Trigonometry
Class 10th Mathematics Part 2 MHB Solution
- If sintegrate heta = 7/25 find the values of cosθ and tanθ.
- sin^2theta /costheta +costheta = sectheta Prove that:
- If tantheta = 3/4 find the values of secθ and cosθ.
- If cottheta = 40/9 find the values of cosecθ and sinθ.
- If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
- If tanθ = 1 then, find the values of sintegrate heta +costheta /sectheta +cosectheta…
- cos^2theta (1+tan^2theta) = 1 Prove that:
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove that:…
- (sec θ - cos θ) (cot θ + tan θ) = tan θ sec θ Prove that:
- cot θ + tan θ = cosec θ sec θ Prove that:
- 1/sectheta -tantheta = sectheta +tantheta Prove that:
- sin^4 θ - cos^4 θ = 1 - 2cos^2 θ Prove that:
- sectheta +tantheta = costheta /1-sintegrate heta Prove that:
- If tantheta + 1/tantheta = 2 then show that tan^2theta + 1/tan^2theta = 2 Prove that:…
- tana/(1+tan^2a)^2 + cota/(1+cot^2a)^2 = sinacosa Prove that:
- sec^4 A (1- sin^4 A) - 2tan^2 A = 1 Prove that:
- tantheta /sectheta -1 = tantheta +sectheta +1/tantheta +sectheta -1 Prove that:…
- A person is standing at a distance of 80m from a church looking at its top. The angle…
- From the top of a lighthouse, an observer looking at a ship makes angle of depression…
- Two buildings are facing each other on a road of width 12 metre. From the top of the…
- Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the…
- A storm broke a tree and the treetop rested 20 m from the base of the tree, making an…
- A kite is flying at a height of 60 m above the ground. The string attached to the kite…
- sinθ cosecθ = ? Choose the correct alternative answer for the following question.A. 1…
- cosec45° = ? Choose the correct alternative answer for the following question.A.…
- 1 + tan^2 θ = ? Choose the correct alternative answer for the following question.A.…
- When we see at a higher level, from the horizontal line,angle formed is....... Choose…
- If sintegrate heta = 11/61 find the values of cosθ using trigonometric identity.…
- If tan θ = 2, find the values of other trigonometric ratios.
- If sectheta = 13/12 find the values of other trigonometric ratios.…
- secθ(1 - sinθ) (secθ + tanθ) = 1 Prove the following.
- (secθ + tanθ) (1 - sinθ) = cosθ Prove the following.
- sec^2 θ + cosec^2 θ = sec^2 θ × cosec^2 θ Prove the following.
- cot^2 θ - tan^2 θ = cosec^2 θ - sec^2 θ Prove the following.
- tan^4 θ + tan^2 θ = sec^4 θ - sec^2 θ Prove the following.
- 1/1-sintegrate heta + 1/1+sintegrate heta = 2sec^2theta Prove the following.…
- sec^6 x - tan^6 x = 1 + 3 sec^2 x × tan^2 x Prove the following.
- tantheta /sectheta +1 = sectheta -1/tantheta Prove the following.…
- tan^3theta -1/tantheta -1 = sec^2theta +tantheta Prove the following.…
- sintegrate heta -costheta +1/sintegrate heta +costheta -1 = 1/sectheta -tantheta Prove…
- A boy standing at a distance of 48 meters from a building observes the top of the…
- From the top of the light house, an observer looks at a ship and finds the angle of…
- Two buildings are in front of each other on a road of width 15 meters. From the top of…
- A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to…
- While landing at an airport, a pilot made an angle of depression of 20°. Average speed…
Practice Set 6.1
Question 1.If 
find the values of cosθ and tanθ.
Answer:
We know that,
sin2θ + cos2θ = 1
Also,
Question 2.
Prove that:
Answer:
Taking LHS
[As, sin2θ + cos2θ = 1]
= sec θ 
= RHS
Proved !
Question 3.
If 
find the values of secθ and cosθ.
Answer:
We know that,
sec2θ= 1 + tan2θ
Also,
Question 4.
If 
find the values of cosecθ and sinθ.
Answer:
We know that,
cosec2θ = 1 + cot2θ
⇒ 
Also,
∴ SinΘ= 
Question 5.
