Solution: (i) We have, sin 20o/ cos 70o = sin (90o – 70o)/ cos 70o = cos 70o/ cos70o = 1 [∵ sin (90 – θ) = cos θ] (ii) We have, cos 19o/ sin 71o = cos (90o – 71o)/ sin 71o = sin 71o/ sin 71o = 1 [∵ cos (90 – θ) = sin θ]

Solution:

(i) We have,

sin 20^{o}/ cos 70^{o} = sin (90^{o} – 70^{o})/ cos 70^{o }= cos 70^{o}/ cos70^{o} = 1 [∵ sin (90 – θ) = cos θ]

(ii) We have,

cos 19^{o}/ sin 71^{o }= cos (90^{o} – 71^{o})/ sin 71^{o }= sin 71^{o}/ sin 71^{o} = 1 [∵ cos (90 – θ) = sin θ]

Solution: Given, cos θ = 12/13…… (1) By definition, we know that cos θ = Base side adjacent to ∠θ / Hypotenuse……. (2) When comparing equation (1) and (2), we get Base side adjacent to ∠θ = 12 and Hypotenuse = 13 From the figure, Base side BC = 12 Hypotenuse AC = 13 Side AB is unknown here, and it caRead more

Solution:

Given, cos θ = 12/13…… (1)

By definition, we know that

cos θ = Base side adjacent to ∠θ / Hypotenuse……. (2)

When comparing equation (1) and (2), we get

Base side adjacent to ∠θ = 12 and Hypotenuse = 13

From the figure,

Base side BC = 12

Hypotenuse AC = 13

Side AB is unknown here, and it can be found by using Pythagoras theorem.

Thus, by applying Pythagoras theorem,

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

13^{2 }= AB^{2} + 12^{2}

Therefore,

AB^{2 }= 13^{2 }– 12^{2}

AB^{2 }= 169 – 144

AB^{2} = 25

AB = √25

AB = 5 …. (3)

Now, we know that

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse

Thus, sin θ = AB/AC [from figure]

⇒ sin θ = 5/13… (4)

And, tan θ = sin θ / cos θ = (5/13) / (12/13)

⇒ tan θ = 12/13… (5)

Taking L.H.S we have

L.H.S = sin θ (1 – tan θ)

Substituting the value of sin θ and tan θ from equation (4) and (5),

Solution: Given, tan θ = a/b And we know by definition that tan θ = opposite side/ adjacent side Thus, by comparison, Opposite side = a and adjacent side = b To find the hypotenuse, we know that by Pythagoras theorem that Hypotenuse2 = opposite side2 + adjacent side2 ⇒ Hypotenuse = √(a2 + b2) So, byRead more

Solution:

Given,

tan θ = a/b

And we know by definition that

tan θ = opposite side/ adjacent side

Thus, by comparison,

Opposite side = a and adjacent side = b

To find the hypotenuse, we know that by Pythagoras theorem that

Solution: Given, 3cot A = 4 ⇒ cot A = 4/3 By definition, tan A = 1/ Cot A = 1/ (4/3) ⇒ tan A = 3/4 Thus, Base side adjacent to ∠A = 4 Perpendicular side opposite to ∠A = 3 In ΔABC, Hypotenuse is unknown. Thus, by applying Pythagoras theorem in ΔABC, We get AC2 = AB2 + BC2 AC2 = 42 + 32 AC2 = 16 + 9Read more

Solution: Given: △PQR is right-angled at Q. PQ = 4cm RQ = 3cm Required to find: sin P, sin R, sec P, sec R Given △PQR, By using Pythagoras theorem to △PQR, we get PR2 = PQ2 +RQ2 Substituting the respective values, PR2 = 42 +32 PR2 = 16 + 9 PR2 = 25 PR = √25 PR = 5 ⇒ Hypotenuse =5 By definition, sinRead more

Solution:

Given:

△PQR is right-angled at Q.

