# Frank Solutions for Chapter 26 Trigonometrical Ratios of Standard Angles Class 9 Mathematics ICSE

### Exercise 27.1

1. Without using tables, evaluate the following:

(i) sin 60° sin 30° + cos 30° cos 60°

(ii) sec 30° cosec 60° + cos 60° sin 30°

(iii) sec 45° sin 45° – sin 30° sec 60°

(iv) sin2 30° sin245° + sin2 60° sin2 90°

(v) tan2 30° + tan2 60° + tan2 45°

(vi) sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°

(vii) cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60°

(viii) cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°

(ix) (sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

(x) 4(sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)

(i) sin 60° sin 30° + cos 30° cos 60°

sin 60° = (√3/2)

sin 30° = (1/2)

cos 30° = (√3/2)

cos 60° = (1/2)

On substituting, we get,

sin 60° sin 30° + cos 30° cos 60° = (√3/2) ×(1/2) + (√3/2)×(1/2)

= (√3/4) + (√3/4)

We get,

= (√3/2)

(ii) sec 30° cosec 60° + cos 60° sin 30°

cos 30° = (√3/2) ⇒ sec 30° = (2/√3)

sin 60° = (√3/2) ⇒ cosec 60° = (2/√3)

cos 60° = (1/2)

sin 30° = (1/2)

On substituting, we get,

sec 30° cosec 60° + cos 60° sin 30° = (2/√3)× (2/√3) + (1/2)×(1/2)

= (4/3) + (1/4)

= (16 + 3)/12

We get,

= 19/12

(iii) sec 45° sin 45° – sin 30° sec 60°

cos 45° = (1/√2) ⇒ sec 45° = √2

sin 45° = (1/√2)

sin 30° = (1/2)

cos 60° = (1/2) ⇒ sec 60° = 2

sec 45° sin 45° – sin 30° sec 60°

On substituting, we get,

= (√2)×(1/√2) – (1/2)×2

= 1 – 1

We get,

= 0

(iv) sin2 30° sin2 45° + sin2 60° sin2 90°

sin 30° = (1/2)

sin 45° = (1/√2)

sin 60° = (√3/2)

sin 90° = 1

On substituting, we get,

sin2 30° sin2 45° + sin2 60° sin2 90° = (1/2)2 (1/√2)2 + (√3/2)2 (1)2

= (1/4)×(1/2) + (3/4) (1)

= (1/8) + (3/4)

= (1 + 6)/8

We get,

= 7/8

(v) tan2 30° + tan2 60° + tan2 45°

tan 30° = (1/√3)

tan 60° = √3

tan 45° = 1

On substituting, we get,

tan2 30° + tan2 60° + tan2 45° = (1 /√3)2 + (√3)2 + 1

= (1/3) + 3 + 1

= (1 + 9 + 3)/3

We get,

= 13/3

(vi) sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°

sin 30° = (1/2)

cos 45° = (1/√2)

tan 30° = (1/√3)

sin 90° = 1

cos 0° = 1

On substituting, we get,

sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°

= (1/2)2 (1/√2)2 + 4 (1 / √3)2 + 1 + 1

= (1/4) (1/2) + (4/3) + 2

= (1/8) + (4/3) + 2

= (3 + 32 + 48)/24

We get,

= 83/24

(vii) cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60°

sin 45° = (1/√2)

cosec 45° = (√2/1)

sin 30° = cos 60° = (1/2)

sec 60° = 2

cos 30° = (√3/2)

sec 30° = (2/√3)

tan 45° = 1

cot 45° = 1

On substituting, we get,

cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60° = (√2/1)2 (2/√3)2 – (1/2)2 – 4 (1)2 + (2)2

= 2×(4/3) – (1/4) – 4 + 4

= (8/3) – (1/4)

= (32 – 3)/12

We get,

= 29/12

(viii) cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°

sin 30° = (1/2)

cosec 30° = 2

cos 60° = (1/2)

sec 60° = 2

cos 45° = (1/√2)

sec 45° = √2

tan 45° = 1

sin 90° = 1

tan 30° = (1/√3)

cot 30° = √3

On substituting, we get,

cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°

= (2)3 (1/2) (1)3 (1)2 (√2)2 (√3)

= 8 x (1/2)×2 ×√ 3

We get,

= 8√3

(ix) (sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

sin 30° = (1/2)

sin 45° = (1/√2)

sin 90° = 1

cos 45° = (1/√2)

cos 60° = (1/2)

(sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

On substituting, we get,

= {(1) + (1/√2) + (1/2)} {(1) – (1/√2) + (1/2)}

= {(3/2) + (1/√2)} {(3/2) – (1/√2)}

= (3/2)2 – (1/√2)2

= (9/4) – (1/2)

We get,

= (9 – 2)/4

= 7/4

(x) 4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)

sin 30° = (1/2)

sin 90° = 1

cos 45° = (1/√2)

cos 60° = (1/2)

On substituting, we get,

4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)

