# Concise Selina Solutions for Class 9 Maths Chapter 23 - Trigonometrical Ratios of Standard Angles

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Exercise 23A Page: 291

1. Find the value of:

(i) sin 30oÂ cosÂ 30o

(ii) tan 30oÂ tan 60o

(iii) cos2Â 60oÂ + sin2Â 30o

(iv) cosec2Â 60oÂ – tan2Â 30o

(v) sin2Â 30oÂ + cos2Â 30oÂ + cot2Â 45o

(vi)Â cos2Â 60oÂ + sec2Â 30oÂ + tan2Â 45o.

Solution:

(i) Given sin 30oÂ cosÂ 30o

By substituting the values, we get

sin 30oÂ cosÂ 30o = Â½ (âˆš3/2)

= âˆš3/4

(ii) Given tan 30oÂ tan 60o

By substituting the values, we get

tan 30oÂ tan 60o = 1/âˆš3 (âˆš3)

= 1

(iii) Given cos2Â 60oÂ + sin2Â 30o

By substituting the values, we get

cos2Â 60oÂ + sin2Â 30o = (Â½)2 +(Â½)2

= Â¼ + Â¼

= Â½

(iv) Given cosec2Â 60oÂ – tan2Â 30o

By substituting the values, we get

cosec2Â 60oÂ – tan2Â 30o = (2/âˆš3)2 â€“ (1/âˆš3)2

= 4/3 â€“ 1/3

= 1

(v) Given sin2Â 30oÂ + cos2Â 30oÂ + cot2Â 45o

By substituting the values, we get

sin2Â 30oÂ + cos2Â 30oÂ + cot2Â 45o = (Â½)2 + (âˆš3/2)2 + 12

= Â¼ + Â¾ + 1

= 2

(vi) Given cos2Â 60oÂ + sec2Â 30oÂ + tan2Â 45o

By substituting the values, we get

cos2Â 60oÂ + sec2Â 30oÂ + tan2Â 45o = (Â½)2 + (2/âˆš3)2 + 12

= Â¼ + 4/3 + 1

= 31/12

2. Find the value of:

(i) tan2Â 30oÂ + tan2Â 45oÂ + tan2Â 60o

(ii)Â

(iii) 3 sin2Â 30oÂ + 2 tan2Â 60oÂ – 5 cos2Â 45o.

Solution:

(i) Given tan2Â 30oÂ + tan2Â 45oÂ + tan2Â 60o

By substituting the values, we get

tan2Â 30oÂ + tan2Â 45oÂ + tan2Â 60o = (1//âˆš3)2 + 12 + (/âˆš3)2

= 1/3 + 1 + 3

= 13/3

= 4 1/3

(ii) Given

Â

By substituting the values, we get

= Â½ + 2/1 â€“ 5/2

= (1 + 4 â€“ 5)/2

= 0

(iii) Given 3 sin2Â 30oÂ + 2 tan2Â 60oÂ – 5 cos2Â 45o.

By substituting the values, we get

3 sin2Â 30oÂ + 2 tan2Â 60oÂ – 5 cos2Â 45o. = 3 (Â½)2 + 2 (âˆš3)2 + 5 (1/âˆš3)2

= Â¾ + 6 â€“ 5/2

= (3 + 24 â€“ 10)/4

= 4 Â¼

3. Prove that:

(i) sin 60oÂ cosÂ 30oÂ +Â cosÂ 60o. sin 30oÂ = 1

(ii)Â cosÂ 30o.Â cosÂ 60oÂ – sin 30o. sin 60oÂ = 0

(iii) cosec2Â 45oÂ – cot2Â 45oÂ = 1

(iv) cos2Â 30oÂ – sin2Â 30oÂ =Â cosÂ 60o.

(v)

(vi) 3 cosec2Â 60oÂ – 2 cot2Â 30oÂ + sec2Â 45oÂ = 0.

