Exercise 8.1

2. In Fig, find tan P – cot R.

Now, here, we have hypotenuse PR = cm, and side PQ = cm.

Now, from the pythagoras theorem, we have

PQ2 + QR2 = PR2RQ2

So, we have,

QR = PR2−QR2

= 132−122

= 25 = cm

We know that, cot R = adjacent side of ∠Ropposite side of ∠R

tan P = opposite side of ∠Padjacent side of ∠P

Hence, we have cot R = QRPQ = 512125

Also, tan P = QRPQ = 512125

So, tan P - cot R = 512 - 512 =

3. If sin A = 34 calculate cos A and tan A

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

Given that:

sin A = 3443

⇒ BCAC = 34−34

Let BC be 3k. Therefore, hypotenuse AC will be 4k where k is a positivenegative integer.

Applying Pythagoras theorem on ∆ABC, we obtain:

AC2 = AB2 + BC2AC2

AB2 = AC2BC2 - BC2

AB2 = 4k2 - 3k2

AB2 = k2 - k2

AB2 = k2

AB = 7k

cos A = side adjacent to ∠Ahypotenuse = ABAC = 7k4k = 74

tan A = side opposite to ∠Aside adjacent to ∠A = BCAB = 3k7k = 37

Thus, cos A= 74 and tan A =37

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

Let us consider a right-angled ΔABC, right-angled at B.

cot A= side adjacent to ∠Aside opposite to ∠A = ABBCACBC

It is given that 15 cot A =

⇒ ABBC = 815158

Let AB be 8k. Therefore, BC will be 15 k where k is a positivenegative integer.

Applying Pythagoras theorem in ΔABC, we obtain.

AC2 = AB2 + BC2

AC2 =8k2 + 15k2

AC2= k2 + k2

AC2 = k2

AC = k

sin A = side opposite to ∠Ahypotenuse = BCAC = 15k / 17k = 1517175

sec A = hypotenuseside adjacent to ∠A = ACAB = 17k / 8k = 178−178

Thus, sin A = 1517 and sec A = 178.

5. Given sec θ = 1312, calculate all other trigonometric ratios.

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

It is given that:

sec θ = hypotenuseside adjacent to ∠θ = ACAB = 13121213

Let AC = k and AB = k, where k is a positive integer.

Applying Pythagoras theorem in Δ ABC, we obtain:

AC2 = AB2 + BC2

BC2 = AC2- AB2

BC2 = 13k2 - 12k2

BC2 = k2 - k2

BC2 = k2

BC = k

sin θ = side opposite to ∠θhypotenuse = BCAC = 513−513

cos θ = side adjacent to ∠θhypotenuse = ABAC = 12131312

tan θ = side opposite to ∠θside adjacent to ∠θ = BCAB = 512−512

cot θ = side adjacent to ∠θside opposite to ∠θ = ABBC = 125512

cosec θ = hypotenuseside opposite to ∠θ = ACBC = 135513

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

In the right-angled triangle ABC as shown below, ∠A and ∠B are acute angles and ∠C is right angle.

cos A = side adjacent to ∠Ahypotenuse = ACABBCAB

cos B = side adjacent to ∠Bhypotenuse = BCABACAB

Given that cos A = cos B

Therefore, ACAB = BCAB

AC =

Hence, ∠A = ∠B (angles opposite to equal sides of a triangle are .)

Let's look into an alternative approach to solve the question.

Let us consider a triangle ABC in which CO ⊥ .