= 4 {(1/2)4 + (1/2)4} – 3 {(1/√2)2 – (1)2}

= 4 {(1/16) + (1/16)} – 3 {(1/2) – 1}

On calculating further, we get,

= 4×(2/16) + 3×(1/2)

= (1/2) + (3/2)

= 4/2

= 2

2. Without using tables, find the value of the following:

(i) (sin 30° – sin 90° + 2 cos 0°)/tan 30° tan 60°

(ii) (sin 30°/sin 45°) + (tan 45°/sec 60°) – (sin 60°/cot 45°) – (cos 30°/sin 90°)

(iii) (tan 45°/cosec 30°) + (sec 60°/cot 45°) – (5 sin 90°/2 cos 0°)

(iv) (tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos 90°)/(cosec 30° + sec 60° – cot2 30°)

(v) (4/cot2 30°) + (1/sin2 60°) – cos2 45°

(i) (sin 30° – sin 90° + 2 cos 0°)/tan 30° tan 60°

= {(1/2) – (1) + 2×1}/(1/√3)×√3

On further calculation, we get,

= {(1/2) – 1 + 2}/1

= (1/2) – 1 + 2

= (1/2) + 1

We get,

= 3/2

(ii) (sin 30°/sin 45°) + (tan 45°/sec 60°) – (sin 60°/cot 45°) – (cos 30°/sin 90°)

= {(1/2)/(1/√2)} + (1/2) – {(√3/2)/1} – {(√3/2)/1}

On calculating further, we get,

= (√2/2) + (1/2) – (√3/2) – (√3/2)

= (√2 + 1 – 2√3)/2

(iii) (tan 45°/cosec 30°) + (sec 60°/cot 45°) – (5 sin 90°/2 cos 0°)

= (1/2) + (2/1) – (5×1)/(2×1)

On further calculation, we get,

= (1/2) + (2/1) – (5/2)

= (1 + 4 – 5)/2

We get,

= 0

(iv) (tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos 90°)/(cosec 30° + sec 60° – cot2 30°)

= {(√3)2 + 4×(1/√2)2 + 3×(2/√3)2 + 5×0}/(2) + (2) – (√3)2

On further calculation, we get,

= {3 + 4×(1/2) + 3×(4/3) + 0}/(2 + 2 – 3)

= (3 + 2 + 4)/(4 – 3)

We get,

= 9

(v) (4/cot2 30°) + (1/sin2 60°) – cos2 45°

= {4/(√3)2} + {1/(√3/2)2} – (1/√2)2

On further calculation, we get,

= (4/3) + {1/(3/4)} – (1/2)

= (4/3) + (4/3) – (1/2)

= (8 + 8 – 3)/6

We get,

= 13/6

3. Prove that:

(a) sin 60° . cos 30° – sin 60°. sin 30° = (1/2)

(b) cos 60°. Cos 30° – sin 60°. sin 30° = 0

(c) sec2 45° – tan2 45° = 1

(d) {(cot 30° + 1)/(cot 30° – 1)}2 = (sec 30° + 1)/(sec 30° – 1)

(a) Consider L.H.S.

sin 60°. cos 30° – cos 60°. sin 30°

= (√3/2) x (√3/2) – (1/2) x (1/2)

On simplification, we get,

= (3/4) – (1/4)

= (3 – 1)/4

= 2/4

We get,

= 1/2

= R.H. S.

Hence, proved

(b) Consider L.H.S.

cos 60°. cos 30° – sin 60°. sin 30°

= (1/2)×(√3/2) – (√3/2)×(1/2)

On calculating further, we get,

= (√3/4) – (√3/4)

We get,

= 0

= R.H.S.

Hence, proved

(c) Consider L.H.S.

sec2 45° – tan2 45°

= (√2)2 – (1)2

= 2 – 1

= 1

= R.H.S.

Hence, proved

(d) Consider L.H.S.

{(cot 30° + 1)/(cot 30° – 1)}2

= {(√3) + 1/(√3) – 1}2

= {(√3 + 1)/(√3 – 1) x (√3 + 1)/(√3 + 1)}2

On further calculation, we get,

= {(√3)2 + (1)2 + 2√3}/{(√3)2 + (1)2 – 2√3}

= (3 + 1 + 2√3)/(3 + 1 – 2√3)

= (4 + 2√3)/(4 – 2√3)

Taking 2 as common, we get,

= 2 (2 + √3)/2 (2 – √3)

= (2 + √3)/(2 – √3)

= {(2/√3) + 1}/{(2/√3) – 1}

= (sec 30° + 1)/(sec 30° – 1)

= R.H.S.

