Ex-5.3 Trigonometric Ratios, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q.1) Evaluate the following : (i) sin20/cos70

Sol (i) : Given that, sin20/cos70

Since sin (90 – Θ) = cosΘ

⇒ sin20/cos70 =  sin(90 − 70)/cos70

⇒ sin20/cos70  =  cos70/cos70

⇒ sin20/cos70  = 1

Therefore sin20/cos70  = 1

(ii) cos19/sin71

Sol.(ii): Given that, cos19/sin71

⇒ cos19/sin71  =  cos(90 − 71)/sin71

⇒ cos19/sin71  =  sin71/sin71

⇒ cos19/sin71  = 1

Since cos(90-Θ) = sin Θ

Therefore cos19/sin71  = 1

(iii) sin21/cos69

Sol.(iii): Given that, sin21/cos69

Since (90-Θ) = cosΘ

⇒ sin21/cos69  =  sin(90 − 69)/cos69

⇒ sin21/cos69  =  cos69/cos69

⇒ sin21/cos69  = 1

(iv) tan10/cot80

Sol.(iv): We are given that, tan10/cot80

Since tan(90-Θ) = cotΘ

⇒ tan10cot80  = tan(90 − 80)/cot80

⇒ tan10/cot80  =

cot80/cot80

⇒ tan10/cot80  = 1

Therefore tan10/cot80  = 1

(v) sec11/cosec79

Sol.(v):

Given that, sec11/cosec79

Since sec(90-Θ) = cosecΘ

⇒ sec11/cosec79  =  sec(90 − 79)/cosec79

⇒ sec11/cosec79  =  cosec79/cosec79

⇒ sec11/cosec79  = 1

Therefore sec11/cosec79  = 1

Q.2: EVALUATE THE FOLLOWING :

(i) (sin49°/cos41°)²  +  (cos41°/sin49°)²

Sol.(i):

We have to find: (sin49°/cos41°)²  +  (cos41°/sin49°)²

Since sec70°/cosec20°  +  sin59°/cos31° sin(90° – Θ) = cosΘ and cos(90° – Θ) = sinΘ

So

= 1 + 1 = 2

So value of (sin49°/cos41°)²  +  (cos41°/sin49°)² is 2

(ii) cos48° – sin42°

Sol.(ii)

We have to find: cos48° – sin42°

Since cos(90° − Θ) = sinΘ.So

cos48° – sin42°  =  cos(90° − 42°) − sin42°

= sin42° – sin42°  = 0

So value of cos48° – sin42° is 0

(iii) cot40°/tan50° − 1/2(cos35°/sin55°)

Sol.(iii)

We have to find:

cot40°/tan50° − 1/2(cos35°/sin55°)

Since cot(90° − Θ) = tanΘ and cos(90° − Θ) = sinΘ

= tan50°/tan50° − 1/2(sin55°/sin55°)

=  1 − 1/2  =  1/2

So value of cot40°/tan50° − 1/2(cos35°/sin55°) is 1/2

(iv)(sin27°/cos63°)² – (cos63°/sin27°)²

Sol(iv)

We have to find: (sin27°/cos63°)² – (cos63°/sin27°)²

Since sin (90° − Θ) = cosΘ and cos (90° − Θ) = sinΘ

= (cos63°/cos63°)² – (sin27°/sin27°)²  = 1-1 = 0

So value of (sin27°/cos63°)2 – (cos63°/sin27°)² is 0

(v)tan35°/cot55° + cot78°/tan12° − 1

Sol.(v)

We have to find:

tan35°/cot55° + cot78°/tan12° − 1

Since tan (90° − Θ) = cotΘ and cot (90° − Θ) = tanΘ  = 1

So value of tan35°/cot55° + cot78°/tan12°is1

(vi) sec70°/cosec20° + sin59°/cos31°

Sol.(vi)

We have to find: sec70°/cosec20° + sin59°/cos31°

Since sec70°/cosec20° + sin59°/cos31°and sec(90° − Θ) = cosecΘ

So

= cosec20°/cosec20° + cos31°/cos31°  = 1 + 1 = 2

So value of sec70°/cosec20° + sin59°/cos31° is 2

(vii) cosec31° − sec59°.

Sol(vii)

We have to find: cosec31° − sec59°

Since cosec(90° − Θ) = secΘ.

So

=  cosec31° − sec59°

=  cosec(90° − 59°) − sec59  = sec59° − sec59°  = 0

So value of cosec31° − sec59° is 0

(viii)(sin72° + cos18°) (sin72° − cos18°)

Sol.(viii)

We have to find:  (sin72° + cos18°) (sin72° − cos18°)

Since sin(90° − Θ) = cosΘ,So

(sin72° + cos18°) (sin72° − cos18°) =  (sin72°)²–(cos18°)²

= [sin(90° − 18°)]² − (cos18°)²

= (cos18°)²– (cos18°)²

= cos²18° − cos²18°  = 0

So value of (sin72° + cos18°) (sin72° − cos18°) is 0.

(ix) sin35°/sin55°–cos35°/cos55°

Sol(ix)

We find :

sin35°/sin55°–cos35°/cos55°

Since sin(90° − Θ) = cosΘ and cos(90° − Θ) = sinΘ

sin35°/sin55°–cos35°/cos55° =  sin(90° − 55°)sin55°–cos(90° − 55°)cos55°  = 1-1 = 0

So value of sin35°/sin55°–cos35°/cos55° is  0

(x) tan48°tan23°tan42°tan67°

Sol.(x)

We have to find tan48°tan23°tan42°tan67°

Since tan(90° − Θ) = cotΘ.