If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
Answer:
5secθ - 12cosecθ = 0
⇒ 5secθ = 12cosecθ
As 
we have,
Also, We know that,
sec2θ= 1 + tan2θ
Also,
Now, again using
Question 6.
If tanθ = 1 then, find the values of 
Answer:
Given,
tan θ = 1
⇒ θ = 45° [as tan 45° = 1]
Also,
= sinθ cosθ
= sin 45° cos 45°
Question 7.
Prove that:
Answer:
Taking LHS
cos2θ(1 + tan2θ)
= cos2θ sec2θ [As, sec2θ = 1 + tan2θ]
= 1
= RHS
Proved !
Question 8.
Prove that:
Answer:
Taking LHS
[(a + b)(a - b) = a2 - b2 ]
[As, sin2θ + cos2θ = 1]
= RHS
Proved !
Question 9.
Prove that:
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
Answer:
Taking LHS
(secθ - cosθ)(cotθ + tanθ)
= secθcotθ + secθtanθ - cosθcotθ - cosθtanθ
Taking LCM of first three terms,
[As, sin2θ + cos2θ = 1]
= tanθsecθ
= RHS
Proved !
Question 10.
Prove that:
cot θ + tan θ = cosec θ sec θ
Answer:
Taking LHS, and putting 
and 
= cotθ + tanθ
[As, sin2θ + cos2θ = 1]
= RHS
Proved !
Question 11.
Prove that:
Answer:
Taking LHS
= secθ + tanθ [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]
= RHS
Proved !
Question 12.
Prove that:
sin4 θ – cos4 θ = 1 – 2cos2 θ
Answer:
L.H.S = sin4θ – cos4θ
= (sin2θ – cos2θ)(sin2θ + cos2θ)
= (sin2θ – cos2θ)
= (1 – cos2θ – cos2θ)
= 1- 2cos2θ
Question 13.
Prove that:
Answer:
Taking RHS
(Multiplying both Numerator and Denominator by 1+SinΘ)
[As, sin2θ + cos2θ = 1]
= secθ + tanθ
= LHS
Proved !
Question 14.
Prove that:
If 
then show that 
Answer:
Given,
Squaring both side,
[(a + b)2 = a2 + b2 + 2ab]
Hence, Proved !
Question 15.
Prove that:
Answer:
Taking RHS
= sinA cos3A + cosA sin3A
= sinAcosA(cos2A + sin2A) [As, sin2θ + cos2θ = 1]
= sinAcosA
= RHS
Proved !
Question 16.
Prove that:
sec4A (1– sin4A) – 2tan2 A = 1
Answer:
Taking LHS
= sec4A(1 - sin4A) - 2tan2A
= sec4A - sin4A sec4A - 2tan2A
= sec4A - tan4A - tan2A - tan2A
= sec4A - tan2A(1 + tan2A) - tan2A
= sec4A - tan2A sec2A - tan2A [As, sec2θ = 1 + tan2θ ]
= sec2A(sec2A - tan2A) - tan2A
= sec2A - tan2A [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]
= 1
= RHS
Proved !
Question 17.
Prove that:
Answer:
Taking RHS
[As sec2θ -1 = tan2θ ]
= LHS
Proved.
Practice Set 6.2
Question 1.A person is standing at a distance of 80m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Answer:
Let 'A' be the person, standing 80 m away from a church BC,
Angle of elevation, ∠BAC = θ = 45°
Clearly, ∆ABC is a right-angled triangle, in which
⇒ BC = 80 m
Therefore, Height of church is 80 m.
Question 2.
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (√3 = 1.73)
Answer:
Let PQ be a light house of height 80 cm such that PQ = 90 m
And R be a ship.
Angle of depression from P to ship R = ∠BPR = 60°
Also, ∠PRQ(say θ) = ∠BPR = 60° [Alternate Angles]
Clearly, PQR is a right-angled triangle.
Now, In ∆PQR
⇒ QR = 30(1.73)
⇒ QR = 51.90 m
Hence, Ship is 51.90 m away from the light house.
Question 3.
