#### Page No 181:

#### Question 1:

In ΔABC right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

#### Answer:

Applying Pythagoras theorem for ΔABC, we obtain

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

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

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ AC = cm = 25 cm

(i) sin A =

cos A =

(ii)

sin C =

cos C =

#### Page No 181:

#### Question 2:

In the given figure find tan P − cot R

#### Answer:

Applying Pythagoras theorem for ΔPQR, 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

tan P − cot R =

#### Page No 181:

#### Question 3:

If sin A =, calculate cos A and tan A.

#### Answer:

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

Given that,

Let BC be 3*k*. Therefore, AC will be 4*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

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

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

16*k* ^{2} − 9*k* ^{2} = AB^{2}

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

AB =

#### Page No 181:

#### Question 4:

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

#### Answer:

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

It is given that,

cot A =

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

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

= 289*k*^{2}

AC = 17*k*

#### Page No 181:

#### Question 5:

Given sec θ =, calculate all other trigonometric ratios.

#### Answer:

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

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

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

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

BC = 5*k*

#### Page No 181:

#### Question 6:

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

∠A = ∠B.

#### Answer:

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

It is given that

cos A = cos B

… (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

By using the converse of B.P.T,

CD||BP

⇒∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠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 ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

**Alternatively,**

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

It is given that,

cos A = cos B

Let

⇒
AD = *k* BD … (1)

And, AC =
*k* BC … (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD^{2}
= AC^{2} − AD^{2} … (3)

And, CD^{2}
= BC^{2} − BD^{2} … (4)

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

AC^{2}
− AD^{2} = BC^{2} − BD^{2}

⇒
(*k *BC)^{2} − (*k* BD)^{2} = BC^{2}
− BD^{2}

⇒
*k*^{2} (BC^{2} − BD^{2}) = BC^{2}
− BD^{2}

⇒
*k*^{2} = 1

⇒ *k*
= 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)

#### Page No 181:

#### Question 7:

If cot θ =, evaluate

(i) (ii) cot^{2}
θ

#### Answer:

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

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

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

= 113*k*^{2}

AC =

(i)

(ii) cot^{2}
θ = (cot θ)^{2}
=
=

#### Page No 181:

#### Question 8:

If 3 cot A = 4, Check whether

#### Answer:

It is given that 3cot A = 4

Or, cot A =

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

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

In ΔABC,

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

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

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

= 25*k*^{2}

AC = 5*k*

cos^{2}
A − sin^{2} A =

∴

#### Page No 181:

#### Question 9:

In ΔABC, right angled at B. If, find the value of

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

(ii) cos A cos C − sin A sin C

#### Answer:

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

In ΔABC,

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

=

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

∴
AC = 2*k*

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

(ii) cos A cos C − sin A sin C

#### Page No 181:

#### Question 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.

#### Answer:

Given that, PR + QR = 25

PQ = 5

Let PR be *x*.

Therefore, QR = 25 − *x*

Applying Pythagoras theorem in ΔPQR, we obtain

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

*x*^{2} = (5)^{2} + (25 − *x*)^{2}

*x*^{2} = 25 + 625 + *x*^{2} − 50*x*

50*x* = 650

*x* = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

#### Page No 181:

#### Question 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 =for some value of angle A.

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

(iv) cot A is the product of cot and A

(v) sin θ =, for some angle θ

#### Answer:

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

But > 1

∴tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii)

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

Applying Pythagoras theorem in ΔABC, we obtain

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

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

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

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

BC = 10.9*k*

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

BC should be such that,

AC − AB < BC < AC + AB

12*k* − 5*k* < BC < 12*k* + 5*k*

7*k *< BC < 17 *k*

However, BC = 10.9*k*. Clearly, such a triangle is possible and
hence, such value of sec A is possible.

Hence, 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 =

We know that in a right-angled triangle,

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

#### Page No 187:

#### Question 1:

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan^{2}45°
+ cos^{2}30° − sin^{2}60°

(iii)

(iv)

(v)

#### Answer:

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan^{2}45°
+ cos^{2}30° − sin^{2}60°

(iii)

(iv)

(v)

#### Page No 187:

#### Question 2:

Choose the correct option and justify your choice.

(i)

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii)

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv)

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

#### Answer:

(i)

Out of the given alternatives, only

Hence, (A) is correct.