Hence, proved

4. Find the value of ‘A’, if

(a) 2 cos A = 1

(b) 2 sin 2A = 1

(c) cosec 3A = (2/√3)

(d) 2 cos 3A = 1

(e) √3 cot A = 1

(f) cot 3A = 1

(a) 2 cos A = 1

⇒ cos A = (1/2)

⇒ cos A = cos 60°

⇒ A = 60°

Therefore, the value of ‘A’ is 60°

(b) 2 sin 2A = 1

⇒ sin 2A = (1/2)

⇒ sin 2A = sin 30°

⇒ 2A = 30°

We get,

A = 15°

Therefore, the value of ‘A’ is 15°

(c) cosec 3A = (2/√3)

⇒ cosec 3A = cosec 60°

⇒ 3A = 60°

We get,

A = 20°

Therefore, the value of ‘A’ is 20°

(d) 2 cos 3A = 1

⇒ cos 3A = (1/2)

⇒ cos 3A = cos 60°

⇒ 3A = 60°

We get,

A = 20°

Therefore, the value of ‘A’ is 20°

(e) √3 cot A = 1

⇒ cot A = (1/√3)

⇒ cot A = cot 60°

⇒ A = 60°

Therefore, the value of ‘A’ is 60°

(f) cot 3A = 1

⇒ cot 3A = cot 45°

⇒ 3A = 45°

We get,

A = 15°

Therefore, the value of ‘A’ is 15°

5. Find the value of ‘A’, if

(a) (1 – cosec A) (2 – sec A) = 0

(b) (2 – cosec 2A) cos 3A = 0

(a) (1 – cosec A) (2 – sec A) = 0

Here,

1 – cosec A = 0 and 2 – sec A = 0

On calculating further, we get,

cosec A = 1 and sec A = 2

⇒ cosec A = cosec 90° and sec A = sec 60°

⇒ A = 90° and A = 60°

(b) (2 – cosec 2A) cos 3A = 0

Here,

2 – cosec 2A = 0 and cos 3A = 0

On further calculation, we get,

⇒ cosec 2A = 2 and cos 3A = 0

⇒ cosec 2A = cosec 30° and cos 3A = cos 90°

We get,

2A = 30° and 3A = 90°

⇒ A = 15° and A = 30°

6. Find the value of ‘x’ in each of the following:
(a)

(b)

(c)

(d)

(a)

From the figure,

We have,

sin 60° = BC/AC

√3/2 = 12/x

On simplification, we get,

x = (2×12)/√3

⇒ x = 24/√3

⇒ x = (8×3)/√3

We get,

x = 8√3

(b)

From the figure,

We have

tan 45° = BC/AB

⇒ 1 = 24/x

We get,

x = 24

(c)

From the figure,

We have,

cos x = AB/AC

⇒ cos x = 12/24

⇒ cos x = 1/2

⇒ cos x = cos 60°

We get,

x = 60°

(d)

From the figure,

We have,

sin x = BC/AC

⇒ sin x = (15/√2)/15

⇒ sin x = (1/√2)

⇒ sin x = sin 45°

We get,

⇒ x = 45°

7. Find the length of AD.

Given: ABC = 60°, DBC = 45° and BC = 24 cm.

In △ABC,

tan 60° = AC/BC

√3 = AC/24

We get,

AC = 24√3 cm

In △DBC,

tan 45° = DC/BC

⇒ 1 = DC/24

We get,

DC = 24 cm

Now,

Substituting the values of AC and DC, we get,

⇒ AD = 24 (√3 – 1) cm

Therefore, the length of AD is 24(√3 – 1) cm

8. Find lengths of diagonals AC and BD. Given AB = 24 cm and BAD = 60°

Since all sides are equal,

∴ The given figure is a rhombus

We know that,

Diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex

Let the diagonals AC and BD intersect each other at point O

Hence,

OA = OC = (1/2) AC

OB = OD = (1/2) BD

∠AOB = 90°

Given

⇒ ∠OAB = 30°

In right-angled △AOB,

sin 30° = OB/AB

= (1/2)

Given AB = 24,

⇒OB/24 = (1/2)

OB = 24/2

We get,

OB = 12 cm

cos 30° = OA/AB

= √3/2

⇒ OA/24 = √3/2

OA = (24√3)/2

We get,

OA = 12√3 cm

Therefore,

Length of diagonal AC = 2×OA = 2×12√3 = 24√3 cm and

Length of diagonal BD = 2×OB = 2×12 = 24 cm

9. Find the length of EC.

CD = 28 cm

⇒ AB = 28 cm

In right △ABE,

tan 30° = BE/AB

⇒ 1/√3 = BE/28

We get,

BE = (28/√3)

In right △ABC,

tan 60° = CB/AB

⇒ √3 = CB/28

We get,

CB = 28√3

Hence,

Length of EC = CB + BE

= 28√3 + (28/√3)

On further calculation, we get,

= (84 + 28)/√3

= (112/√3)

Hence, the length of EC is (112/√3) cm

10. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 1.5 cm each and are perpendicular to AB. Given that AED = 45° and ACD = 30°. Find:

(a) AB

(b) AC

(c) AE

(1/√3) = 1.5/DC (given AD = 1.5 cm)

⇒ DC = 1.5√3

Here,

AB || DC and AD ⊥ EC, ABCD is a parallelogram

Therefore, opposite sides are equal

⇒ AB = DC = 1.5√3 cm

⇒ (1/2) = 1.5/AC (given AD = 1.5 cm)

