**1. In ∆ABC right angled at B, AB = 24 cm, BC = 7 m. Determine **

**(i) sin A, ****cos A **

**(ii) sin C, cos C **

Solution:

(i)

Applying Pythagoras theorem for ∆ABC,

we obtain AC^{2} = AB^{2} + BC^{2}

= (24 cm)^{2} + (7 cm)^{2}

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ AC = √(625)cm = 25 cm

We know that,

sinA =

^{Opposite side}/_{hypotenuse}

sinA = ^{BC}/_{AC } = ^{7}/_{25 }

cosA =

^{ Adjacent side}/_{hypotenuse}

cosA = ^{AB}/_{AC } = ^{24}/_{25 }

(ii)

sinC =

^{Opposite side}/_{hypotenuse}

= ^{24}/25

cosC =

^{ Adjacent side}/_{hypotenuse}

= ^{7}/25

**2. In the given figure find tan P − cot R **

Solution:

Applying Pythagoras theorem for ∆PQR,

we obtain PR^{2} = PQ^{2} + QR^{2}

(13 cm)^{2} = (12 cm)^{2} + QR^{2}

169 cm^{2} = 144 cm^{2} + QR^{2}

25 cm^{2} = QR^{2}

QR = 5 cm

tanP = ^{opposite side}/_{Adjacent side}

tanP = ^{5}/_{12}

cotR = ^{adjacent side}/_{Opposite side}

cotR = ^{5}/_{12}

tan P − cot R = ^{5}/_{12} – ^{5}/_{12 }= 0

**3. If sin A = ¾ , calculate cos A and tan A. **

Solution:

Let ∆ABC be a right-angled triangle, right-angled at point B.

Given that, Let BC be 3k.

sinA = ^{3}/_{4}

cosA = ^{3}/_{4}

Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC,

we obtain AC^{2} = AB^{2} + BC^{2}

(4k)^{2} = AB^{2} + (3k)^{2}

16k^{2} − 9k^{2} = AB^{2}

7k^{2} = AB^{2}

AB = √7k

**4. Given 15 cot A = 8. Find sin A and sec A **

Solution:

Consider a right-angled triangle, right-angled at B.

cotR = ^{adjacent side}/_{Opposite side}

cotR = ^{AB}/BC

It is given that,

cotA = 8/15

Let AB be 8k.

Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC,

we obtain AC^{2} = AB^{2} + BC^{2}

= (8k)^{ 2} + (15k)^{ 2}

= 64k^{2} + 225k^{2}

= 289k^{2}

AC = 17k

SinA = (opposite side)/(Hypotenuse side) = BC/AC = 15k/17k = 15/17

secA = hypotenuse/(adjacent side) = AC/AB = 15k/8k = 15/8

**5. Given sec θ = 13/12 , calculate all other trigonometric ratios**

Solution:

Consider a right-angle triangle ∆ABC, right-angled at point B.

sec**θ** = hypotenuse/(adjacent side) = AC/AB = 13/12

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

(AC)^{2} = (AB)^{2} + (BC)^{2}

(13k)^{2} = (12k)^{2} + (BC)^{2}

169k^{2} = 144k^{2} + BC^{2}

25k^{2} = BC^{2}

BC = 5k

sinθ = opposie side to θ)/hypotenuse = BC/AC = 5k/13k = 5/13

cosθ = (adjacent side to θ)/hypotenuse = AB/AC = 12k/13k = 12/13

tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 5k/12k = 5/12

cotθ = adjacent side to θ/opposite side to θ = AB/BC = 12k/5k = 12/5

cosecθ = hypotenuse/adjacent side to θ = AC/BC = 13k/12k = 13/12

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

Solution:

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that ,

cos A = cos B

AD/AC = BD/BC ————(1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

AD/BD = AC/BC

⇒AD/BD = AC/CP [By construction we have BC = CP]————(2)

By using the converse of B.P.T, CD||BP

⇒ ∠ACD = ∠CPB (Corresponding angles)———–(3)

∠BCD = ∠CBP (Alternate interior angles)———-(4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle)——(5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD————- (6)

In ∆CAD and ∆CBD, ∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴ ∠CAD = ∠CBD

⇒∠A = ∠B

**7. If cot θ = 7/8 , evaluate **

**(i) [(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)] **

**(ii) cot²θ**

Solution:

Let us consider a right triangle ABC, right-angled at point B.

cotθ = adjacent side to θ/opposite side to θ = BC/AB = 7/8

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

AC^{2} = AB^{2} + BC^{2}

= (8k)^{2} + (7k)^{2}

= 64k^{2} + 49k^{2}

= 113k^{2}

AC = √113k

sinθ = opposie side to θ)/hypotenuse = AB/AC = 8k/√113k = 8/√113

cosθ = (adjacent side to θ)/hypotenuse = BC/AC = 7k/√113k = 7/√113

(i) **[(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)] = (1-sin²θ)/(1-cos²θ)**

= [1 – (8/√113)²]/[1 – (7/√113)²]

= [1 – (64/113)]/[1 – (49/113)]

= [(49/113)]/[(64/113)]

= 49/64

(ii)cot**²θ = cotθ² = (7/8)² = 49/64**

**8. If 3 cot A = 4, Check whether [(1-tan²A)/(1+tan²A)] = cos²A – sin²A or not**

Solution:

It is given that 3cot A = 4 Or, cot A = 4/3

Consider a right triangle ABC, right-angled at point B.

cotA = adjacent side to A/opposite side to A = AB/BC = 4/3

If AB is 4k, then BC will be 3k, where k is a positive integer.

