of 2tan 30°/1+tan^2 30°**

Given:

2tan 30°/1 + tan^2 30°

tan 30 = 1/√3

Now substitute the value in the above equation, we get

2tan 30°/1 + tan^2 30° = [2(1/√3)] / [1 + (1/√3)^2]

= [2(1/√3)] / [1 + 1/3]

= (2/√3) / (4/3)

= 6/4√3

= √3/2

= sin 60°

Q.7(#Introduction To Trigonometry)

**Evaluate: sin 18°/cos 72°**

To simplify this, convert the sin function into cos function

We know that 18° is written as 90° – 18°, which is equal to the cos 72°.

sin 18°/cos 72° = sin (90° – 18°) /cos 72°

Substitute the value, we get

= cos 72° /cos 72°

= 1