⇒ AC = 2×1.5

We get,

AC = 3 cm

⇒ (1/√2) = 1.5/AE (given AD = 1.5 cm)

We get,

AE = 1.5√2 cm

11. Evaluate the following:

(a) sin 62°/cos 28°

(b) sec 34°/cosec 56°

(c) tan 12°/cot 78°

(d) sin 25° cos 43°/sin 47° cos 65°

(e) sec 32° cot 26°/tan 64° cosec 58°

(f) cos 34° cos 33°/ sin 57° sin 56°

(a) sin 62°/cos 28°

This can be written as,

= sin (90° – 28°)/cos 28°

= cos 28°/cos 28°

We get,

= 1

(b) sec 34°/cosec 56°

This can be written as,

= sec (90° – 56°)/cosec 56°

= cosec 56°/cosec 56°

We get,

= 1

(c) tan 12°/cot 78°

This can be written as,

= tan (90° – 78°)/cot 78°

= cot 78°/cot 78°

We get,

= 1

(d) sin 25° cos 43°/sin 47° cos 65°

This can be written as,

= sin (90° – 65°) cos (90° – 47°)/sin 47° cos 65°

= cos 65° sin 47°/sin 47° cos 65°

We get,

= 1

(e) sec 32° cot 26°/tan 64° cosec 58°

This can be written as,

= sec (90° – 58°) cot (90° – 64°)/tan 64° cosec 58°

= cosec 58° tan 64° / tan 64° cosec 58°

We get,

= 1

(f) cos 34° cos 33°/sin 57° sin 56°

This can be written as,

= cos (90° – 56°) cos (90° – 57°)/sin 57° sin 56°

= sin 56° sin 57°/sin 57° sin 56°

We get,

= 1

12. Evaluate the following:

(a) sin 31° – cos 59°

(b) cot 27° – tan 63°

(c) cosec 54° – sec 36°

(d) sin 28° sec 62° + tan 49° tan 41°

(e) sec 16° tan 28° – cot 62° cosec 74°

(f) sin 22° cos 44° – sin 46° cos 68°

(a) sin 31° – cos 59°

= sin (90° – 59°) – cos 59°

= cos 59° – cos 59°

We get,

= 0

(b) cot 27° – tan 63°

= cot (90° – 63°) – tan 63°

= tan 63° – tan 63°

We get,

= 0

(c) cosec 54° – sec 36°

= cosec (90° – 36°) – sec 36°

= sec 36° – sec 36°

We get,

= 0

(d) sin 28° sec 62° + tan 49° tan 41°

= sin 28° sec (90° – 28°) + tan 49° tan (90° – 49°)

= sin 28° cosec 28° + tan 49° cot 49°

= sin 28° ×(1/sin 28°) + tan 49°×(1/tan 49°)

= 1 + 1

We get,

= 2

(e) sec 16° tan 28° – cot 62° cosec 74°

= sec (90° – 74°) tan (90° – 62°) – cot 62° cosec 74°

= cosec 74° cot 62° – cot 62° cosec 74°

We get,

= 0

(f) sin 22° cos 44° – sin 46° cos 68°

= sin (90° – 68°) cos (90° – 46°) – sin 46° cos 68°

= cos 68° sin 46° – sin 46° cos 68°

We get,

= 0

13. Evaluate the following:

(a) (sin 36°/cos 54°) + (sec 31°/cosec 59°)

(b) (tan 42°/cot 48°) – (cos 33°/sin 57°)

(c) (2 sin 28°/cos 62°) + (3 cot 49°/tan 41°)

(d) (5 sec 68°/cosec 22°) + (3 sin 52° sec 38°)/(cot 51° cot 39°)

(a) (sin 36°/cos 54°) + (sec 31°/cosec 59°)

This can be written as,

= {sin (90° – 54°)/cos 54°} + {sec (90° – 59°)/cosec 59°}

= (cos 54°/cos 54°) + (cosec 59°/cosec 59°)

= 1 + 1

We get,

= 2

(b) (tan 42°/cot 48°) – (cos 33°/sin 57°)

This can be written as,

= {tan (90° – 48°)/cot 48°} – {cos (90° – 57°)/sin 57°}

= (cot 48°/cot 48°) – (sin 57°/sin 57°)

= 1 – 1

We get,

= 0

(c) (2 sin 28°/cos 62°) + (3 cot 49°/tan 41°)

This can be written as,

= {2 sin (90° – 62°)/cos 62°} + {3 cot (90° – 41°)/tan 41°}

= (2 cos 62°/cos 62°) + (3 tan 41°/ tan 41°)

= 2 + 3

We get,

= 5

(d) (5 sec 68°/cosec 22°) + (3 sin 52° sec 38°)/(cot 51° cot 39°)

This can be written as,

= {5 sec (90° – 22°)/cosec 22°} + {3 sin 52° sec (90° – 52°)/cot 51° cot (90° – 51°)}