PQ = 4cm

RQ = 3cm

Required to find: sin P, sin R, sec P, sec R

Given △PQR,

By using Pythagoras theorem to △PQR, we get

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

Substituting the respective values,

PR^{2} = 4^{2} +3^{2}

PR^{2} = 16 + 9

PR^{2} = 25

PR = √25

PR = 5

⇒ Hypotenuse =5

By definition,

sin P = Perpendicular side opposite to angle P/ Hypotenuse

sin P = RQ/ PR

⇒ sin P = 3/5

And,

sin R = Perpendicular side opposite to angle R/ Hypotenuse

Solution: Given that, sin A = 9/41 …………. (1) Required to find: cos A, tan A By definition, we know that sin A = Perpendicular/ Hypotenuse……………(2) On Comparing eq. (1) and (2), we get Perpendicular side = 9 and Hypotenuse = 41 Let’s construct △ABC as shown below, And, here the length of base AB is unRead more

Solution: By using Pythagoras theorem in △PQR, we have PR2 = PQ2 + QR2 Putting the length of given side PR and PQ in the above equation, 132 = 122 + QR2 QR2 = 132 – 122 QR2 = 169 – 144 QR2 = 25 QR = √25 = 5 By definition, tan P = Perpendicular side opposite to P/ Base side adjacent to angle P tan PRead more

Solution:

By using Pythagoras theorem in △PQR, we have

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

Putting the length of given side PR and PQ in the above equation,

13^{2 }= 12^{2} + QR^{2}

QR^{2} = 13^{2} – 12^{2}

QR^{2} = 169 – 144

QR^{2 }= 25

QR = √25 = 5

By definition,

tan P = Perpendicular side opposite to P/ Base side adjacent to angle P

tan P = QR/PQ

tan P = 5/12 ………. (1)

And,

cot R= Base/Perpendicular

cot R= QR/PQ

cot R= 5/12 …. (2)

When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations should also be equal.

## 1. Evaluate the following: (i) sin 20o/ cos 70o (ii) cos 19o/ sin 71o

Solution: (i) We have, sin 20o/ cos 70o = sin (90o – 70o)/ cos 70o = cos 70o/ cos70o = 1 [∵ sin (90 – θ) = cos θ] (ii) We have, cos 19o/ sin 71o = cos (90o – 71o)/ sin 71o = sin 71o/ sin 71o = 1 [∵ cos (90 – θ) = sin θ]

Solution:(i) We have,

sin 20

^{o}/ cos 70^{o}= sin (90^{o}– 70^{o})/ cos 70^{o }= cos 70^{o}/ cos70^{o}= 1 [∵ sin (90 – θ) = cos θ](ii) We have,

cos 19

See less^{o}/ sin 71^{o }= cos (90^{o}– 71^{o})/ sin 71^{o }= sin 71^{o}/ sin 71^{o}= 1 [∵ cos (90 – θ) = sin θ]## Evaluate each of the following: 1. sin 45∘ sin 30∘ + cos 45∘ cos 30∘

Solution:

See lessSolution:## 14. If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156

Solution: Given, cos θ = 12/13…… (1) By definition, we know that cos θ = Base side adjacent to ∠θ / Hypotenuse……. (2) When comparing equation (1) and (2), we get Base side adjacent to ∠θ = 12 and Hypotenuse = 13 From the figure, Base side BC = 12 Hypotenuse AC = 13 Side AB is unknown here, and it caRead more

Solution:Given, cos θ = 12/13…… (1)

By definition, we know that

cos θ = Base side adjacent to ∠θ / Hypotenuse……. (2)

When comparing equation (1) and (2), we get

Base side adjacent to ∠θ = 12 and Hypotenuse = 13

From the figure,

Base side BC = 12

Hypotenuse AC = 13

Side AB is unknown here, and it can be found by using Pythagoras theorem.

Thus, by applying Pythagoras theorem,

AC

^{2 }= AB^{2}+ BC^{2}13

^{2 }= AB^{2}+ 12^{2}Therefore,

AB

^{2 }= 13^{2 }– 12^{2}AB

^{2 }= 169 – 144AB

^{2}= 25AB = √25

AB = 5 …. (3)

Now, we know that

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse

Thus, sin θ = AB/AC [from figure]

⇒ sin θ = 5/13… (4)

And, tan θ = sin θ / cos θ = (5/13) / (12/13)

⇒ tan θ = 12/13… (5)

Taking L.H.S we have

L.H.S = sin θ (1 – tan θ)

