Frank Solutions Class 9 Maths Chapter 27 Trigonometrical Ratios of Standard Angles

Frank Solutions for Class 9 Maths Chapter 27 Trigonometrical Ratios of Standard Angles are provided here. This chapter plays a vital role in the Class 9 syllabus, as it is continued in higher classes as well. Students who aim to secure better academic scores are advised to practise Frank Solutions on a regular basis.

Chapter 27 provides basic information about the Trigonometrical Ratios of Standard Angles. The solutions are formulated with the intention of helping students boost their exam preparation. It also builds the confidence of students to solve more complex problems within a short duration. Students can easily access Frank Solutions for Class 9 Maths Chapter 27 Trigonometrical Ratios of Standard Angles PDF from the link given below.

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1. Without using tables, evaluate the following:

(i) sin 60⁰ sin 30⁰ + cos 30⁰ cos 60⁰

(ii) sec 30⁰ cosec 60⁰ + cos 60⁰ sin 30⁰

(iii) sec 45⁰ sin 45⁰ – sin 30⁰ sec 60⁰

(iv) sin² 30⁰ sin²45⁰ + sin² 60⁰ sin² 90⁰

(v) tan² 30⁰ + tan² 60⁰ + tan² 45⁰

(vi) sin² 30⁰ cos² 45⁰ + 4 tan² 30⁰ + sin² 90⁰ + cos² 0⁰

(vii) cosec² 45⁰ sec² 30⁰ – sin² 30⁰ – 4 cot² 45⁰ + sec² 60⁰

(viii) cosec³ 30⁰ cos 60⁰ tan³ 45⁰ sin² 90⁰ sec² 45⁰ cot 30⁰

(ix) (sin 90⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ – cos 45⁰ + cos 60⁰)

(x) 4(sin⁴ 30⁰ + cos⁴ 60⁰) – 3 (cos² 45⁰ – sin² 90⁰)

Solution:

(i) sin 60⁰ sin 30⁰ + cos 30⁰ cos 60⁰

sin 60⁰ = (√3 / 2)

sin 30⁰ = (1 / 2)

cos 30⁰ = (√3 / 2)

cos 60⁰ = (1 / 2)

On substituting, we get,

sin 60⁰ sin 30⁰ + cos 30⁰ cos 60⁰ = (√3 / 2) x (1 / 2) + (√3 / 2) x (1/ 2)

= (√3 / 4) + (√3 / 4)

We get,

= (√3 / 2)

(ii) sec 30⁰ cosec 60⁰ + cos 60⁰ sin 30⁰

cos 30⁰ = (√3 / 2) ⇒ sec 30⁰ = (2 / √3)

sin 60⁰ = (√3 / 2) ⇒ cosec 60⁰ = (2 /√3)

cos 60⁰ = (1 / 2)

sin 30⁰ = (1 / 2)

On substituting, we get,

sec 30⁰ cosec 60⁰ + cos 60⁰ sin 30⁰ = (2 / √3) x (2 / √3) + (1 / 2) x (1 / 2)

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

= (16 + 3) / 12

We get,

= 19 / 12

(iii) sec 45⁰ sin 45⁰ – sin 30⁰ sec 60⁰

cos 45⁰ = (1 / √2) ⇒ sec 45⁰ = √2

sin 45⁰ = (1 / √2)

sin 30⁰ = (1 / 2)

cos 60⁰ = (1 / 2) ⇒ sec 60⁰ = 2

sec 45⁰ sin 45⁰ – sin 30⁰ sec 60⁰

On substituting, we get,

= (√2) x (1 / √2) – (1 / 2) x 2

= 1 – 1

We get,

= 0

(iv) sin² 30⁰ sin² 45⁰ + sin² 60⁰ sin² 90⁰

sin 30⁰ = (1 / 2)

sin 45⁰ = (1 / √2)

sin 60⁰ = (√3 / 2)

sin 90⁰ = 1

On substituting, we get,

sin² 30⁰ sin² 45⁰ + sin² 60⁰ sin² 90⁰ = (1 / 2)² (1 / √2)² + (√3 / 2)² (1)²

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

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

= (1 + 6) / 8

We get,

= 7 / 8

(v) tan² 30⁰ + tan² 60⁰ + tan² 45⁰

tan 30⁰ = (1 / √3)