= (5 cosec 22°/cosec 22°) + (3 sin 52° cosec 52°/cot 51° tan 51°)

= 5 + {3 sin 52° ×(1/sin 52°)/cot 51° ×(1/cot 51°)}

= 5 + 3/1

= 5 + 3

We get,

= 8

14. Express each of the following in terms of trigonometric ratios of angles between 0°and 45°

(a) sin 65° + cot 59°

(b) cos 72° – cos 88°

(c) cosec 64° + sec 70°

(d) tan 77° – cot 63° + sin 57°

(e) sin 53° + sec 66° – sin 50°

(f) cos 84° + cosec 69° – cot 68°

(a) sin 65° + cot 59°

This can be written as,

= sin (90° – 25°) + cot (90° – 31°)

We get,

= cos 25° + tan 31°

(b) cos 72° – cos 88°

This can be written as,

= cos (90° – 18°) – cos (90° – 2°)

We get,

= sin 18° – sin 2°

(c) cosec 64° + sec 70°

This can be written as,

= cosec (90° – 26°) + sec (90° – 20°)

We get,

= sec 26° + cosec 20°

(d) tan 77° – cot 63° + sin 57°

This can be written as,

= tan (90° – 13°) – cot (90° – 27°) + sin (90° – 33°)

We get,

= cot 13° – tan 27° + cos 33°

(e) sin 53° + sec 66° – sin 50°

This can be written as,

= sin (90° – 37°) + sec (90° – 24°) – sin (90° – 40°)

We get,

= cos 37° + cosec 24° – cos 40°

(f) cos 84° + cosec 69° – cot 68°

This can be written as,

= cos (90° – 6°) + cosec (90° – 21°) – cot (90° – 22°)

We get,

= sin 6° + sec 21° – tan 22°

15. Evaluate the following:

(a) sin 35° sin 45° sec 55° sec 45°

(b) cot 20° cot 40° cot 45° cot 50° cot 70°

(c) cos 39° cos 48° cos 60° cosec 42° cosec 51°

(d) sin (35° + Î¸) – cos (55° – Î¸) – tan (42° + Î¸) + cot (48° – Î¸)

(e) tan (78° + Î¸) + cosec (42° + Î¸) – cot (12° – Î¸) – sec (48° – Î¸)

(f) (3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + {(4 tan 23° tan 37° tan 67° tan 53°)/(cos 17° cos 67° cosec 73° cosec 23°)}

(g) (sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)

(h) {(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°)} + {(3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)}

(i) {(3 sin2 40°)/(4 cos2 50°)} – {(cosec2 28°)/(4 sec2 62°)} + {(cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 45° sin 65° sec 75°)}

(j) {(5 cot 5° cot 15° cot 25° cot 35° cot 45°) / (7 tan 45° tan 55° tan 65° tan 75° tan 85°)} + {(2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)}

(a) sin 35° sin 45° sec 55° sec 45°

This can be written as,

= sin (90° – 55°)×(1/√2)×(1/cos 55°)×(√2)

= cos 55° ×(1/cos 55°)×(1/√2)×(√2)

We get,

= 1

(b) cot 20° cot 40° cot 45° cot 50° cot 70°

This can be written as,

= cot (90° – 70°)×cot (90° – 50°)× 1×cot 50° cot 70°

= tan 70° × tan 50° × cot 50° × cot 70°

= tan 70° × cot 70° × tan 50° × cot 50°

= tan 70° × (1/tan 70°) × tan 50° × (1/tan 50°)

We get,

= 1

(c) cos 39° cos 48° cos 60° cosec 42° cosec 51°

This can be written as,

= cos (90° – 51°) ×cos (90° – 42°) ×(1/2) ×(1/sin 42°)× (1/sin 51°)

= sin 51° ×sin 42° × (1/2)×(1/sin 42°)×(1/sin 51°)

We get,

= 1/2

(d) sin (35° + Î¸) – cos (55° – Î¸) – tan (42° + Î¸) + cot (48° – Î¸)

This can be written as,

= sin {90° – (55° – Î¸)} – cos (55° – Î¸) – tan {90° – (48° – Î¸)} + cot (48° – Î¸)

= cos (55° – Î¸) – cos (55° – Î¸) – cot (48° – Î¸) + cot (48° – Î¸)

We get,

= 0

(e) tan (78° + Î¸) + cosec (42° + Î¸) – cot (12° – Î¸) – sec (48° – Î¸)

This can be written as,

= tan {90° – (12° – Î¸)} + cosec {90° – (48° – Î¸)} – cot (12° – Î¸) – sec (48° – Î¸)

= cot (12° – Î¸) + sec (48° – Î¸) – cot (12° – Î¸) – sec (48° – Î¸)

We get,

= 0

(f) (3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + {(4 tan 23° tan 37° tan 67° tan 53°)/(cos 17° cos 67° cosec 73° cosec 23°)}