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Answer:
Let AB and CD be two building, with
AB = 10 m
And angle of elevation from top of AB to top of CD = ∠CAP = 60°
Width of road = BD = 12 m
Clearly, ABDP is a rectangle
With
AB = PD = 10 m
BD = AP = 12 m
And APC is a right-angled triangle, In ∆APC
⇒ CP = 12√3 m
Also,
CD = CP + PD = (12√3 + 10) m
Hence, height of other building is (10 + 12√3 m).
Question 4.
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.
Answer:
Let AB and CD be two poles, and AC be the wire joining their top.
and
Angle made by wire with horizontal = ∠CAP = θ [say]
Given,
AB = 7 m [Let AB be smaller pole]
CD = 18 m
AC = 22 m
Clearly, APDB is a rectangle with
AB = PD = 7 m
Also,
CP = CD - PD = 18 - 7 = 11 m
In ∆APC
⇒ θ = 30° 
Question 5.
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Answer:
Let AB be a tree, and C is the point of break.
Height of tree = BC + AC
As, the treetop is rested 20 m from the base, making an angle of 60° with the horizontal.
In ∆ABC
AB = 20 m
∠ABC, θ = 60°
Now,
⇒ AC = 20 tan 60°
⇒ AC = 20√3 m
⇒ BC = 40 m
Height of tree = BC + AC = (40 + 20√3) meters.
Question 6.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (√3 = 1.73 )
Answer:
Let AC be a string and kite is flying at point A, with height
AB = 60 m
Angle make by string with horizontal, θ = ∠ACB = 60°
Clearly, ABC is a right-angled triangle.
In ∆ABC
⇒ AC = 40(1.73)
⇒ AC = 6.92 m
Hence, length of string is 6.92 m
Problem Set 6
Question 1.Choose the correct alternative answer for the following question.
sinθ cosecθ = ?
A. 1
B. 0
C. 
D. 
Answer:
We know,
⇒ sinθ cosecθ = 1
Question 2.
Choose the correct alternative answer for the following question.
cosec45° = ?
A. 
B.
C. 
D. 
Answer:
As, cosec45° = √2
Question 3.
Choose the correct alternative answer for the following question.
1 + tan2θ = ?
A. cot2θ
B. cosec2θ
C. sec2θ
D. tan2θ
Answer:
We know that,
1 + tan2θ = sec2θ
Question 4.
Choose the correct alternative answer for the following question.
When we see at a higher level, from the horizontal line,angle formed is.......
A. angle of elevation.
B. angle of depression.
C. 0
D. straight angle.
Answer:
When we see at a higher level, from the horizontal line, the angle formed is known as angle of elevation.
Question 5.
If 
find the values of cosθ using trigonometric identity.
Answer:
As,
sin2θ + cos2θ = 1
Question 6.
If tan θ = 2, find the values of other trigonometric ratios.
Answer:
We know that,
sec2θ= 1 + tan2θ
⇒ sec2θ = 1 + (2)2
⇒ sec2θ = 5
⇒ sec θ = √5 …[1]
Also,
…[2]
Now, using
…[3]
Also,
…[4]
….[5]
Question 7.
If 
find the values of other trigonometric ratios.
Answer:
We know that,
sec2θ= 1 + tan2θ
⇒ tan2θ = sec2θ - 1
… [1]
Also,
… [2]
Now, using
… [3]
Also,
… [4]
….[5]
Question 8.
Prove the following.
secθ(1 - sinθ) (secθ + tanθ) = 1
Answer:
Taking LHS
secθ(1 - sinθ) (secθ + tanθ)
[(a + b)(a - b) = a2 - b2]
[ sin2θ + cos2θ = 1]
= 1
= RHS
Proved !
Question 9.
Prove the following.
(secθ + tanθ) (1 - sinθ) = cosθ
Answer:
Taking LHS
(1 - sinθ) (secθ + tanθ)
[(a + b)(a - b) = a2 - b2]
[ sin2θ + cos2θ = 1]
= cos θ
= RHS
Proved !
Question 10.
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Answer:
Taking LHS
Sec2θ + cosec2θ
[ sin2θ + cos2θ = 1]
= sec2θ × cosec2θ
= RHS
Proved !
Question 11.
Prove the following.
cot2θ - tan2θ = cosec2θ - sec2θ
Answer:
Taking LHS
cot2θ - tan2θ
[ Now, cosec2θ - 1 = cot2θ and sec2θ - 1 = tan2θ]
= cosec2θ - 1 - (sec2θ - 1)
= cosec2θ - sec2θ
= RHS
Proved !