tan 60⁰ = √3

tan 45⁰ = 1

On substituting, we get,

tan² 30⁰ + tan² 60⁰ + tan² 45⁰ = (1 / √3)² + (√3)² + 1

= (1 / 3) + 3 + 1

= (1 + 9 + 3) / 3

We get,

= 13 / 3

(vi) sin² 30⁰ cos² 45⁰ + 4 tan² 30⁰ + sin² 90⁰ + cos² 0⁰

sin 30⁰ = (1 / 2)

cos 45⁰ = (1 / √2)

tan 30⁰ = (1 / √3)

sin 90⁰ = 1

cos 0⁰ = 1

On substituting, we get,

sin² 30⁰ cos² 45⁰ + 4 tan² 30⁰ + sin² 90⁰ + cos² 0⁰

= (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 45⁰ sec² 30⁰ – sin² 30⁰ – 4 cot² 45⁰ + sec² 60⁰

sin 45⁰ = (1 / √2)

cosec 45⁰ = (√2 / 1)

sin 30⁰ = cos 60⁰ = (1 / 2)

sec 60⁰ = 2

cos 30⁰ = (√3 / 2)

sec 30⁰ = (2 / √3)

tan 45⁰ = 1

cot 45⁰ = 1

On substituting, we get,

cosec² 45⁰ sec² 30⁰ – sin² 30⁰ – 4 cot² 45⁰ + sec² 60⁰ = (√2 / 1)² (2 / √3)² – (1 / 2)² – 4 (1)² + (2)²

= 2 x (4 / 3) – (1 / 4) – 4 + 4

= (8 / 3) – (1 / 4)

= (32 – 3) / 12

We get,

= 29 / 12

(viii) cosec³ 30⁰ cos 60⁰ tan³ 45⁰ sin² 90⁰ sec² 45⁰ cot 30⁰

sin 30⁰ = (1 / 2)

cosec 30⁰ = 2

cos 60⁰ = (1 / 2)

sec 60⁰ = 2

cos 45⁰ = (1 / √2)

sec 45⁰ = √2

tan 45⁰ = 1

sin 90⁰ = 1

tan 30⁰ = (1 / √3)

cot 30⁰ = √3

On substituting, we get,

cosec³ 30⁰ cos 60⁰ tan³ 45⁰ sin² 90⁰ sec² 45⁰ cot 30⁰

= (2)³ (1 / 2) (1)³ (1)² (√2)² (√3)

= 8 x (1 / 2) x 2 x √ 3

We get,

= 8√3

(ix) (sin 90⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ – cos 45⁰ + cos 60⁰)

sin 30⁰ = (1 / 2)

sin 45⁰ = (1 / √2)

sin 90⁰ = 1

cos 45⁰ = (1 / √2)

cos 60⁰ = (1 / 2)

(sin 90⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ – cos 45⁰ + cos 60⁰)

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 30⁰ + cos⁴ 60⁰) – 3 (cos² 45⁰ – sin² 90⁰)

sin 30⁰ = (1 / 2)

sin 90⁰ = 1

cos 45⁰ = (1 / √2)

cos 60⁰ = (1 / 2)

On substituting, we get,

4 (sin⁴ 30⁰ + cos⁴ 60⁰) – 3 (cos² 45⁰ – sin² 90⁰)

= 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 x (2 / 16) + 3 x (1 / 2)

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

= 4 / 2

We get,

= 2

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

(i) (sin 30⁰ – sin 90⁰ + 2 cos 0⁰) / tan 30⁰ tan 60⁰

(ii) (sin 30⁰ / sin 45⁰) + (tan 45⁰ / sec 60⁰) – (sin 60⁰ / cot 45⁰) – (cos 30⁰ / sin 90⁰)

(iii) (tan 45⁰ / cosec 30⁰) + (sec 60⁰ / cot 45⁰) – (5 sin 90⁰ / 2 cos 0⁰)

(iv) (tan² 60⁰ + 4 cos² 45⁰ + 3 sec² 30⁰ + 5 cos 90⁰) / (cosec 30⁰ + sec 60⁰ – cot² 30⁰)