Substituting the value of sin θ and tan θ from equation (4) and (5),

We get,

See less

## 9. If tan θ = a/b, find the value of (cos θ + sin θ)/ (cos θ – sin θ)

Solution: Given, tan θ = a/b And we know by definition that tan θ = opposite side/ adjacent side Thus, by comparison, Opposite side = a and adjacent side = b To find the hypotenuse, we know that by Pythagoras theorem that Hypotenuse2 = opposite side2 + adjacent side2 ⇒ Hypotenuse = √(a2 + b2) So, byRead more

Solution:Given,

tan θ = a/b

And we know by definition that

tan θ = opposite side/ adjacent side

Thus, by comparison,

Opposite side = a and adjacent side = b

To find the hypotenuse, we know that by Pythagoras theorem that

Hypotenuse

^{2}= opposite side^{2}+ adjacent side^{2}⇒ Hypotenuse = √(a

^{2}+ b^{2})So, by definition

sin θ = opposite side/ Hypotenuse

sin θ = a/ √(a

^{2}+ b^{2})And,

cos θ = adjacent side/ Hypotenuse

cos θ = b/ √(a

^{2}+ b^{2})Now,

After substituting for cos θ and sin θ, we have

∴

Hence, proved.

See less## 9. If tan θ = a/b, find the value of (cos θ + sin θ)/ (cos θ – sin θ)

Solution: Given, 3cot A = 4 ⇒ cot A = 4/3 By definition, tan A = 1/ Cot A = 1/ (4/3) ⇒ tan A = 3/4 Thus, Base side adjacent to ∠A = 4 Perpendicular side opposite to ∠A = 3 In ΔABC, Hypotenuse is unknown. Thus, by applying Pythagoras theorem in ΔABC, We get AC2 = AB2 + BC2 AC2 = 42 + 32 AC2 = 16 + 9Read more

Solution:Given,

3cot A = 4

⇒ cot A = 4/3

By definition,

tan A = 1/ Cot A = 1/ (4/3)

⇒ tan A = 3/4

Thus,

Base side adjacent to ∠A = 4

Perpendicular side opposite to ∠A = 3

In ΔABC, Hypotenuse is unknown.

Thus, by applying Pythagoras theorem in ΔABC,

We get

AC

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

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

^{2}= 16 + 9AC

^{2}= 25AC = √25

AC = 5

Hence, hypotenuse = 5

Now, we can find that

sin A = opposite side to ∠A/ Hypotenuse = 3/5

And,

cos A = adjacent side to ∠A/ Hypotenuse = 4/5

Taking the LHS,

Thus, LHS = 7/25

Now, taking RHS,

See less## 7. If cot θ = 7/8, evaluate (i) (1+sin θ)(1–sin θ)/ (1+cos θ)(1–cos θ) (ii) cot2 θ

Solution: (i) Required to evaluate: , given = cot θ = 7/8 Taking the numerator, we have (1+sin θ)(1–sin θ) = 1 – sin2 θ [Since, (a+b)(a-b) = a2 – b2] Similarly, (1+cos θ)(1–cos θ) = 1 – cos2 θ We know that, sin2 θ + cos2 θ = 1 ⇒ 1 – cos2 θ = sin2 θ And, 1 – sin2 θ = cos2 θ Thus, (1+sin θ)(1 –sin θ)Read more

Solution:(i)Required to evaluate:, given = cot θ = 7/8

Taking the numerator, we have

(1+sin θ)(1–sin θ) = 1 – sin

^{2}θ [Since, (a+b)(a-b) = a^{2}– b^{2}]Similarly,

(1+cos θ)(1–cos θ) = 1 – cos

^{2}θWe know that,

sin

^{2}θ + cos^{2}θ = 1⇒ 1 – cos

^{2}θ = sin^{2}θAnd,

1 – sin

^{2}θ = cos^{2}θThus,

(1+sin θ)(1 –sin θ) = 1 – sin

^{2}θ = cos^{2}θ(1+cos θ)(1–cos θ) = 1 – cos

^{2}θ = sin^{2}θ⇒

= cos

^{2}θ/ sin^{2}θ= (cos θ/sin θ)

^{2}And, we know that (cos θ/sin θ) = cot θ

⇒

=(cot θ)^{2}= (7/8)

^{2}= 49/ 64

(ii)Given,cot θ = 7/8

So, by squaring on both sides we get

(cot θ)

^{2}= (7/8)^{2}∴ cot θ

See less^{2}= 49/64## 6. In △PQR, right-angled at Q, PQ = 4cm and RQ = 3 cm. Find the value of sin P, sin R, sec P and sec R.

