Introduction to Trigonometry

Solution NCERT Exercise 8.1 part2

Question (4) Given 15 cot A = 8, find sin A and sec A.

Solution

Let, ABC is a triangle, in which, ∠ B = 90^o

Given 15 cot A = 8

In the given image of the right angle triangle, the acute angle A is being considered.

For the acute angle A, the side opposite to the ∠ A is the perpendicular (p)

The side adjacent to the ∠ A is the base (b)

And, the side opposite to the right angle B is the hypotenuse (h)

Here, in the given triangle ABC

Base (b) = AB

Perpendicular (p) = BC

And, the side AC = hypotenuse (p)

∴ cot A = ⁸/₁₅

we know that

cot A = ^{Base (b)}/_{Perpendicular(p)}

⇒ cot A = ⁸/₁₅ = ^Base(b)/_{Perpendicular(p)}

∴ b (base) = AB = 8 k and

p (perpendicular) = BC = 15 k

where k is a positive integer,

Calculation of hypotenuse

Now, According to Pythagoras theorem, we know that

(Hypotenuse)² = (Perpendicular)² + (Base)²

Thus, after substituting the values of perpendicular (p) and base (b) in the above expressions, we get

⇒ AC² = (15k)² +(8k)²

⇒ AC²=225k² + 64k²

⇒ AC²=289k²

∴ AC = 289 k²

⇒ AC = 17k = hypotenuse (h)

Calculation of the trigonometric ratio for sin A

Now, we know that

sin A = ^{Perpendicular (p)}/_{Hypotenuse(h)}

After substituting values of 'Perpendicular(p)' and 'Hypotenuse(h)' in the above expression, we get

sin A = ^{15 k}/_{17 k}

⇒ sin A = ¹⁵/₁₇

Calculation of the trigonometric ratio for sec A

Again we know that

sec A = ^{Hypotenuse(h)}/_Base(b)

After substituting values of 'hypotenuse(h)' and 'base(b)' in the above expression, we get

⇒ sec A = ^{17 k}/_{8 k}

⇒ sec A = ¹⁷/₈

Thus, Sin A = ¹⁵/₁₇ and sec A = ¹⁷/₈ Answer

Alternate Method to solve the question "Given 15 cot A = 8, find sin A and sec A"

Given, 15 cot A = 8

∴ cot A = ⁸/₁₅

⇒ ¹/_{tan A} = ⁸/₁₅

⇒ tan A = ¹⁵/₈ - - - - -(i)

Calculation of trigonometric ratio of sec A from tan A

Now, we know that

sec² A = 1 + tan² A

After substituting value of tan A from equation (i), we get

sec² A = 1 + (¹⁵/₈)²

⇒ sec² A = 1 + ²²⁵/₆₄

⇒ sec² A = ^{64 + 225}/₆₄

⇒ sec² A = ²⁸⁹/₆₄

⇒ sec A = ²⁸⁹/₆₄

⇒ sec A = ¹⁷/₈

Calculation of trigonometric ratio of sin A from sec A

we have sec A = ¹⁷/₈

⇒ ¹/_{cos A} = ¹⁷/₈

[∵ sec A = ¹/_cosA]

⇒ cos A = ⁸/₁₇ - - - - (ii)

Now, we know that

sin² A = 1 – cos² A

After substituting value of cos A from the equation (ii), we get

sin² A = 1 – (⁸/₁₇)²

⇒ sin² A = 1 – ⁶⁴/₂₈₉

⇒ sin² A = ^{289 – 64}/₂₈₉

⇒ sin² A = ²²⁵/₂₈₉

⇒ sin A = ²²⁵/₂₈₉

⇒ sin A = ¹⁵/₁₇

Thus, sin A = ¹⁵/₁₇ and sec A = ¹⁷/₈ Answer

Question (5) Given, sec θ = ¹³⁄₁₂ calculate all other trigonometric ratios.

Solution:

[Strategy to solve the question: The ratio of two sides of a right-angled triangle is given. Thus after calculating the third side, using Pythagoras theorem, all other trigonometric ratios can be calculated.]