So

tan48°tan23°tan42°tan67° =  tan(90° − 42°)tan(90° − 67°)tan42°tan67°

= cot42°cot67°tan42°tan67°

= (tan67°cot67°)(tan42°cot42°)  = 1×1 = 1

So value of tan48°tan23°tan42°tan67° is 1

(xi)sec50°sin40° + cos40°cosec50°

Sol.(xi)

We have to find sec50°sin40° + cos40°cosec50°

Since cos(90° − Θ) = sinΘ,sec(90° − Θ) = cosecΘ and sinΘ.cosecΘ = 1.

So

sec50°sin40° + cos40°cosec50°  =  sec(90° − 40°)sin40°  + cos(90° − 50°)cosec50°  = 1 + 1 = 2

So value of sec50°sin40° + cos40°cosec50° is 2.

Q.3) Express cos75° + cot75° in terms of angle between 0⁰ and 30°.

Sol. 3 :

Given that: cos75° + cot75°

= cos75° + cot75°

=  cos(90° − 15°) + cot(90° − 15°)

= sin 15° + tan 15°

Hence the correct answer is sin 15° + tan 15°

Q.4) If  sin3A = cos(A – 26⁰), where 3A is an acute angle, find the value of A.

Sol.4:

We are given 3A is an acute angle

We have: sin3A = cos(A-26°)

⇒ sin3A = sin(90°-(A-26°))

⇒ sin3A = sin(116°-A)

⇒ 3A = 116°-A

⇒ 4A = 116°

⇒ A = 29°

Hence thecorrect answer is 29°

Q.5)If A, B, C are the interior angles of a triangle ABC, prove that,

Sol.5:

(i)We have to prove: tan = cotB/2

Since we know that in triangle ABC

A + B + C = 180

⇒ C + A = 180°-B

Hence proved

(ii)We have to prove : sin = cosA/2

Since we know that in triangle ABC

A + B + C = 180

⇒ B + C = 180°-A

sin= cosA/2

Hence proved

Q.6)Prove that :

(i)tan20°tan35°tan45°tan55°tan70°  = 1

(ii)sin48°.sec48° + cos48°.cosec42°  = 2

(iii) sin70°/cos20° + cosec20°/sec70° − 2cos70°.cosec20° = 0

(iv) cos80°/sin10° + cos59°.cosec31° = 2

 Sol.6:

(i)Therefore

tan20°tan35°tan45°tan55°tan70°

= tan(90°–70°) tan(90°–55°) tan45°tan55°tan70°

= cot70°cot55°tan45°tan55°tan70°

=  (tan70°cot70°)(tan55°cot55°)tan45° 

= 1x1x1 = 1

Hence proved

(ii) We will simplify the left hand side

sin48°.sec48° + cos48°.cosec42° = sin48°. sec(90° − 48°)cos48°.cosec(90° − 48°)

= sin48°.cos48° + cos48°.sin48° 

= 1 + 1 = 2

Hence proved

(iii) We have,sin70°/cos20° + cosec20°/sec70° − 2cos70°.cosec20° = 0

So we will calculate left hand side

sin70°/cos20° + cosec20°/sec70° − 2cos70°.cosec20° = 0 =  sin70°/cos20° + cos70°sin20° − 2cos70°.cosec(90° − 70°)

–2cos70°.cosec(90° − 70°)

= cos20°/cos20° + sin20°/sin20° − 2×1  = 1 + 1-2 = 2-2 = 0

Hence proved

(iv)We have cos80°/sin10° + cos59°.cosec31° = 2

We will simplify the left hand side

cos80°/sin10° + cos59°.cosec31° =   + cos59°.cosec(90° − 59°)

= sin10°/sin10° + cos59°.sec59°  = 1 + 1 = 2

Hence proved.

Question 7.

If A,B,C are the interior of triangle ABC , show that

(i) sin = cosA/2

Sol. 

A + B + C = 180⁰

B + C = 180⁰– A/2

LHS = RHS

(ii)cos(90° − A/2) = sinA/2

LHS = RHS

Question 8

If 2Θ + 45°and30 − Θ are acute angles , find the degree measure of

 Θ satisfying sin(20 + 45°) = cos(30° + Θ)

Sol. Here 20 + 45° = sin(60° + Θ)

We know that ,((90° − Θ) =  cos(Θ)

=  sin(2Θ + 45°) = sin(90° − (30° − Θ))

=  sin(2Θ + 45°) = sin(90° − 30° − Θ)

=  sin(2Θ + 45°) = sin(60° + Θ)

On equating sin of angle of we get,

=   2Θ + 45° = 60° + Θ  =  Θ = 15°

Question 9

If Θ is appositive acute angle such that secΘ = csc60° , find

2cos2Θ − 1

Sol. 

We know that, sec(90° − Θ) = csc²Θ

=  sec(Θ) = sec(90° − 60°)

=  Θ = 30°  =  2cos²Θ − 1

=  2cos²30 − 1

=  2(√3/2)² − 1

=  2(3/4) − 1  = (3/2) − 1  =  (1/2)

Q10.If sin3Θ = cos(Θ − 6°) where 3ΘandΘ − 6^circ acute angles, find the value of Θ.

Sol:     

We have, sin3Θ = cos(Θ − 6°)

cos(90° + 3Θ) = cos(Θ − 6°)

90°–3Θ = Θ–6°

− 3Θ − Θ = 6° − 90°

− 4Θ = 96°

Θ = − 96°/ − 4 = 24°

Q11.If sec2A = csc(A − 42°) where 2A  is acute angle, find the value of A.

Sol:we know that sec(90 − 3Theta) = cscΘ

sec2A = sec(90–(A − 42))

sec2A = sec(90–A + 42))

sec2A = sec(132–A))

Now equating both the angles we get

2A = 132 – A

A =   = 132/3

A = 44