Question 1. Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A
Solution:
(i) sin A
We know that
cosec2A = 1 + cot2A
1/sin2A = 1 + cot2A
sin2A = 1/(1 + cot2A)
sin A = 1/(1+cot2A)1/2
(ii) sec A
sec2A = 1 + tan2A
Sec2A = 1 + 1/cot2A
sec2A = (cot2A + 1) / cot2A
sec A = (cot2A + 1)1/2 / cot A
(iii) tan A
tan A = 1 / cot A
tan A = cot -1 A
Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
(i) cos A
cos A = 1/sec A
(ii) sin A
We know that
sin2A = 1 - cos2A
Also , cos2A = 1 / sec2A
sin2A = 1 - 1 / sec2A
sin2A = (sec2A - 1) / sec2A
sin A = (sec2A - 1)1/2 / sec A
(iii) tan A
We know that
tan2A + 1 = sec2A
tan A = (sec2A - 1)½
(iv) cosec A
We know
cosec A = 1/ sinA
cosec A = sec A / (sec2A - 1)½
(v) cot A
We know
cot A = cos A / sin A
cot A = (1/sec A) / ((sec2A - 1)1/2 / sec A)
cot A = 1 / (sec2A - 1)1/2
Question 3. Evaluate:
(i) (sin2 63° + sin2 27°)/(cos2 17° + cos2 73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
(i) ([sin(90-27)]2 + sin2 27) / ([cos(90-73)]2 + cos2 73)
We know that
sin(90-x) = cos x
cos(90-x) = sin x
(cos2(27) + sin2 27) / (sin2(73) + cos2 73)
Using
sin2A + cos2A = 1
1/1 = 1
(ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]
Using
sin(90-x) = cos x
cos(90-x) = sin x
= [sin 25 * sin 25] + [cos 25 * cos 25]
= sin2 25 + cos2 25
= 1
Question 4. Choose the correct option. Justify your choice.
Solution:
(i) 9 sec2 A – 9 tan2 A
(A) 1 (B) 9 (C) 8 (D) 0
Using sec2A - tan2A = 1
9 (sec2A - tan2A ) = 9(1)
Ans (B)
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) –1
Simplifying all ratios
= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ - 1/sinθ)
= ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ - 1 )/sinθ)
= ((cosθ + sinθ)2 - 1) / (sinθ cosθ)
= (1 + 2*cosθ*sinθ - 1) / (sinθ cosθ)
= 2
Ans (C)
(iii) (sec A + tan A) * (1 – sin A)
(A) sec A (B) sin A (C) cosec A (D) cos A
Simplifying sec A and tan A
= (1/cos A + sin A/cos A)*(1 - sin A)
= ((1 + sin A)/cos A)*(1 - sin A)
= (1 - sin2A)/cos A
= cos2A / cos A
= cos A
Ans (D)
(iv) (1 + tan2A) / (1 + cot2A)
(A) sec2A (B) –1 (C) cot2A (D) tan2A
Simplifying tan A and cot A
= (1 + (sin2A / cos2A)) / (1 + (cos2A / sin2A))
= ((cos2A + sin2A) / cos2A) / ((cos2A + sin2A) / sin2A)
= sin2A / cos2A
= tan2A
Ans (D)
Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Solution:
(i) (cosec θ – cot θ)2 = (1 - cosθ) / (1 + cosθ)
Solving LHS
Simplifying cosec θ and cot θ
= (1-cos θ)2 / sin2θ
= (1-cos θ)2 / (1-cos2θ)
Using a2 - b2 = (a+b)*(a-b)
= (1-cos θ)2 / [(1-cos θ)*(1+cos θ)]
= (1-cos θ) / (1+cos θ) = RHS
Hence Proved
(ii) (cos A / (1+sin A) + ((1+sin A) / cos A) = 2 sec A
Solving LHS
Taking LCM
= (cos2A + (1+sin A)2) / ((1+sin A) cos A)
= (cos2A + 1 + sin2A + 2 sin A ) / ((1 + sin A)*cos A)
Using sin2A + cos2A = 1
= (2 + 2*sin A) / ((1+sin A)*cos A)
= (2*(1 + sin A)) / ((1 + sin A)*cos A)
= 2 / cos A
= 2 sec A = RHS
Hence Proved
(iii) (tan θ / (1 - cot θ)) + (cot θ / (1 - tan θ)) = 1 + sec θ*cosec θ
Solving LHS
Changing tan θ and cot θ in terms of sin θ and cos θ and simplifying
= ((sin2θ) / (cos θ *(sin θ-cos θ))) + ((cos2θ ) / (sin θ *(sin θ-cos θ)))
= (1 / (sin θ-cos θ)) * [(sin3θ - cos3θ) / (sin θ * cos θ)]
= (1 / (sin θ - cos θ)) * [ ((sin θ - cos θ) * ( sin2θ + cos2θ + sin θ * cos θ ))/(sin θ *cos θ)]
= (1+sin θ*cos θ) / (sin θ*cos θ)
= sec θ*cosec θ + 1 = RHS
Hence Proved
(iv) (1 + sec A) / sec A = sin2A / (1 - cos A)
Solving LHS