(ii)

Hence, (D) is correct.

(iii)Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, (A) is correct.

(iv)

Out of the given alternatives, only tan 60°

Hence, (C) is correct.

#### Page No 187:

#### Question 3:

If and;

0° < A + B ≤ 90°, A > B find A and B.

#### Answer:

⇒

⇒ A + B = 60 … (1)

⇒ tan (A − B) = tan30

⇒ A − B = 30 … (2)

On adding both equations, we obtain

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

#### Page No 187:

#### Question 4:

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

(i) sin (A + B) = sin A + sin B

(ii) The value of sinθ increases as θ increases

(iii) The value of cos θ increases as θ increases

(iv) sinθ = cos θ for all values of θ

(v) cot A is not defined for A = 0°

#### Answer:

(i) sin (A + B) = sin A + sin B

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as

sin 0° = 0

sin 90° = 1

Hence, the given statement is true.

(iii) cos 0° = 1

cos90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°.

Hence, the given statement is false.

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As

It is not true for all other values of θ.

As and ,

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As ,

= undefined

Hence, the given statement is true.

#### Page No 189:

#### Question 1:

Evaluate

(I)

(II)

(III) cos 48° − sin 42°

(IV)cosec 31° − sec 59°

#### Answer:

(I)

(II)

(III)cos 48° − sin 42° = cos (90°− 42°) − sin 42°

= sin 42° − sin 42°

= 0

(IV) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°

= sec 59° − sec 59°

= 0

#### Page No 189:

#### Question 2:

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0

#### Answer:

(I) tan 48° tan 23° tan 42° tan 67°

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

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 1

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

#### Page No 189:

#### Question 3:

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

#### Answer:

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

#### Page No 189:

#### Question 4:

If tan A = cot B, prove that A + B = 90°

#### Answer:

Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

#### Page No 189:

#### Question 5:

If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

#### Answer:

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

#### Page No 190:

#### Question 6:

If A, Band C are interior angles of a triangle ABC then show that

#### Answer:

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° − ∠A

#### Page No 190:

#### Question 7:

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

#### Answer:

sin 67° + cos 75°

**= **sin (90° − 23°) + cos (90° − 15°)

**= **cos 23° + sin 15°

#### Page No 193:

#### Question 1:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

#### Answer:

We know that,

will always be positive as we are adding two positive quantities.

Therefore,

We know that,

However,

Therefore,

Also,

#### Page No 193:

#### Question 2:

Write all the other trigonometric ratios of ∠A in terms of sec A.

#### Answer:

We know that,

Also, sin^{2} A + cos^{2} A = 1

sin^{2} A = 1 − cos^{2} A

tan^{2}A + 1 = sec^{2}A

tan^{2}A = sec^{2}A − 1

#### Page No 193:

#### Question 3:

Evaluate

(i)

(ii) sin25° cos65° + cos25° sin65°

#### Answer:

(i)

(As sin^{2}A + cos^{2}A = 1)

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin^{2}25° + cos^{2}25°

= 1 (As sin^{2}A + cos^{2}A = 1)

#### Page No 193:

#### Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec^{2} A − 9 tan^{2} A =

(A) 1

(B) 9

(C) 8

(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

(A) 0

(B) 1

(C) 2

(D) −1

(iii) (secA + tanA) (1 − sinA) =

(A) secA

(B) sinA

(C) cosecA

(D) cosA

(iv)

(A) sec^{2 }A

(B) −1

(C) cot^{2 }A

(D) tan^{2 }A

#### Answer:

(i) 9 sec^{2}A − 9 tan^{2}A

= 9 (sec^{2}A − tan^{2}A__)__

= 9 (1) [As sec^{2} A − tan^{2} A = 1]

= 9

Hence, alternative (B) is correct.

(ii)

(1 + tan θ + sec θ) (1 + cot θ − cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 − sinA)

= cosA

Hence, alternative (D) is correct.

(iv)

Hence, alternative (D) is correct.

#### Page No 193:

#### Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

#### Answer:

(i)

(ii)

(iii)

= secθ cosec θ +

= R.H.S.

(iv)

= R.H.S

(v)

Using the identity cosec^{2} = 1 + cot^{2},

L.H.S =

= cosec A + cot A

= R.H.S

(vi)

(vii)

(viii)

(ix)

Hence, L.H.S = R.H.S

(x)

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