(v) (4 / cot² 30⁰) + (1 / sin² 60⁰) – cos² 45⁰

Solution:

(i) (sin 30⁰ – sin 90⁰ + 2 cos 0⁰) / tan 30⁰ tan 60⁰

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

On further calculation, we get,

= {(1 / 2) – 1 + 2} / 1

= (1 / 2) – 1 + 2

= (1 / 2) + 1

We get,

= 3 / 2

(ii) (sin 30⁰ / sin 45⁰) + (tan 45⁰ / sec 60⁰) – (sin 60⁰ / cot 45⁰) – (cos 30⁰ / sin 90⁰)

= {(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 45⁰ / cosec 30⁰) + (sec 60⁰ / cot 45⁰) – (5 sin 90⁰ / 2 cos 0⁰)

= (1 / 2) + (2 / 1) – (5 x 1) / (2 x 1)

On further calculation, we get,

= (1 / 2) + (2 / 1) – (5 / 2)

= (1 + 4 – 5) / 2

We get,

= 0

(iv) (tan² 60⁰ + 4 cos² 45⁰ + 3 sec² 30⁰ + 5 cos 90⁰) / (cosec 30⁰ + sec 60⁰ – cot² 30⁰)

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

On further calculation, we get,

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

= (3 + 2 + 4) / (4 – 3)

We get,

= 9

(v) (4 / cot² 30⁰) + (1 / sin² 60⁰) – cos² 45⁰

= {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 60⁰. cos 30⁰ – sin 60⁰. sin 30⁰ = (1 / 2)

(b) cos 60⁰. Cos 30⁰ – sin 60⁰. sin 30⁰ = 0

(c) sec² 45⁰ – tan² 45⁰ = 1

(d) {(cot 30⁰ + 1) / (cot 30⁰ – 1)}² = (sec 30⁰ + 1) / (sec 30⁰ – 1)

Solution:

(a) Consider L.H.S.

sin 60⁰. cos 30⁰ – cos 60⁰. sin 30⁰

= (√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 60⁰. cos 30⁰ – sin 60⁰. sin 30⁰

= (1 / 2) x (√3 / 2) – (√3 / 2) x (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 45⁰ – tan² 45⁰

= (√2)² – (1)²

= 2 – 1

= 1

= R.H.S.

Hence, proved

(d) Consider L.H.S.

{(cot 30⁰ + 1) / (cot 30⁰ – 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 30⁰ + 1) / (sec 30⁰ – 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

Solution:

(a) 2 cos A = 1

cos A = (1 / 2)

cos A = cos 60⁰

A = 60⁰

Therefore, the value of ‘A’ is 60⁰

(b) 2 sin 2A = 1

sin 2A = (1 / 2)

sin 2A = sin 30⁰

2A = 30⁰

We get,

A = 15⁰

Therefore, the value of ‘A’ is 15⁰

(c) cosec 3A = (2 / √3)

cosec 3A = cosec 60⁰

3A = 60⁰

We get,

A = 20⁰

Therefore, the value of ‘A’ is 20⁰

(d) 2 cos 3A = 1

cos 3A = (1 / 2)

cos 3A = cos 60⁰

3A = 60⁰

We get,

A = 20⁰

Therefore, the value of ‘A’ is 20⁰

(e) √3 cot A = 1

cot A = (1 / √3)

cot A = cot 60⁰

A = 60⁰

Therefore, the value of ‘A’ is 60⁰

(f) cot 3A = 1

cot 3A = cot 45⁰

3A = 45⁰

We get,

A = 15⁰

Therefore, the value of ‘A’ is 15⁰

5. Find the value of ‘A’, if

(a) (1 – cosec A) (2 – sec A) = 0

(b) (2 – cosec 2A) cos 3A = 0

Solution:

(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 90⁰ and sec A = sec 60⁰

A = 90⁰ and A = 60⁰

(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 30⁰ and cos 3A = cos 90⁰

We get,

2A = 30⁰ and 3A = 90⁰

A = 15⁰ and A = 30⁰

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

(a)

(b)

(c)

(d)

Solution:

(a)

From the figure,

We have,

sin 60⁰ = BC / AC

√3 / 2 = 12 / x

On simplification, we get,

x = (2 x 12) / √3

x = 24 / √3

x = (8 x 3) / √3

We get,

x = 8√3

(b)

From the figure,

We have

tan 45⁰ = 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 60⁰

We get,

x = 60⁰

(d)

From the figure,

We have,

sin x = BC / AC

sin x = (15 / √2) / 15

sin x = (1 / √2)

sin x = sin 45⁰

We get,

x = 45⁰

7. Find the length of AD.

Given: ∠ABC = 60⁰, ∠DBC = 45⁰ and BC = 24 cm.