Solution: Given: △PQR is right-angled at Q. PQ = 4cm RQ = 3cm Required to find: sin P, sin R, sec P, sec R Given △PQR, By using Pythagoras theorem to △PQR, we get PR2 = PQ2 +RQ2 Substituting the respective values, PR2 = 42 +32 PR2 = 16 + 9 PR2 = 25 PR = √25 PR = 5 ⇒ Hypotenuse =5 By definition, sinRead more

Solution:Given:

△PQR is right-angled at Q.

PQ = 4cm

RQ = 3cm

Required to find: sin P, sin R, sec P, sec R

Given △PQR,

By using Pythagoras theorem to △PQR, we get

PR

^{2}= PQ^{2}+RQ^{2}Substituting the respective values,

PR

^{2}= 4^{2}+3^{2}PR

^{2}= 16 + 9PR

^{2}= 25PR = √25

PR = 5

⇒ Hypotenuse =5

By definition,

sin P = Perpendicular side opposite to angle P/ Hypotenuse

sin P = RQ/ PR

⇒ sin P = 3/5

And,

sin R = Perpendicular side opposite to angle R/ Hypotenuse

sin R = PQ/ PR

⇒ sin R = 4/5

And,

sec P=1/cos P

secP = Hypotenuse/ Base side adjacent to ∠P

sec P = PR/ PQ

⇒ sec P = 5/4

Now,

sec R = 1/cos R

secR = Hypotenuse/ Base side adjacent to ∠R

sec R = PR/ RQ

⇒ sec R = 5/3

See less## 4. If sin A = 9/41, compute cos A and tan A.

Solution: Given that, sin A = 9/41 …………. (1) Required to find: cos A, tan A By definition, we know that sin A = Perpendicular/ Hypotenuse……………(2) On Comparing eq. (1) and (2), we get Perpendicular side = 9 and Hypotenuse = 41 Let’s construct △ABC as shown below, And, here the length of base AB is unRead more

Solution:Given that, sin A = 9/41 …………. (1)

Required to find: cos A, tan A

By definition, we know that

sin A = Perpendicular/ Hypotenuse……………(2)

On Comparing eq. (1) and (2), we get

Perpendicular side = 9 and Hypotenuse = 41

Let’s construct △ABC as shown below,

And, here the length of base AB is unknown.

Thus, by using Pythagoras theorem in △ABC, we get

AC

^{2}= AB^{2}+ BC^{2}41

^{2}= AB^{2}+ 9^{2}AB

^{2}= 41^{2}– 9^{2}AB

^{2}= 168 – 81AB= 1600

AB = √1600

AB = 40

⇒ Base of triangle ABC, AB = 40

We know that,

cos A = Base/ Hypotenuse

cos A =AB/AC

cos A =40/41

And,

tan A = Perpendicular/ Base

tan A = BC/AB

tan A = 9/40

See less## 3. In fig. 5.37, find tan P and cot R. Is tan P = cot R?

Solution: By using Pythagoras theorem in △PQR, we have PR2 = PQ2 + QR2 Putting the length of given side PR and PQ in the above equation, 132 = 122 + QR2 QR2 = 132 – 122 QR2 = 169 – 144 QR2 = 25 QR = √25 = 5 By definition, tan P = Perpendicular side opposite to P/ Base side adjacent to angle P tan PRead more

Solution:By using Pythagoras theorem in △PQR, we have

PR

^{2}= PQ^{2}+ QR^{2}Putting the length of given side PR and PQ in the above equation,

13

^{2 }= 12^{2}+ QR^{2}QR

^{2}= 13^{2}– 12^{2}QR

^{2}= 169 – 144QR

^{2 }= 25QR = √25 = 5

By definition,

tan P = Perpendicular side opposite to P/ Base side adjacent to angle P

tan P = QR/PQ

tan P = 5/12 ………. (1)

And,

cot R= Base/Perpendicular

cot R= QR/PQ

cot R= 5/12 …. (2)

When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations should also be equal.

∴ tan P = cot R

See lessYes, tan P = cot R = 5/12