It is given that cos A = cos Bcos C

AOAC = BOBCAOBC

AOBO = ACBCAOBC

Let AOBO = ACBC= k

AO = k × ...(i)

AC = k × ...(ii)

By applying Pythagoras theorem in ΔCAO and ΔCBO, we get

AC2 = AO2BC2 + CO2 BO2(from ΔCAO)

CO2 = AC2BC2 - AO2 ...(iii)

BC2 = BO2AO2 + CO2 (from ΔCBO)

CO2= BC2AC2 - BO2 ...(iv)

From equation (iii) and equation (iv), we get

AC2 - AO2 = BC2 - BO2AO2

kBC2 - kBO2 = BC2AC2 - BO2 [From equation (i) and (ii)]

k2 BC2 - k2 BO2 = BC2- BO2AO2

k2 (BC2 - BO2) = BC2 - BO2

k2 = BC2−BO2BC2−BO2=

k =

Putting this value in equation (ii) we obtain,

AC =

Thus, ∠A = ∠B (angles opposite to equal sides of triangle are equal.)

7. If cot θ = 78, evaluate: (i) (1 + sin θ)(1 - sin θ)(1 + cos θ)(1 - cos θ) (ii) cot2θ

Consider ΔABC as shown below where angle B is a right angle.

cot θ = side adjacent to θside opposite to θ = ABBC = 7887

Let AB = 7k and BC = 8k, where k is a positivenegative integer.

By applying Pythagoras theorem in Δ ABC, we get

AC2 = AB2 + BC2AC2

= 7k2 + 8k2

= k2 + k2

= k2

AC = 113k2

= 113k

Therefore, sin θ = side opposite to θhypotenuse = BCAC = 8k113k = 8113

cos θ = side adjacent to θhypotenuse = ABAC = 7k113k = 7113

8. If 3 cot A = 4, check whether 1−tan2A1+tan2A = cos2 A - sin2 A or not

Let ΔABC be a right-angled triangle where angle B is a right angle.

cot A = side adjacent to ∠Aside opposite to ∠A = ABBC = 4334

Let AB = 4k and BC = 3k, where k is a integer.

By applying the Pythagoras theorem in ΔABC, we get,

AC2 = AB2 + BC2

= 4k2 + 3k2

= k2 + k2

= k2

AC = 25k2

= k

Therefore,

tan A = side opposite to ∠Aside adjacent to ∠A = BCAB = 3k4k = 3443

sin A = side opposite to ∠Ahypotenuse = BCAC = 3k5k = 35−35

cos A = side adjacent to ∠Ahypotenuse = ABAC = 4k5k = 4553

L.H.S = 1−tan2A1+tan2A

= 1−3421+342

= 1−9161+916

= 16−916+9

= 725−725

R.H.S = cos2 A - sin2 A

452 - 352

= 1625 - 925625

= 16−925

= 725−725

Therefore, 1−tan2A1+tan2A = cos2 A - sin2 A

9. In the triangle ABC right-angled at B, if tan A = 1/√3, find the value of:

Let ΔABC be a right-angled triangle such that tan A = 13

tan A = side opposite to ∠A side adjacent to ∠A = BCAB = 13

Let BC = k and AB = √3 k, where k is a integer.

By applying Pythagoras theorem in ΔABC, we have

AC2 = AB2 + BC2AC2

= 3k2 + k2

= 3k2 + k2

= 4k2

AC = k

Therefore, sin A = side opposite to ∠Ahypotenuse = BCAC = 12−12

cos A = side adjacent to ∠Ahypotenuse = ABAC = 3223

sin C = side opposite to ∠Chypotenuse = ABAC = 3223

cos C = side adjacent to ∠Chypotenuse = BCAC = 12−12

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

Given ∆PQR is right-angled at Q.

PQ = cm

PR + QR = cm

Let PR = x cm

Therefore,

QR = cm - PR

= (25 - ) cm

By applying Pythagoras theorem in ∆ PQR, we obtain.

PR2 = PQ2 + QR2

x2 = 52 + 25−x2

x2 = + - x + x2

50x =

x = 65050 =

Therefore, PR = cm

QR = (25 - 13) cm = cm

By substituting the values obtained above in the trigonometric ratios below we get,

sin P = side opposite to angle Phypotenuse = QRPR = 12131312

cos P = side adjacent to angle Phypotenuse = PQPR = 513−513

tan P = side opposite to angle Pside adjacent to angle P = QRPQ = 125512

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