Question 12.
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Answer:
Taking LHS
tan4θ + tan2θ
= tan2θ( tan2θ + 1)
= (sec2θ - 1)(sec2θ) [1 + tan2θ = sec2θ]
= sec4θ - sec2θ
= RHS
Proved !
Question 13.
Prove the following.
Answer:
Taking LHS
[ sin2θ + cos2θ = 1]
= 2 sec2θ
= RHS
= Proved
Question 14.
Prove the following.
sec6x – tan6x = 1 + 3 sec2x × tan2x
Answer:
Taking LHS
sec6x - tan6x
= (sec2x)3 - (tan2x)3
= (sec2x - tan2x)(sec4x + tan2x sec2x + tan4x)
[As, a3 - b3 = (a - b)(a2 + ab + b2)]
= sec4x + tan4x + tan2x sec2x + 2tan2x sec2x - 2tan2x sec2x
[As, sec2θ - tan2θ = 1]
= sec4x + tan4x - 2tan2x sec2x + 3tan2x sec2x
= (sec2x - tan2x)2 + 3tan2x sec2x [a2 + b2 - 2ab = (a - b)2]
= 12 + 3tan2x sec2x
= 1 + 3tan2x sec2x
= RHS
Proved.
Question 15.
Prove the following.
Answer:
Taking LHS
[tan2θ = sec2θ - 1]
= RHS
Proved !
Question 16.
Prove the following.
Answer:
Taking LHS
[a3 - b3 = (a - b)(a2 + ab + b2)]
= tan2θ + tanθ + 1
= sec2θ + tanθ [1 + tan2θ = sec2θ]
= RHS
Proved.
Question 17.
Prove the following.
Answer:
Taking LHS
Dividing numerator and denominator by cosθ
[As sec2θ - tan2θ = 1]
= RHS
Proved.
Question 18.
A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Answer:
Let 'R' be the person, standing 48 m away from a building PQ,
Angle of elevation, ∠PRQ = θ = 30°
Clearly, ∆ABC is a right-angled triangle, in which
Therefore, Height of church is 16√3 m.
Question 19.
From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.
Answer:
Let PQ be a light house of height 80 cm such that PQ = 100 m
And R be a ship.
Angle of depression from P to ship R = ∠BPR = 30°
Also, ∠PRQ(say θ) = ∠BPR = 30° [Alternate Angles]
Clearly, PQR is a right-angled triangle.
Now, In ∆PQR
⇒ QR = 100√3 m
Hence, Ship is 100√3 m away from the light house.
Question 20.
Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?
Answer:
Let AB and CD be two building, with
AB = 12 m
And angle of elevation from top of AB to top of CD = ∠CAP = 30°
Width of road = BD = 15 m
Clearly, ABDP is a rectangle
With
AB = PD = 12 m
BD = AP = 15 m
And APC is a right-angled triangle, In ∆APC
⇒ CP = 5√3 m
Also,
CD = CP + PD = (5√3 + 12) m
Hence, height of other building is (12 + 5√3 m).
Question 21.
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Answer:
Let AB be the ladder, i.e. AB = 20 m and PQRS be the platform.
In the above figure, clearly
PQ = RS = CD = height of platform = 2 m
Maximum height ladder can reach = BC + CD
Now,
Maximum value of ∠BAC = 70°
In right-angled triangle ABC,
⇒ BC = 18.8 m
Maximum height ladder can reach = BC + CD
= 18.8 + 2 = 20.8 meters
Question 22.
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)
Answer:
Let the initial position of plain be P and after landing it position be R.
Now,
Angle of depression while landing, ∠CPR = 20°
Also, ∠ CPR = ∠PRQ (say θ) = 20° [Alternate Angles]
Now,
Speed of plane = 200 km / hr
Time taken for landing = 54 seconds
Distance travelled in landing = PR
Also, distance = speed × time
Now, In ∆PQR
⇒ PQ = 3000 × sin 20°
⇒ PQ = 3000(0.342)
⇒ PQ = 1026 m
So, Plane was at a height of 1026 m at the start of landing.