This can be written as,

= {3 sin (90° – 53°)/cos 53°} – {5 cosec (90° – 51°)/sec 51°} + [{4 tan (90° – 67°) tan (90° – 53°) x (1/cot 67°) x (1/cot 53°)}]/{cos (90° – 73°) cos (90° – 23°) ×(1/sin 73°) x (1/sin 23°)}

= (3 cos 53°/cos 53°) – (5 sec 51°/sec 51°) + [4 cot 67° cot 53°×(1/cot 67°) x (1/cot 53°]/{sin 73° sin 23° ×(1/sin 73°)×(1/sin 23°)}

On calculating further, we get,

= 3 – 5 + 4

= 2

(g) (sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)

= 0 ×sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° ×0)

(∵ sin 0° = 0 and cos 90° = 0)

We get,

= 0

(h) {(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°)} + {(3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)}

This can be written as,

= {2 sin (90° – 65°) sin (90° – 55°) sec 55° sec 65°}/{5 tan (90° – 61°)×1× tan 61°} + {3 cos (90° – 70°) cos (90° – 40°) cot (90° – 20°) cot (90° – 50°)}/(5 tan 20° tan 50° sin 70° sin 40°)

= {2 cos 65° cos 55°× (1/cos 55°)× (1/cos 65°)}/{5 cot 61° ×1× (1/cot 61°)} + {3 sin 70° sin 40° tan 20° tan 50°}/(5 tan 20° tan 50° sin 70° sin 40°)

On calculating further, we get,

= (2/5) + (3/5)

= (2 + 3)/5

= 5/5

= 1

(i) {(3 sin2 40°)/(4 cos2 50°)} – {(cosec2 28°)/(4 sec2 62°)} + {(cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 45° sin 65° sec 75°)}

= {3 sin2 (90° – 50°)/4 cos2 50°} – {cosec2 (90° – 62°)/4 sec2 62°} + {cos (90° – 80°) cos 25° ×(1/ √2) ×(1/sin 80°}/{2 sin (90° – 75°) ×(1/√2) ×sin (90°– 25°)× (1/cos 75°)

= (3 cos2 50°/4 cos2 50°) – (sec2 62°/4 sec2 62°) + (sin 80°×cos 25°×(1/sin 80°)/{2 cos 75° x cos 25° x (1/cos 75°)}

On further calculation, we get,

= (3/4) – (1/4) + (1/2)

= (1/2) + (1/2)

= 1

(j) {(5 cot 5° cot 15° cot 25° cot 35° cot 45°)/(7 tan 45° tan 55° tan 65° tan 75° tan 85°)} + {(2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)}

= {5 cot (90° – 85°) cot (90° – 75°) cot (90° – 65°) cot (90° – 55°) ×1}/(7×1× tan 55° tan 65° tan 75° tan 85°) + {2 cosec (90° – 78°) cosec (90° – 66°) cos 78° cos 66°}/{7 sin (90° – 76°) sin (90° – 67°) sec 76° sec 67°}

= (5 tan 85° tan 75° tan 65° tan 55°)/(7× tan 55° tan 65° tan 75° tan 85°) + {(2 sec 78° sec 66° ×(1/sec 78°)×(1/sec 66°)}/{7 cos 76° cos 67°×(1/cos 76°) ×(1/cos 67°)}

On further calculation, we get,

= (5/7) + (2/7)

= (5 + 2)/7

= 7/7

= 1

16. (a) Solve for ‘Î¸’ :

sin Î¸/3 = 1

(b) Solve for ‘Î¸’:

cot2 (Î¸ – 5)° = 3

(c) Solve for ‘Î¸‘ ;

sec (Î¸/2 + 10°) = 2/√3

(a)

(b)
(c)

17. Find the value of x in the following:

(i) 2 sin 3x = √3

(ii) 2 sin x/2 = 1

(iii) √3 sin x = cos x

(iv) tan x = sin 45° cos 45° + sin 30°

(v) √3 tan 2x = cos 60° + sin 45° cos 45°

(vi) cos 2x = cos 60° cos 30° + sin 60° sin 30°

(i)

(ii)
(iii)
(iv)
(v)
(vi)

18. If sin Î¸ = cos Î¸ and 0° < Î¸ < 90°, find the value of ‘Î¸’.

19. If tan Î¸ = cot Î¸ and 0° ≤ Î¸ ≤ 90°, find the value of ‘Î¸’.

20. If √2 = 1.414 and √3 = 1.732, find the value of the following correct to two decimal places.

21. (a) If
Î¸ = 30°, verify that tan 2Î¸ = (2 tan Î¸)/(1 – tan2 Î¸)

(b) If Î¸ = 30°, verify that: sin 2Î¸ = 2 tan Î¸/(1 + tan2 Î¸)

(c) cos 2Î¸ = (1 – tan2 Î¸)/(1 + tan2 Î¸) = cos4 Î¸ – sin4 Î¸ = 2 cos2 Î¸ - 1

= 1 – 2 sin2 Î¸

If A = 30°, verify that:

(d) If Î¸ = 30°, verify that :

sin 3Î¸ = 4 sin Î¸. sin (60° + Î¸)

(e) If Î¸ = 30°, verify that:

1 – sin 2Î¸ = (sin Î¸ – cos Î¸)2

(a)

(b)
(c)
(d)
(e)

22. Evaluate the following:

(a) (sin 3Î¸ – 2 sin 4Î¸)/(cos 3 Î¸ – 2 cos 4 Î¸), when 2 Î¸ = 30°

(b) (1 – cos Î¸)(1 + cos Î¸)/(1 – sin Î¸)(1 + sin Î¸), if Î¸ = 30°

(a)

(b)

23. If Î¸ = 15° , find the value of: 3/2 cos 3Î¸ – sin 6Î¸ + 3 sin (5Î¸ + 15°)- 2 tan2 3Î¸

24. If A = B = 60° , verify that:

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

(ii) sin (A – B) = sin A cos B - cos A sin B

(iii) tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

(i)

(ii)

(iii)

25. (a) If A = 30° and B = 60°, verify that:

sin (A + B) = sin A cos B + cos A sin B

(b) If A = 30° and B = 60°, verify that :

cos (A + B) = cos A cos B – sin A sin B

(c) If A = 30° B = 60°, verify that :

sin (A + B)/(cos A. cos B) = tan A + tan B

(d) If A = 30° and B = 60°, verify that:

sin (A + B)/(sin A. sin B) = cot B – cot A

(a)

(b)
(c)

(d)

26. (a) If A = B = 45°, verify that

Sin (A – B) = sin A. cos B = cos A. sin B

(b) If A = B = 45°, verify that

Cos (A – B) = cos A. cos B + sin A. sin B

(a)

(b)

27. If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the value of sin 15° and cos 15°.

28. (a) If Î¸ < 90°, find the value of:

sin2 Î¸ + cos2 Î¸

(b) If Î¸ < 90°, find the value of :

tan2 Î¸ – 1/cos2 Î¸

(c) If √3 sec 2Î¸ = 2 and Î¸ < 90°, find the value of cos2 (30°+ Î¸) + sin2 (45° - Î¸)

(a)

(b)

(c)

29. In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find

(a) cos Î¸

(b) sin2 Î¸ – cos2 Î¸

(c) Use tan Î¸ to find the value of RQ

(a)
(b)
(c)

30. Find the value of:
If 3 tan2 Î¸ – 1 = 0, find the value

(a) cos 2Î¸

(b) sin 2Î¸

(a) cos 2Î¸ = cos 2 × 30° = cos 60° = 1/2

(b) sin 3Î¸ = sin 3 × 30° = sin 90° = 1

31. If sin (A + B) = 1 and cos (A – B) = 1, find A and B.

32. If tan (A – B) = 1/√3 and tan (A + B) = √3, find A and B.

33. If sin (A – B) = 1/2 and cos (A + B) = 1/2, find A and B.

34. In ABC right angled at B, A = C. Find the value of :

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

(ii) sin A sin B + cos A cos B

(i)

(ii)

35. If tan A = 1/2, tan B = 1/3 and tan (A + B) = (tan A + tan B)/(1 – tan A tan B), find A + B.

### Exercise 27.2

1. In a right triangle ABC, right angled at C, if B = 60° and AB = 15 units, find the remaining angles and sides.

2. If ABC is a right triangle such that C = 90°, A = 45° and BC = 7 units, find B, AB and AC.

3. In a rectangle ABCD, AB = 20 cm, BAC = 60°, calculate side BC and diagonals AC and BD.

4. Find:

(a) BC

(c) AC

(a)

(b)
(c)

5. (a) Find the value ‘x’, if:

(b) Find the value ‘x’, if:

(c) Find the value ‘x’, if:

(d) Find the value ‘x’, if:

(e) Find the value ‘x’, if:

(f) Find the value ‘x’, if:

(a)

(b)
(c)
(d)
(e)
(f)

6. (a) Find the value of ‘y’ if √3 = 1.723.

(b) Find the value of ‘y’ if √3 = 1.723.

(a)

(b)

7. In the given figure, if tan Î¸ = 5/13, tan Î± = 3/5 and RS = 12 m, find the value of ‘h’.

8. (a) Find x and y, in each of the following figure:

(b) Find x and y, in each of the following figure:

(a)

(b)

9. If tan x° = 5/12. tan y° = 3/4 and AB = 48 m; find the length CD.

10. In a right-angled triangle ABC; B = 90°. Find the magnitude of angle A, if :

(a) AB is √3 times of BC.

(b) BC is √3 times of BC.

Consider the following figure,

(a)
(b)

11. A ladder is placed against a vertical tower. If the ladder makes an angle of 30° with the ground and reaches upto a height of 18 m of the tower, find the length of the ladder.