Solution:

In △ABC,

tan 60⁰ = AC / BC

√3 = AC / 24

We get,

AC = 24√3 cm

In △DBC,

tan 45⁰ = DC / BC

1 = DC / 24

We get,

DC = 24 cm

Now,

AC = AD + DC

AD = AC – DC

Substituting the values of AC and DC, we get,

AD = 24√3 – 24

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 = 60⁰

Solution:

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 = 90⁰

Given

∠BAD = 60⁰

⇒ ∠OAB = (1/ 2) ∠BAD

⇒ ∠OAB = 30⁰

In right-angled △AOB,

sin 30⁰ = OB / AB

= (1 / 2)

Given AB = 24,

⇒OB / 24 = (1 / 2)

OB = 24 / 2

We get,

OB = 12 cm

cos 30⁰ = OA / AB

= √3 / 2

⇒OA / 24 = √3 / 2

OA = (24√3) / 2

We get,

OA = 12√3 cm

Therefore,

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

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

9. Find the length of EC.

Solution:

CD = 28 cm

⇒ AB = 28 cm

In the right △ABE,

tan 30⁰ = BE / AB

1 / √3 = BE / 28

We get,

BE = (28 / √3)

In the right △ABC,

tan 60⁰ = 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 = 45⁰ and ∠ACD = 30⁰. Find:

(a) AB

(b) AC

(c) AE

Solution:

(a) In right △ADC,

tan 30⁰ = AD / DC

(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

(b) In the right △ADC,

sin 30⁰ = AD / AC

(1 / 2) = 1.5 / AC (given AD = 1.5 cm)

AC = 2 x 1.5

We get,

AC = 3 cm

(c) In the right △ADE,

sin 45⁰ = AD / AE

(1 / √2) = 1.5 / AE (given AD = 1.5 cm)

We get,

AE = 1.5√2 cm

11. Evaluate the following:

(a) sin 62⁰ / cos 28⁰

(b) sec 34⁰ / cosec 56⁰

(c) tan 12⁰ / cot 78⁰

(d) sin 25⁰ cos 43⁰ / sin 47⁰ cos 65⁰

(e) sec 32⁰ cot 26⁰ / tan 64⁰ cosec 58⁰

(f) cos 34⁰ cos 33⁰ / sin 57⁰ sin 56⁰

Solution:

(a) sin 62⁰ / cos 28⁰

This can be written as,

= sin (90⁰ – 28⁰) / cos 28⁰

= cos 28⁰ / cos 28⁰

We get,

= 1

(b) sec 34⁰ / cosec 56⁰

This can be written as,

= sec (90⁰ – 56⁰) / cosec 56⁰

= cosec 56⁰ / cosec 56⁰

We get,

= 1

(c) tan 12⁰ / cot 78⁰

This can be written as,

= tan (90⁰ – 78⁰) / cot 78⁰

= cot 78⁰ / cot 78⁰

We get,

= 1

(d) sin 25⁰ cos 43⁰ / sin 47⁰ cos 65⁰

This can be written as,

= sin (90⁰ – 65⁰) cos (90⁰ – 47⁰) / sin 47⁰ cos 65⁰

= cos 65⁰ sin 47⁰ / sin 47⁰ cos 65⁰

We get,

= 1

(e) sec 32⁰ cot 26⁰ / tan 64⁰ cosec 58⁰

This can be written as,

= sec (90⁰ – 58⁰) cot (90⁰ – 64⁰) / tan 64⁰ cosec 58⁰

= cosec 58⁰ tan 64⁰ / tan 64⁰ cosec 58⁰

We get,

= 1

(f) cos 34⁰ cos 33⁰ / sin 57⁰ sin 56⁰

This can be written as,

= cos (90⁰ – 56⁰) cos (90⁰ – 57⁰) / sin 57⁰ sin 56⁰

= sin 56⁰ sin 57⁰ / sin 57⁰ sin 56⁰

We get,

= 1

12. Evaluate the following:

(a) sin 31⁰ – cos 59⁰

(b) cot 27⁰ – tan 63⁰

(c) cosec 54⁰ – sec 36⁰

(d) sin 28⁰ sec 62⁰ + tan 49⁰ tan 41⁰

(e) sec 16⁰ tan 28⁰ – cot 62⁰ cosec 74⁰

(f) sin 22⁰ cos 44⁰ – sin 46⁰ cos 68⁰

Solution:

(a) sin 31⁰ – cos 59⁰

= sin (90⁰ – 59⁰) – cos 59⁰

= cos 59⁰ – cos 59⁰

We get,

= 0

(b) cot 27⁰ – tan 63⁰

= cot (90⁰ – 63⁰) – tan 63⁰

= tan 63⁰ – tan 63⁰

We get,

= 0

(c) cosec 54⁰ – sec 36⁰

= cosec (90⁰ – 36⁰) – sec 36⁰

= sec 36⁰ – sec 36⁰

We get,

= 0

(d) sin 28⁰ sec 62⁰ + tan 49⁰ tan 41⁰

= sin 28⁰ sec (90⁰ – 28⁰) + tan 49⁰ tan (90⁰ – 49⁰)

= sin 28⁰ cosec 28⁰ + tan 49⁰ cot 49⁰

= sin 28⁰ x (1 / sin 28⁰) + tan 49⁰ x (1 / tan 49⁰)

= 1 + 1

We get,

= 2

(e) sec 16⁰ tan 28⁰ – cot 62⁰ cosec 74⁰

= sec (90⁰ – 74⁰) tan (90⁰ – 62⁰) – cot 62⁰ cosec 74⁰

= cosec 74⁰ cot 62⁰ – cot 62⁰ cosec 74⁰

We get,

= 0

(f) sin 22⁰ cos 44⁰ – sin 46⁰ cos 68⁰

= sin (90⁰ – 68⁰) cos (90⁰ – 46⁰) – sin 46⁰ cos 68⁰

= cos 68⁰ sin 46⁰ – sin 46⁰ cos 68⁰

We get,

= 0

13. Evaluate the following:

(a) (sin 36⁰ / cos 54⁰) + (sec 31⁰ / cosec 59⁰)

(b) (tan 42⁰ / cot 48⁰) – (cos 33⁰ / sin 57⁰)

(c) (2 sin 28⁰ / cos 62⁰) + (3 cot 49⁰ / tan 41⁰)

(d) (5 sec 68⁰ / cosec 22⁰) + (3 sin 52⁰ sec 38⁰) / (cot 51⁰ cot 39⁰)

Solution:

(a) (sin 36⁰ / cos 54⁰) + (sec 31⁰ / cosec 59⁰)

This can be written as,

= {sin (90⁰ – 54⁰) / cos 54⁰} + {sec (90⁰ – 59⁰) / cosec 59⁰}

= (cos 54⁰ / cos 54⁰) + (cosec 59⁰ / cosec 59⁰)

= 1 + 1

We get,

= 2

(b) (tan 42⁰ / cot 48⁰) – (cos 33⁰ / sin 57⁰)

This can be written as,

= {tan (90⁰ – 48⁰) / cot 48⁰} – {cos (90⁰ – 57⁰) / sin 57⁰}

= (cot 48⁰ / cot 48⁰) – (sin 57⁰ / sin 57⁰)

= 1 – 1

We get,

= 0

(c) (2 sin 28⁰ / cos 62⁰) + (3 cot 49⁰ / tan 41⁰)

This can be written as,

= {2 sin (90⁰ – 62⁰) / cos 62⁰} + {3 cot (90⁰ – 41⁰) / tan 41⁰}

= (2 cos 62⁰ / cos 62⁰) + (3 tan 41⁰ / tan 41⁰)

= 2 + 3

We get,

= 5

(d) (5 sec 68⁰ / cosec 22⁰) + (3 sin 52⁰ sec 38⁰) / (cot 51⁰ cot 39⁰)