Solution:

(i) Given sin 60oÂ cosÂ 30oÂ +Â cosÂ 60o. sin 30o

LHS = sin 60oÂ cosÂ 30oÂ +Â cosÂ 60o. sin 30o

Now we have to prove that RHS = 1

= (âˆš3/2) (âˆš3/2) + Â½ Â½

= Â¾ + Â¼

= 1

= RHS

(ii) Given cosÂ 30o.Â cosÂ 60oÂ – sin 30o. sin 60oÂ = 0

LHS = cosÂ 30o.Â cosÂ 60oÂ – sin 30o. sin 60o

= (âˆš3/2) Â½ – Â½ (âˆš3/2)

= (âˆš3/4) – (âˆš3/4)

= 0

= RHS

(iii) Given cosec2Â 45oÂ – cot2Â 45oÂ = 1

LHS = cosec2Â 45oÂ – cot2Â 45oÂ = 1

= (âˆš2)2 â€“ 12

= 2 â€“ 1

= 1

= RHS

(iv) Given cos2Â 30oÂ – sin2Â 30oÂ =Â cosÂ 60o.

LHS = cos2Â 30oÂ – sin2Â 30oÂ =Â cosÂ 60o.

= (âˆš3/2)2 â€“ (Â½)2

= Â¾ – Â¼

= Â½

= cos 60o

= RHS

(vi) Given 3 cosec2Â 60oÂ – 2 cot2Â 30oÂ + sec2Â 45oÂ = 0.

LHS = 3 cosec2Â 60oÂ – 2 cot2Â 30oÂ + sec2Â 45oÂ = 0.

= 3 (2/âˆš3)2 â€“ 2 (âˆš3)2 + (âˆš2)2

= 4 â€“ 6 + 2

= 0

= RHS

4. Prove that

Solution:

(ii) RHS =

Now substituting the values, we get

= (1 â€“ 1/3)/ (1 + 1/3)

= Â½

Consider LHS

cos (2 Ã— 30o)

= cos 60o

= Â½

Therefore, LHS = RHS

= âˆš3

LHS,

tan (2 Ã— 30o)

= tan 60o

= âˆš3

Therefore, LHS = RHS

5. ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios:

(i) sin 45o

(ii)Â cosÂ 45o

(iii) tan 45o

Solution:

Given that AB = BC = x

(i) sin 45o = AB/AC

= x/xâˆš2

= 1/âˆš2

(ii) cos 45o = BC/AC

= x/xâˆš2

= 1/âˆš2

(iii) tan 45o = AB/BC

= x/x

= 1

6. Prove that:

(i)Â sinÂ 60oÂ = 2 sin 30oÂ cosÂ 30o.

(ii) 4 (sin4Â 30oÂ + cos4Â 60o) – 3 (cos2Â 45oÂ – sin2Â 90o) = 2

Solution:

(i) LHS = sin 60o

= âˆš3/2

RHS = 2 sin 30oÂ cosÂ 30o

= 2 (âˆš3/2) (Â½)

= âˆš3/2

Therefore LHS = RHS

(ii) LHS = 4 (sin4Â 30oÂ + cos4Â 60o) – 3 (cos2Â 45oÂ – sin2Â 90o)

Now by substituting the values we get

= 4[(Â½)4 + (Â½)4] â€“ 3 [(1/âˆš2)2 + 14]

= 4(1/16 + 1/16) â€“ 3 (Â½ – 1)

= 8/16 + 3/2

= 2

LHS = RHS

7. (i) If sin x =Â cosÂ x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0oÂ â‰¤ AÂ â‰¤ 90o, state the value of A.

(iii) If tanÂ Î¸= cotÂ Î¸ and 0oÂ â‰¤ Î¸ â‰¤ 90o, state the value of Î¸.

(iv) If sin x =Â cosÂ y; write the relation between x and y, if both the angles x and y are acute.