In ∆ABC, (AC)^{2} = (AB)^{2} + (BC)^{2}

= (4k)^{2} + (3k)^{2}

= 16k^{2 }+ 9k^{2}

= 25k^{2}

AC = 5k

sinA = opposie side to A)/hypotenuse = BC/AC = 3k/5k = 3/5

cosA = (adjacent side to A)/hypotenuse = AB/AC = 4k/5k = 4/5

tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 3k/4k = 3/4

To check **[(1-tan²A)/(1+tan²A)] = cos²A – sin²A**

**LHS, [(1-tan²A)/(1+tan²A)] = [1-(3/4)²/(1+3/4)²] **

= [1-(9/16)]/[1+(9/16)]

= [(7/16)]/[(25/16)]

= 7/25,LHS

RHS, cos²A – sin²A

= (3/5)² – (4/5)²

= (9/25) – (16/25)

= [9-16]/25

= 7/25,RHS

Thus, LHS = RHS

Hence, **[(1-tan²A)/(1+tan²A)] = cos²A – sin²A**

**9. In ∆ABC, right angled at B. If tanA = 1/√3 , find the value of **

**(i) sin A cos C + cos A sin C **

**(ii) cos A cos C − sin A sin C **

Solution:

Given **tanA = 1/√3**

i.e., BC/AB **= 1/√3**

If BC is k, then AB will be , where k is a positive integer.

In ∆ABC, AC^{2} = AB^{2} + BC^{2}

AC^{2 }= (√3k)^{2 }+ k^{2}

AC^{2 }= 3k^{2 }+ k^{2} = 4k^{2}

AC = √(4k^{2}) = 2k

sinA= opposie side to A)/hypotenuse = BC/AC = k/2k = 1/2

cosA = (adjacent side to A)/hypotenuse = AB/AC = √3k/2k = √3/2

sinC= opposie side to C/hypotenuse = AB/AC = √3k/2k = √3/2

cosC = (adjacent side to C)/hypotenuse = BC/AC = k/2k = 1/2

To find, **(i) sin A cos C + cos A sin C**

=^{ 1}/_{2} x ^{1}/_{2} + ^{√3}/_{2} x ^{√3}/_{2}

= ^{1}/_{4} + ^{3}/_{4}

= ^{4}/_{4}

= 1

to find **(ii) cos A cos C − sin A sin C **

=^{ √3}/_{2} x ^{1}/_{2} – ^{1}/_{2} x ^{√3}/_{2}

=^{ √3}/_{4} – ^{√3}/_{4}

= 0

**10. In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. **

Solution:

Given that, PR + QR = 25

PQ = 5

Let PR be x.

Therefore, QR = 25 − x

Applying Pythagoras theorem in ∆PQR, we obtain

PR^{2} = PQ^{2} + QR^{2}

x^{2} = (5)^{2} + (25 − x)^{2}

x^{2} = 25 + 625 + x^{2 }− 50x

50x = 650

x = 13 Therefore, PR = 13 cm , QR = (25 − 13) cm = 12 cm

sinP = opposie side to P)/hypotenuse = RQ/PR = 12/13

cosP = (adjacent side to P)/hypotenuse = PQ/PR = 5/13

tanP = opposite side tp P/adjacent side to P = RQ/PQ = 12/5

**11. State whether the following are true or false. Justify your answer. **

**(i) The value of tan A is always less than 1. **

**(ii) sec A = 12/5 for some value of angle A. **

**(iii) cos A is the abbreviation used for the cosec of angle A. **

**(iv) cot A is the product of cot and A (v) sin θ = 4/3 , for some angle θ **

Solution:

(i) Consider a ∆ABC, right-angled at B.

tanA = opposite side tp A/adjacent side to A = 12/5

But 12/5 > 1

Therefore, tanA > 1

So, tan A < 1 is not always true. Hence, the given statement is false

(ii) secA = 12/5

sec A = AC/AB = 12/5

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

AC^{2} = AB^{2} + BC^{2 }

(12k)^{2} = (5k)^{2} + BC^{2}

144k^{2} = 25k^{2} + BC^{2}

BC^{2} = 119k^{2}

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k.

Clearly, such a triangle is possible and hence, such value of sec A is possible. Therefore, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false.

(v) sin θ = 4/3

We know that in a right-angled triangle,

sinθ = opposie side to θ)/hypotenuse

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible. Hence, the given statement is false.

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