This can be written as,

= {5 sec (90⁰ – 22⁰) / cosec 22⁰} + {3 sin 52⁰ sec (90⁰ – 52⁰) / cot 51⁰ cot (90⁰ – 51⁰)}

= (5 cosec 22⁰ / cosec 22⁰) + (3 sin 52⁰ cosec 52⁰ / cot 51⁰ tan 51⁰)

= 5 + {3 sin 52⁰ x (1 / sin 52⁰) / cot 51⁰ x (1 / cot 51⁰)}

= 5 + 3 / 1

= 5 + 3

We get,

= 8

14. Express each of the following in terms of trigonometric ratios of angles between 0⁰ and 45⁰

(a) sin 65⁰ + cot 59⁰

(b) cos 72⁰ – cos 88⁰

(c) cosec 64⁰ + sec 70⁰

(d) tan 77⁰ – cot 63⁰ + sin 57⁰

(e) sin 53⁰ + sec 66⁰ – sin 50⁰

(f) cos 84⁰ + cosec 69⁰ – cot 68⁰

Solution:

(a) sin 65⁰ + cot 59⁰

This can be written as,

= sin (90⁰ – 25⁰) + cot (90⁰ – 31⁰)

We get,

= cos 25⁰ + tan 31⁰

(b) cos 72⁰ – cos 88⁰

This can be written as,

= cos (90⁰ – 18⁰) – cos (90⁰ – 2⁰)

We get,

= sin 18⁰ – sin 2⁰

(c) cosec 64⁰ + sec 70⁰

This can be written as,

= cosec (90⁰ – 26⁰) + sec (90⁰ – 20⁰)

We get,

= sec 26⁰ + cosec 20⁰

(d) tan 77⁰ – cot 63⁰ + sin 57⁰

This can be written as,

= tan (90⁰ – 13⁰) – cot (90⁰ – 27⁰) + sin (90⁰ – 33⁰)

We get,

= cot 13⁰ – tan 27⁰ + cos 33⁰

(e) sin 53⁰ + sec 66⁰ – sin 50⁰

This can be written as,

= sin (90⁰ – 37⁰) + sec (90⁰ – 24⁰) – sin (90⁰ – 40⁰)

We get,

= cos 37⁰ + cosec 24⁰ – cos 40⁰

(f) cos 84⁰ + cosec 69⁰ – cot 68⁰

This can be written as,

= cos (90⁰ – 6⁰) + cosec (90⁰ – 21⁰) – cot (90⁰ – 22⁰)

We get,

= sin 6⁰ + sec 21⁰ – tan 22⁰

15. Evaluate the following:

(a) sin 35⁰ sin 45⁰ sec 55⁰ sec 45⁰

(b) cot 20⁰ cot 40⁰ cot 45⁰ cot 50⁰ cot 70⁰

(c) cos 39⁰ cos 48⁰ cos 60⁰ cosec 42⁰ cosec 51⁰

(d) sin (35⁰ + θ) – cos (55⁰ – θ) – tan (42⁰ + θ) + cot (48⁰ – θ)

(e) tan (78⁰ + θ) + cosec (42⁰ + θ) – cot (12⁰ – θ) – sec (48⁰ – θ)

(f) (3 sin 37⁰ / cos 53⁰) – (5 cosec 39⁰ / sec 51⁰) + {(4 tan 23⁰ tan 37⁰ tan 67⁰ tan 53⁰) / (cos 17⁰ cos 67⁰ cosec 73⁰ cosec 23⁰)}

(g) (sin 0⁰ sin 35⁰ sin 55⁰ sin 75⁰) / (cos 22⁰ cos 64⁰ cos 68⁰ cos 90⁰)

(h) {(2 sin 25⁰ sin 35⁰ sec 55⁰ sec 65⁰) / (5 tan 29⁰ tan 45⁰ tan 61⁰)} + {(3 cos 20⁰ cos 50⁰ cot 70⁰ cot 40⁰) / (5 tan 20⁰ tan 50⁰ sin 70⁰ sin 40⁰)}

(i) {(3 sin² 40⁰) / (4 cos² 50⁰)} – {(cosec² 28⁰) / (4 sec² 62⁰)} + {(cos 10⁰ cos 25⁰ cos 45⁰ cosec 80⁰) / (2 sin 15⁰ sin 45⁰ sin 65⁰ sec 75⁰)}