Solution:

(i) The angle, x is acute and hence we have, 0 < x

We know that

Cos2x + sin2 x = 1

Since cos x = sin x

Above equation will become

2 sin2 x = 1

Sin x = 1/âˆš2

Therefore, x = 45o

(ii) sec A = cosec A

Cos A = sin A

Cos2 A = sin2 A

Cos2x + sin2 x = 1

Above equation will become

Cos2 A = 1 â€“ cos2 A

2 cos2 A = 1

Cos A = 1/âˆš2

A = 45o

(iii) tanÂ Î¸ = cotÂ Î¸

tanÂ Î¸ = 1/tanÂ Î¸

tan2Â Î¸ = 1

tanÂ Î¸ = 1

tanÂ Î¸ = tan 45o

Î¸ = 45o

(iv) sin x = cos y = sin (90o â€“ y)

If x and y are acute angles

x = 90o â€“ y

which implies,

x + y = 90o

hence x and y are complementary angles.

8. (i) If sin x =Â cosÂ y, then x + y =Â 45o;Â write true of false.

(ii)Â secÂ Î¸. CotÂ Î¸ = cosec Î¸; write true or false.

(iii) For any angleÂ Î¸, state the value of:

Sin2Â Î¸ + cos2 Î¸.

Solution:

(i) sin x = cos y = sin (Ï€/2 â€“ y)

If x and y acute angles,

x = (Ï€/2 â€“ y)

x + y = Ï€/2

x + y = 45o is false

(ii) secÂ Î¸. CotÂ Î¸ = 1/ cosÂ Î¸. cosÂ Î¸/ sinÂ Î¸

= cosec Î¸

secÂ Î¸. CotÂ Î¸ = cosec Î¸

is true.

(iii) Sin2Â Î¸ + cos2 Î¸ = Sin2Â Î¸ + 1 – sin2 Î¸.

= 1

9. State for any acute angleÂ Î¸Â whether:

(i) sinÂ Î¸Â increases or decreases asÂ Î¸Â increases:

(ii)Â cosÂ Î¸Â increases or decreases asÂ Î¸Â increases.

(iii)Â tanÂ Î¸Â increases or decreases asÂ Î¸Â decreases.

Solution:

(i) For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means “opposite/hypotenuse” gets larger or increases.

(ii)For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means “base/hypotenuse” gets smaller or decreases.

(iii)For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means “opposite /base” gets decreases.

10. IfÂ âˆš3Â = 1.732, find (correct to two decimal place) the value of each of the following:

(i) sin 60oÂ

(ii)Â 2/ tan 30o

Solution:

(i) sin 60oÂ = âˆš3Â /2

= 1.732/2

= 0.87

(ii) 2/ tan 30o = 2/ (1/âˆš3)

= 2âˆš3Â

= 2 (1.732)

= 3.46

Exercise 23b Page: 293

1. Given A = 60oÂ and B = 30o, prove that:

(i) sin (A + B) = sin AÂ cosÂ B +Â cosÂ A sin B

(ii)Â cosÂ (A + B) =Â cosÂ AÂ cosÂ B – sin A sin B

(iii)Â cosÂ (A – B) =Â cosÂ AÂ cosÂ B + sin A sin B

Solution:

(i) Given A = 60oÂ and B = 30o

LHS = sin (A + B)

= sin (60o + 30o)

= sin 90o

= 1

RHS = sin A cos B + cos A sin B

= sin 60o cos 30o + cos 60o sin 30o

= âˆš3/2 (âˆš3/2) + Â½ Â½

= Â¾ + Â¼

= 1

LHS = RHS

(ii) LHS = cos (A + B)

= cos (60o + 30o)

= cos 90o

= 0

RHS = cos A cos B – sin A sin B

RHS = cos 60o cos 30o – sin 60o sin 30o

= Â½ (âˆš3/2) – (âˆš3/2) Â½

= âˆšÂ¾ – âˆš3/4

= 0

LHS = RHS

(iii) LHS = cos (A – B)

= cos (60o – 30o)

= cos 30o

= âˆš3/2

RHS = cos A cos B + sin A sin B

RHS = cos 60o cos 30o + sin 60o sin 30o

= Â½ (âˆš3/2) + (âˆš3/2) Â½

= âˆšÂ¾ + âˆš3/4

= âˆš3/2

LHS = RHS

(iv) LHS = tan (A – B)

= tan (60o – 30o)

= tan 30o

= 1/âˆš3

= (âˆš3 â€“ 1/âˆš3)/ 1 + âˆš3 (1/âˆš3)

= 2/ 2 âˆš3

= 1/âˆš3

Therefore, LHS = RHS

2. If A =30o, then prove that:

(i) sin 2A = 2sin AÂ cosÂ A =Â

(ii)Â cosÂ 2A = cos2A – sin2A

=Â

(iii) 2 cos2Â A – 1 = 1 – 2 sin2A

(iv)Â sinÂ 3A = 3 sin A – 4 sin3A.