Given, sec θ = ¹³⁄₁₂

We know that,

sec θ = ^{Hypotenuse(h)}⁄_Base(b)

⇒ ^h⁄_b = ¹³⁄₁₂

∴ Hypotenuse (h) = 13 k

and, Base (b) = 12 k

we know that according to Pythagoras Theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

After substituting the values of hypotenuse (h) and Base (b) in the given expression of Pythagoras Theorem

⇒ (13k)² = p² + (12k)²

⇒ 169 k² = p² + 144 k²

⇒ p² = 169 k² – 144 k²

⇒ p² = 25 k²

⇒ p = (25k²)

∴ Perpendicular (p) = 5 k

Now, we have, Hypotenuse (h) = 13 k

Perpendicular (p) = 5 k, and

Base (b) = 12 k

Calculation of ratio of sin A

Now, we know that

sin θ = ^{Perpendicular(p)}⁄_{Hypotenuse(h)}

After substituting the values of Perpendicular (p) and hypotenuse (h) in the given expression

⇒ sin θ = ^{5 k}⁄_{13 k}

⇒ sin θ = ⁵⁄₁₃ - - - - (i)

Calculation of the trigonometric ratio of cosec A

We know that

cosec θ = ^{Hypotenuse(h)}⁄_{Perpendicular(p)}

After substituting the values of Perpendicular (p) and hypotenuse (h) in the given expression

⇒ cosec θ = ^13k⁄_5k

⇒ cosec θ = ¹³⁄₅ - - - - (ii)

Calculation of the trigonometric ratio of cos A

We know that

cos θ = ^{Base (b)}⁄_{Hypotenuse (h)}

After substituting values of 'Base (b)' and 'Hypotenuse(h)', we get

cos θ = ^{12 k}⁄_{13 k}

⇒ cos θ = ¹²⁄₁₃

Calculation of the trigonometric ratio of tan A

We know that

tan θ = ^{Perpendicular(p)}⁄_Base(b)

After substituting values of 'Perpendicular(p)' and 'Base(b)', we get

tan θ = ^{5 k}⁄_{12 k}

⇒ tan θ = ⁵⁄₁₂

Calculation of the trigonometric ratio of cot A

we know that

cot θ = ^Base(b)⁄_{Perpendicular(p)}

After substituting values of'Base(b)' and 'Perpendicular(p)' in the above expression, we get

cot θ = ^{12 k}⁄_{5 k}

⇒ cot θ = ¹²⁄₅

Thus ,

sin θ = ⁵⁄₁₃, cosec θ = ¹³⁄₅, cos θ = ¹²⁄₁₃, tan θ = ⁵⁄₁₂, and cot θ = ¹²⁄₅ Answer

Alternate method to solve the question "Given, sec θ = ¹³⁄₁₂ calculate all other trigonometric ratios."

Given, sec θ = ¹³⁄₁₂

Calculation of the trigonometric ratio of cos θ

We know that

cos θ = ¹⁄ _{sec θ}

After substituting value of sec θ we get

⇒ cos θ = ¹⁄_13/12

⇒ cos θ = ¹²⁄₃ - - - -(i)

Calculation of the trigonometric ratio of sin θ

We know that

sin² θ = 1 – cos² θ

After substituting values of cos θ from equation (i) we get

⇒ sin² θ = 1 – (¹²⁄₁₃)²

⇒ sin² θ = 1 – ¹⁴⁴⁄₁₆₉

⇒ sin² θ = ^{169 – 144}⁄₁₆₉

⇒ sin² θ = ²⁵⁄₁₆₉

⇒ sin θ = ²⁵⁄₁₆₉

⇒ sin θ = 5/13 - - - -(ii)

Calculation of the trigonometric ratio of cosec θ

We know that,

cosec θ = ¹⁄_{sin θ}

After substituting value of sin θ from equation(ii)

We get,

cosec θ = ¹⁄_5/13

⇒ cosec θ = ¹³⁄₅ - - - - (iii)

Calculation of the trigonometric ratio of tan θ

We know that

tan θ = ^{sin θ}⁄_{cos θ}

After substituting value of sin θ and cos θ from equation (i) and (ii)we get

tan θ = ^5/13⁄_12/13

tan θ = ⁵⁄₁₃ × ¹³⁄₁₂

⇒ tan θ = ⁵⁄₁₂ - - - (iv)

Calculation of the trigonometric ratio of cot θ

We know that

cot θ = ¹⁄_{tan θ}

After substituting value of tan θ from equation (iv), we get,

cot θ = ¹⁄_5/12

⇒ cot θ = ¹²⁄₅

Thus ,

sin θ = ⁵⁄₁₃, cosec θ = ¹³⁄₅, cos θ = ¹²⁄₁₃, tan θ = ⁵⁄₁₂, and cot θ = ¹²⁄₅ Answer

Question (6) If ∠ A and ∠ B acute angles such that cos A = cos B, then show that ∠ A = ∠ B

Solution:

We know that in a right-angle triangle, for an acute angle

The side adjacent to the acute angle is called Base (b)

And, the side opposite the acute angle is called Perpendicular (p)