(j) {(5 cot 5⁰ cot 15⁰ cot 25⁰ cot 35⁰ cot 45⁰) / (7 tan 45⁰ tan 55⁰ tan 65⁰ tan 75⁰ tan 85⁰)} + {(2 cosec 12⁰ cosec 24⁰ cos 78⁰ cos 66⁰) / (7 sin 14⁰ sin 23⁰ sec 76⁰ sec 67⁰)}

Solution:

(a) sin 35⁰ sin 45⁰ sec 55⁰ sec 45⁰

This can be written as,

= sin (90⁰ – 55⁰) x (1 / √2) x (1 / cos 55⁰) x (√2)

= cos 55⁰ x (1 / cos 55⁰) x (1 / √2) x (√2)

We get,

= 1

(b) cot 20⁰ cot 40⁰ cot 45⁰ cot 50⁰ cot 70⁰

This can be written as,

= cot (90⁰ – 70⁰) x cot (90⁰ – 50⁰) x 1 x cot 50⁰ cot 70⁰

= tan 70⁰ x tan 50⁰ x cot 50⁰ x cot 70⁰

= tan 70⁰ x cot 70⁰ x tan 50⁰ x cot 50⁰

= tan 70⁰ x (1 / tan 70⁰) x tan 50⁰ x (1 / tan 50⁰)

We get,

= 1

(c) cos 39⁰ cos 48⁰ cos 60⁰ cosec 42⁰ cosec 51⁰

This can be written as,

= cos (90⁰ – 51⁰) x cos (90⁰ – 42⁰) x (1 / 2) x (1 / sin 42⁰) x (1 / sin 51⁰)

= sin 51⁰ x sin 42⁰ x (1 / 2) x (1 / sin 42⁰) x (1 / sin 51⁰)

We get,

= 1 / 2

(d) sin (35⁰ + θ) – cos (55⁰ – θ) – tan (42⁰ + θ) + cot (48⁰ – θ)

This can be written as,

= sin {90⁰ – (55⁰ – θ)} – cos (55⁰ – θ) – tan {90⁰ – (48⁰ – θ)} + cot (48⁰ – θ)

= cos (55⁰ – θ) – cos (55⁰ – θ) – cot (48⁰ – θ) + cot (48⁰ – θ)

We get,

= 0

(e) tan (78⁰ + θ) + cosec (42⁰ + θ) – cot (12⁰ – θ) – sec (48⁰ – θ)

This can be written as,

= tan {90⁰ – (12⁰ – θ)} + cosec {90⁰ – (48⁰ – θ)} – cot (12⁰ – θ) – sec (48⁰ – θ)

= cot (12⁰ – θ) + sec (48⁰ – θ) – cot (12⁰ – θ) – sec (48⁰ – θ)

We get,

= 0

(f) (3 sin 37⁰ / cos 53⁰) – (5 cosec 39⁰ / sec 51⁰) + {(4 tan 23⁰ tan 37⁰ tan 67⁰ tan 53⁰) / (cos 17⁰ cos 67⁰ cosec 73⁰ cosec 23⁰)}

This can be written as,

= {3 sin (90⁰ – 53⁰) / cos 53⁰} – {5 cosec (90⁰ – 51⁰) / sec 51⁰} + [{4 tan (90⁰ – 67⁰) tan (90⁰ – 53⁰) x (1 / cot 67⁰) x (1 / cot 53⁰)}] / {cos (90⁰ – 73⁰) cos (90⁰ – 23⁰) x (1 / sin 73⁰) x (1 / sin 23⁰)}

= (3 cos 53⁰ / cos 53⁰) – (5 sec 51⁰ / sec 51⁰) + [4 cot 67⁰ cot 53⁰ x (1 / cot 67⁰) x (1 / cot 53⁰] / {sin 73⁰ sin 23⁰ x (1 / sin 73⁰) x (1 / sin 23⁰)}

On calculating further, we get,

= 3 – 5 + 4

= 2

(g) (sin 0⁰ sin 35⁰ sin 55⁰ sin 75⁰) / (cos 22⁰ cos 64⁰ cos 68⁰ cos 90⁰)