Solution:

(i) Given A = 30o

Sin 2A = sin 2(30o)

= sin 60o

= âˆš3/2

2 sin A cos A = 2 sin 30o cos 30o

= 2 (Â½) (âˆš3/2)

= âˆš3/2

Now,

(ii) cos 2A = cos 2 (300)

= cos 60o

= Â½

Cos2 A â€“ sin2 A = cos2 30o â€“ sin2 30o

= Â¾ – Â¼

= Â½

= 2/4

= Â½

(iii) 2 cos2 A â€“ 1 = 2 cos2 30o â€“ 1

= 2 (Â¾) â€“ 1

= 3/2 â€“ 1

= Â½

1 â€“ 2 sin2 A = 1 â€“ 2 sin2 30o

= 1 â€“ 2 (Â¼)

= Â½

2 cos2 A â€“ 1 = 1 â€“ sin2 A

(iv) sin 3A = sin 3 (30o)

= sin 90o

= 1

3 sin A â€“ 4 sin3 A = 3 sin 30o â€“ 4 sin3 30o

= 3 (Â½) â€“ 4 (Â½)3

= 3/2 â€“ Â½

= 1

Therefore,

Sin 3A = 3 sin A â€“ 4 sin3 A

3. If A = B = 45o, show that:

(i) sin (A – B) = sin AÂ cosÂ B –Â cosÂ A sin B

(ii)Â cosÂ (A + B) =Â cosÂ AÂ cosÂ B – sin A sin B

Solution:

Given that A = B = 45o

(i) LHS = sin (A â€“ B)

= sin (45o â€“ 45o)

= sin 0o

= 0

RHS = sin A cos B â€“ cos A sin B

= sin 45o cos 45o â€“ cos 45o sin 45o

= 1/âˆš2 (1/âˆš2) â€“ 1/âˆš2 (1/âˆš2)

= 0

Therefore, LHS = RHS

(ii) LHS = cos (A + B)

= cos (45o + 45o)

= cos 90o

= 0

RHS = cos A cos B â€“ sin A sin B

= cos 45o cos 45o â€“ sin 45o sin 45o

= 1/âˆš2 (1/âˆš2) â€“ 1/âˆš2 (1/âˆš2)

= 0

Therefore, LHS = RHS

4. If A = 30o; show that:

(i) sin 3 A = 4 sin A sin (60oÂ – A) sin (60oÂ + A)

(ii) (sin A –Â cosÂ A)2Â = 1 – sin 2A

(iii)Â cosÂ 2A = cos4Â A – sin4Â A

(iv)Â

(v)Â = 2Â cosÂ A.

(vi) 4Â cosÂ AÂ cosÂ (60oÂ – A).Â cosÂ (60oÂ + A) =Â cosÂ 3A

(vii)Â

Solution:

Given that A = 30o

(i) According to the question we have

LHS = sin 3A

= sin 3 (30o)

= sin 90o

= 1

RHS = 4 sin A sin (60o â€“ A) sin (60o + A)

= 4 sin A sin (60o â€“ 30o) sin (60o + 30o)

= 4 (Â½) (Â½) (1)

= 1

LHS = RHS

(ii) According to the question we have

LHS = (sin A â€“ cos A)2

= (sin 30o â€“ cos 30o)2

= (Â½ – âˆš3/2)2

= Â¼ + Â¾ – âˆš3/2

= 1 – âˆš3/2

= (2 – âˆš3)/2

RHS = 1 â€“ sin 2A

= 1 â€“ sin 2 (30o)