And, the side opposite to the right angle (90^o) is known as Hypotenuse (h)

Let ABC is a right-angled triangle

In this triangle, the side opposite the right angle is AB

Thus, AB = Hypotenuse (h)

And, for acute ∠ A

Side adjacent to the ∠ A = Base = AC

And, side opposite to the ∠ A = Perpendicular = BC

We know that

cos A = ^{Base (b)}/_{Hypotenuse(h)}

⇒ cos A = ^AC/_AB -------------- (i)

Now, for acute ∠ B

Side adjacent to the ∠ B = Base = BC

And, side opposite to the ∠ B = Perpendicular = AC

Thus, cos B = ^{Base (b)}/_{Hypotenuse(h)}

⇒ cos B = ^BC/_AB -----(ii)

Now, as given in the question

cos A – cos B

∴ After substituting values of cos A and cos B from equation (i) and (ii), we get

^AC/_AB = ^BC/_AB

After cross multiplication

⇒ AC = ^{BC × AB}/_AB

⇒ AC = BC

Since two sides AC and BC are equal. Thus, The given triangle is an isosceles triangle.

And we know that in a right-angle isosceles triangle two opposite angles other than the right angle are equal.

Thus in the given triangle .

∠ A = ∠ B Proved

Question (7) If cot θ = ⁷/₈ evaluate

(i) ^{(1 + sin θ ) (1 – sin θ )}/_{(1 + cos θ ) (1 – cos θ)}

(ii) cot² θ

Solution:

Given, cot θ = ⁷/₈

We know that cot θ = ^{Base (b)}/_{Perpendicular (p)}

After substituting the value of cot θ

⇒ ⁷/₈ = ^{Base (b)}/_{Perpendicular(p)}

∴ Base (b) = 7k

And, Perpendicular (p) = 8 k

Now, According to Pythagoras Theorem, we know that

[Hypotenuse(h)]² = [Perpendicular(p)]² + [Base(b)]²

⇒ h² = (8 k)² + (7 k)²

⇒ h² = 64 k² + 49 k²

⇒ h² = 113k²

⇒ Hypotenuse(h) = √113 k

Calculation of trigonometric ratio for sinθ

Now,

we know that

sin θ = ^{Perpendicular(p)}/_{Hypotenuse(h)}

After substituting values of 'Perpendicular(p)' and 'Hypotenuse(h)' we get

sin θ = ^8k/_{√113 k}

⇒ sin θ = ⁸/_√113 --------- (i)

Calculation of trigonometric ratio for cosθ

Now, we know that

cos θ = ^Base(b)/_{Hypotenuse(h)}

After substituting values of 'Base(b)' and 'Hypotenuse(h)' we get

cos θ = ^{7 k}/_{√113 k}

⇒ cos θ = ⁷/_√113 --------(ii)

Evaluation of the given question (7)

(i) ^{(1 + sin θ)(1 – sin θ )}/_{(1 + cos θ )(1 – cos θ)}

Solution

= ^{1 – sin2 θ}/_{1 – cos² θ}

[∵ (a+b) (a-b) = a² – b²]

After substituting the values of sin θ and cos θ from equation (i) and (ii) we get

= ^{1 – (8/_√113)2}/_{1 – (⁷/√113)²}

= ^{1 – 64/₁₁₃}/_{1 – ⁴⁹/113}

= ^{113 – 64 /₁₁₃}/_{^{113 – 49} /113}

= ^49/₁₁₃/_⁶⁴/113

= ⁴⁹/₁₁₃ × ¹¹³/₆₄

= ⁴⁹/₆₄ Answer

Alternate method to evaluate ^{(1 + sin θ)(1 – sin θ)}/_{(1 + cos θ )(1 – cos θ)}

= ^{1 – sin2 θ}/_{1 – cos² θ}

[∵ (a+b) (a-b) = a² – b²]

= ^cos2θ/_sin²θ

[∵ 1 – sin²θ = cos²θ and 1 – cos²θ = sin²θ]

⇒ cot²θ

[∵ ^cos2θ/_(sin²θ = cot²θ ]

Now, after substituting the value of cot θ = 7/8, which has been given in the question, we get

⇒ (7/8)² = 49/64

Thus,

^{(1 + sin θ)(1-sin θ )}/_{(1+cos θ )(1-cos θ )} = ⁴⁹/₆₄ Answer

Evaluation of given question (7) (ii) cot²θ

Now, after substituting the value of cot θ = 7/8, (as given in the question)we get

⇒ (7/8)² = 49/64

Thus, cot²θ = ⁴⁹/₆₄ Answer

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