= 0 x sin 35⁰ sin 55⁰ sin 75⁰) / (cos 22⁰ cos 64⁰ cos 68⁰ x 0)

(∵ sin 0⁰ = 0 and cos 90⁰ = 0)

We get,

= 0

(h) {(2 sin 25⁰ sin 35⁰ sec 55⁰ sec 65⁰) / (5 tan 29⁰ tan 45⁰ tan 61⁰)} + {(3 cos 20⁰ cos 50⁰ cot 70⁰ cot 40⁰) / (5 tan 20⁰ tan 50⁰ sin 70⁰ sin 40⁰)}

This can be written as,

= {2 sin (90⁰ – 65⁰) sin (90⁰ – 55⁰) sec 55⁰ sec 65⁰} / {5 tan (90⁰ – 61⁰) x 1 x tan 61⁰} + {3 cos (90⁰ – 70⁰) cos (90⁰ – 40⁰) cot (90⁰ – 20⁰) cot (90⁰ – 50⁰)} / (5 tan 20⁰ tan 50⁰ sin 70⁰ sin 40⁰)

= {2 cos 65⁰ cos 55⁰ x (1 / cos 55⁰) x (1 / cos 65⁰)} / {5 cot 61⁰ x 1 x (1 / cot 61⁰)} + {3 sin 70⁰ sin 40⁰ tan 20⁰ tan 50⁰} / (5 tan 20⁰ tan 50⁰ sin 70⁰ sin 40⁰)

On calculating further, we get,

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

= (2 + 3) / 5

= 5 / 5

= 1

(i) {(3 sin² 40⁰) / (4 cos² 50⁰)} – {(cosec² 28⁰) / (4 sec² 62⁰)} + {(cos 10⁰ cos 25⁰ cos 45⁰ cosec 80⁰) / (2 sin 15⁰ sin 45⁰ sin 65⁰ sec 75⁰)}

= {3 sin² (90⁰ – 50⁰) / 4 cos² 50⁰} – {cosec² (90⁰ – 62⁰) / 4 sec² 62⁰} + {cos (90⁰ – 80⁰) cos 25⁰ x (1 / √2) x (1 / sin 80⁰} / {2 sin (90⁰ – 75⁰) x (1 / √2) x sin (90⁰ – 25⁰) x (1 / cos 75⁰)

= 3 cos² 50⁰ / 4 cos² 50⁰) – (sec² 62⁰ / 4 sec² 62⁰) + (sin 80⁰ x cos 25⁰ x (1 / sin 80⁰) / {2 cos 75⁰ x cos 25⁰ x (1 / cos 75⁰)}

On further calculation, we get,

= (3 / 4) – (1 / 4) + (1 / 2)

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

= 1

(j) {(5 cot 5⁰ cot 15⁰ cot 25⁰ cot 35⁰ cot 45⁰) / (7 tan 45⁰ tan 55⁰ tan 65⁰ tan 75⁰ tan 85⁰)} + {(2 cosec 12⁰ cosec 24⁰ cos 78⁰ cos 66⁰) / (7 sin 14⁰ sin 23⁰ sec 76⁰ sec 67⁰)}

= {5 cot (90⁰ – 85⁰) cot (90⁰ – 75⁰) cot (90⁰ – 65⁰) cot (90⁰ – 55⁰) x 1} / (7 x 1 x tan 55⁰ tan 65⁰ tan 75⁰ tan 85⁰) + {2 cosec (90⁰ – 78⁰) cosec (90⁰ – 66⁰) cos 78⁰ cos 66⁰} / {7 sin (90⁰ – 76⁰) sin (90⁰ – 67⁰) sec 76⁰ sec 67⁰}

= (5 tan 85⁰ tan 75⁰ tan 65⁰ tan 55⁰) / (7 x tan 55⁰ tan 65⁰ tan 75⁰ tan 85⁰) + {(2 sec 78⁰ sec 66⁰ x (1 / sec 78⁰) x (1 / sec 66⁰)} / {7 cos 76⁰ cos 67⁰ x (1 / cos 76⁰) x (1 / cos 67⁰)}

On further calculation, we get,

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

= (5 + 2) / 7

= 7 / 7

= 1