= 1 â€“ sin 60o

= 1 – âˆš3/2

= (2 – âˆš3)/2

Therefore, LHS = RHS

(iii) According to the question we have

LHS = cos 2A

= cos 2 (30o)

= cos 60o

= Â½

RHS = cos4 A â€“ sin4 A

= cos4 30o â€“ sin4 30o

= (âˆš3/2)4 â€“ (Â½)4

= 9/16 â€“ 1/16

= Â½

LHS = RHS

(iv) According to the question we have

LHS = (1 â€“ cos 2A)/ sin 2A

= 1 â€“ cos 2(30o)/ sin 2 (30o)

= 1 â€“ Â½/ (âˆš3/2)

= 1/âˆš3

RHS = tan A

= tan 30o

= 1/âˆš3

LHS = RHS

(v) According to the question we have

= 2âˆš3/2

= âˆš3

RHS = 2 cos A

= 2 cos 30o

= 2 (âˆš3/2)

= âˆš3

(vi) According to the question we have

LHS = 4 cos A cos (60o â€“ A) cos (60o + A)

= 4 cos A cos (60o â€“ 30o) cos (60o + 30o)

= 4 cos 30o cos 30o cos 90o

= 4 (âˆš3/2) (âˆš3/2) 0

= 0

RHS = cos 3A

= cos 3 (30o)

= cos 90o

= 0

LHS = RHS

(vii) According to the question we have

= Â¾ + 9/4

= 12/4

= 3

= RHS

Hence the proof

Exercise 23b Page: 297

1. Solve the following equations for A, if:

(i) 2 sin A = 1Â

(ii) 2Â cosÂ 2 A = 1

(iii) sin 3Â A =Â âˆš3/2Â

(iv) sec 2 A = 2

(v)Â âˆš3Â tan A = 1Â

(vi) tan 3 A = 1

(vii) 2 sin 3 A = 1Â

(viii)Â âˆš3Â cot 2 A = 1

Solution:

(i) Given 2 sin A = 1

Sin A = Â½

Sin A = sin 30o

Therefore, A = 30o

(ii) Given 2 cos 2A = 1

Cos 2A = Â½

Cos 2A = cos 60o

2A = 60o

A = 30o

(iii) Given sin 3A = âˆš3/2

Sin 3A = sin 60o

3A = 60o

A = 20o

(iv) Given sec 2A = 2

Sec 2A = sec 60o

2A = 60o

A = 30o

(v) Given âˆš3 tan A = 1

Tan A = 1/âˆš3

Tan A = tan 30o

A = 30o

(vi) Given tan 3A = 1

Tan 3A = tan 45o

3A = 45o

A = 15o

(vii) Given 2 sin 3A = 1

Sin 3A = Â½

Sin 3A = sin 30o

3A = 30o

A = 10o

(viii) Given âˆš3 cot 3A = 1

Cot 2A = 1/âˆš3

Cot 2A = cot 60o

2A = 60o

A = 30o

2. Calculate the value of A, if:

(i) (sin A – 1) (2Â cosÂ A – 1) = 0

(ii) (tan A – 1) (cosec 3A – 1) = 0

(iii) (sec 2A – 1) (cosec 3A – 1) = 0

(iv)Â cosÂ 3A. (2 sin 2A – 1) = 0

(v) (cosec 2A – 2) (cot 3A – 1) = 0

Solution:

(i) Given (sin A – 1) (2Â cosÂ A – 1) = 0

It can be written as

(sin A – 1) = 0 and (2Â cosÂ A – 1) = 0

Sin A = 1 and cos A = Â½

Sin A = sin 90o and cos A = cos 60o

A = 90o and A = 60o

(ii) Given (tan A – 1) (cosecÂ 3A – 1) = 0

It can be written as

(tan A – 1) = 0 and (cosecÂ 3A – 1) = 0

tan A = 1 and cosec 3A = 1

tan A = tan 45o and cosec 3A = cosec 90o

A = 45o and A = 30o

(iii) Given (sec 2A – 1) (cosecÂ 3A – 1) = 0

It can be written as

(sec 2A – 1) = 0 and (cosecÂ 3A – 1) = 0

sec 2A = 1 and cosec 3A = 1

sec 2A = sec 60o and cosec 3A = cosec 90o

A = 0o and A = 30o

(iv) Given cos 3A (2 sin 2A â€“ 1) = 0

It can be written as

Cos 3A = 0 and 2 sin 2A â€“ 1 = 0

Cos 3A = cos 90o and sin 2A = Â½

3A = 90o and sin 2A = sin 30o

A = 30o and 2A = 30o which implies A = 15o

(v) Given (cosec 2A – 2) (cotÂ 3A – 1) = 0

It can be written as

(cosec 2A – 2) = 0 and (cotÂ 3A – 1) = 0

cosec 2A = 2 and cot 3A = 1

cosec 2A = cosec 30o and cot 3A = cot 45o

2A = 30o and 3A = 45o

A = 15o and A = 15o

3. If 2 sin xoÂ – 1 = 0 and xoÂ is an acute angle; find:

(i)Â sinÂ xoÂ

(ii) xoÂ

(iii)Â cosÂ xoÂ and tan xo.

Solution:

(i) Given 2 sin xoÂ – 1 = 0

2 sin xoÂ = 1

sin xoÂ = Â½

(ii) we have sin xoÂ = Â½

sin xoÂ = sin 30o

xo = 30o

(iii) we have xo = 30o

Cos xo = cos 30o = âˆš3/2

Tan xo = tan 30o = 1/âˆš3

4. If 4 cos2Â xoÂ – 1 = 0 and 0Â Â xoÂ Â 90o, find:

(i) xoÂ

(ii) sin2Â xoÂ + cos2Â xo

(iii)Â

Solution:

(i) Given 4 cos2Â xoÂ – 1 = 0

4 cos2Â xoÂ = 0

cos2Â xoÂ = (Â½)2

cos xo = Â½

cos xo = cos 60o

xo = 60o

(ii) sin2 xo + cos2 x0 = sin2 60o + cos2 60o

= (âˆš3/2)2 + (Â½)2

= Â¾ + Â¼

= 1

(iii) Given

= 1/ (Â½)2 â€“ (âˆš3)2

= 4 â€“ 3

= 1

5. If 4 sin2Â Î¸Â – 1= 0 and angleÂ Î¸Â is less than 90o, find the value ofÂ Î¸Â and hence the value of cos2Â Î¸Â + tan2 Î¸.

Solution:

Given 4 sin2Â Î¸Â – 1= 0

sin2Â Î¸ = Â¼

sin Î¸ = Â½

sin Î¸ = sin 30o

Î¸ = 30o

cos2Â Î¸Â + tan2 Î¸ = cos2Â 30oÂ + tan2 30o

= (âˆš3/2)2 + (1/âˆš3)2

= Â¾ + 1/3

= (9 + 4)/12

= 13/12

6. If sin 3A = 1 and 0Â â‰¤Â AÂ â‰¤Â 90o, find:

(i) sin AÂ

(ii)Â cosÂ 2A

(iii) tan2A â€“Â 1/cos2 A

Solution:

Given sin 3A = 1

Sin 3A = sin 90o

3A = 90o

A = 30o

(i) according to the question we have,

Sin A = sin 30o

Sin A = Â½

(ii) cos 2A = cos 2(30o)

= cos 60o

= Â½

(iii) According to the question we have,

= 1/3 â€“ 4/3

= -3/3

= -1

## Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

The Chapter 23, Trigonometrical Ratios of Standard Angles is composed of 3 exercises and the solutions given here contain answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

23.1 Trigonometrical Ratios of Angles 30o and 60o.

23.2 Trigonometrical Ratios of Angle 45o

23.3 Solving a trigonometric equation

## Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

In Chapter 23 of Class 9, the students are taught about the Trigonometrical Ratios of Standard Angles. The chapter also includes the evaluation of an expression involving trigonometric ratios and problems related to it. Study Chapter 23 of Selina textbook to understand more about Trigonometrical Ratios of Standard Angles. Learn the Selina Solutions for Class 9 effectively to come out with flying colours in the examinations.