12. The perimeter of a rhombus is 100 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.

Consider the following figure,

13. In the given figure; B = 90°, ADB = 45° and AB = 24 m. Find the length of CD.

14. In the given figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically upward and then 20 km at 60° to the vertical. PQ represents the first stage of the journey and QR the second. S is a point vertically below R on the horizontal level as P, find:

(a) the height of the rocket when it is at point R.

(b) the horizontal distance of point S from P.

(a)

(b)

### Exercise 27.3

1. (a) Evaluate the following:

sin 62°/cos 28°

(b) Evaluate the following:

sec 34°/cosec 56°

(c) Evaluate the following:

tan 12°/cot 78°

(d) Evaluate the following:

sin 25° cos 43°

sin 47° cos 65°

(e) Evaluate the following:

sec 32° cot 26°/(tan 64° cosec58°)

(f) Evaluate the following:

(cos 34° cos 33°)/(sin 57° sin 56°)

(a)

(b)

(c)

(d)

(e)
(f)

2. (a) Evaluate the following:

sin 31° - cos 59°

(b) Evaluate the following:

cot 27° - tan 63°

(c) Evaluate the following:

cosec 54° - sec 36°

(d) Evaluate the following:

sin 28° sec 62° + tan 49° tan 41°

(e) Evaluate the following:

sec 16° tan 28° - cot 62° cosec 74°

(f) Evaluate the following:

sin 22° cos 44° - sin 46° cos 68°

(a)

(b)
(c)
(d)
(e)
(f)

3. (a) Evaluate the following:

sin 36°/cos 54° + sec 31°/cosec 59°

(b) Evaluate the following:

tan 42°/cot 48° - cos 33°/sin 57°

(c) Evaluate the following:

2 sin 28°/cos 62° + 3 cot 49°/tan 41°

(d) Evaluate the following:

5 sec 68°/cosec 22° + (3 sin 52° sec 38°)/(cot 51° cot 39°)

(a)

(b)

(c)
(d)

4. (a) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

sin 65° + cot 59°

(b) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

cos 72° - cos 88°

(c) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

cosec 64° + sec 70°

(d) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

tan 77° - cot 63° + sin 57°

(e) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

sin 53° + sec 66° - sin 50°

(f) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:

cos 84° + cosec 69° - cot 68°

(a)

(b)

(c)
(d)
(e)
(f)

5. (a) Evaluate the following:

sin 35° sin 45° sec 55° sec 45°

(b) Evaluate the following:

cot 20° cot 40° cot 45° cot 50° cot 70°

(c) Evaluate the following:

cos 39° cos 48°cos 60° cosec 42° cosec 51°

(d) Evaluate the following:

sin (35° + Î¸) – cos (55° - Î¸) – tan (42° + Î¸) + cot (48° - Î¸)

(e) Evaluate the following:

tan (78° + Î¸) + cosec (42° + Î¸) – cot (12° - Î¸) – sec (48° - Î¸)

(f) Evaluate the following:

(3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + 4 tan 23° tan 37° tan 67° tan 53°/cos 17° cos 67° cosec 73° cosec 23°)

(g) Evaluate the following:

(sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)

(h) Evaluate the following:

(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°) + (3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)

(i) Evaluate the following:

(3 sin2 40° /4 cos2 50°) – (cosec2 28°/ 4 sec2 62°) + (cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 25° sin 45° sin 65° sec 75°)

(j) Evaluate the following:

(5 cot 5° cot 15° cot 25° cot 35° cot 45°)/(7 tan 45° tan 55° tan 65° tan 75° tan 85°) + (2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)

(a)

(b)

(c)
(d)
(e)
(f)
(g)
(h)

(i)
(j)

6. If cos 3Î¸ = sin (Î¸ - 34°), find the value of ° if 3Î¸ is an acute angle.

7. If 4Î¸ = cot (Î¸ + 20°), find the value of ° if 4° is an acute angle.

8. If sec 2Î¸ = cosec 3Î¸, find the value of Î¸ if it is known that both 2Î¸ and 3Î¸ are acute angles.

9. If sin (Î¸ - 15°) = cos (Î¸ - 25°), find the value of (Î¸ - 15°) and (Î¸ - 25°) are acute angles.

10. If A, B and C are interior angles of ABC, Prove that sin (A + B)/2 = cos C/2

11. If P, Q and R are the interior angles of PQR, prove that cot (Q + R)/2 = tan P/2

12. If cos Î¸ = sin 60° and Î¸ is an acute angle find the value of 1 – 2 sin2 Î¸

13. If sec Î¸ = cosec 30° and Î¸ is an acute angle, find the value of 4 sin2Î¸ – 2 cos2Î¸.

14. Prove the following :

(a) tan Î¸ tan (90° – Î¸) = cot Î¸ cot (90° – Î¸)

(b) Prove the following :

sin 58° sec 32° + cos 58° cosec 32° = 2

(c) Prove the following :

tan (90° - Î¸) cot Î¸/(cosec2 Î¸) = cos2 Î¸

(d) Prove the following :

sin2 30° + cos2 30° = 1/2 sec 60°