= cos A + 1
Solving RHS
= (1 - cos2A) / (1 - cos A)
= (1 - cos A) * (1 + cos A) / (1 - cos A)
= 1 + cos A = RHS
Hence Proved
(v) (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot A using the identity cosec2A = 1 + cot2A
Solving LHS
Multiplying numerator and denominator by (cot A - 1 + cosec A)
= (cot2A + 1 + cosec2A - 2*cot A - 2*cosec A + 2*cot A*cosec A) / (cot2A - (1 + cosec2A - 2*cosec A))
= (2*cosec2A - 2*cot A - 2*cosec A + 2*cot A*cosec A) / (cot2A - 1 - cosec2A + 2*cosec A)
= (2* cosec A *(cosec A + cot A) - 2*(cosec A + cot A)) / (cot2A - 1 - cosec2A + 2*cosec A)
= ((cosec A + cot A) * (2*cosec A - 2 )) / (2*cosec A - 2)
= cosec A + cot A = RHS
Hence Proved
(vi) [(1 + sin A) / (1 - sin A)]½ = sec A + tan A
Solving LHS
Multiplying numerator and denominator by (1+sinA)
= [((1 + sin A)*(1 + sin A)) / ((1 - sin A)*(1 + sin A))]½
= (1 + sin A) / (1 - sin2A)½
= (1 + sin A) / (cos2A)1/2
= (1 + sin A) / (cos A)
= sec A + tan A = RHS
Hence Proved
(vii) (sin θ - 2 sin3θ) / (2 cos3θ - cos θ) = tan θ
Solving LHS
= (sin θ * (1 - 2*sin2θ)) / (cos θ * (2*cos2θ - 1))
= (sin θ * (1 - 2*sin2θ )) / (cos θ * (2*(1 - sin2θ) - 1))
= (sin θ *(1 - 2*sin2θ)) / (cos θ * (1 - 2*sin2θ))
= tan θ = RHS
Hence Proved
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
Solving LHS
= sin2A + cosec2A + 2*sin A *cosec A + cos2A + sec2A + 2*cos A *sec A
We know that cosec A = 1 / sin A
= 1 + 1 + cot2A + 1 + tan2A + 2 + 2
= 7 + tan2A + cot2A = RHS
Hence Proved
(ix) (cosec A – sin A)*(sec A – cos A) = 1 / (tan A + cot A)
Solving LHS
= ((1/sin A) - sin A) * ((1/cos A) - cos A)
= ((1 - sin2A) / sin A) * ((1 - cos2A) / cos A)
= (cos2A * sin2A) / (sin A * cos A)
= sin A * cos A
Solving RHS
Simplifying tan A and cot A
= (sin A * cos A) / ( sin2A + cos2A)
= sin A * cos A = RHS
Hence Proved
(x) (1 + tan2A) / (1 + cot2A ) = [(1 - tan A) / (1 - cot A)]2 = tan2A
Solving LHS
Changing cot A = 1 / tan A
= (tan2A * (1 + tan2A)) / (1 + tan2A) = tan2A = RHS
= [(1 - tan A) / (1 - cot A)]2 = (1 + tan2A - 2*tan A) / (1 + cot2A - 2*cot A)
= (sec2A - 2*tan A) / (cosec2A - 2*cot A)
Solving this we get
= tan2A
Hence Proved
Practice Questions - Introduction To Trigonometry
1. A ladder 10 meters long rests against a vertical wall. If the angle between the ladder and the ground is 60°, find the height of the point where the ladder touches the wall.
2. From the top of a 75 m high lighthouse, the angle of depression of a ship is observed to be 30°. Calculate the distance of the ship from the lighthouse.
3. A tree casts a shadow 20 meters long on the ground. The angle of elevation of the sun is 45°. Find the height of the tree.
4. The angle of elevation of the top of a tower from a point 30 meters away from its base is 60°. Find the height of the tower.
5. A kite is flying at a height of 60 meters above the ground. The string attached to the kite is 100 meters long. Find the angle of elevation of the kite.
6. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower are 30° and 60° respectively. If the tower is 50 meters high, find the distance of the point from the tower.
7. A person observes the angle of elevation of a hill as 60°. He walks 100 meters towards the hill and finds the angle of elevation has changed to 75°. Find the height of the hill.
8. The angle of elevation of an airplane from a point on the ground is 60°. After 15 seconds, the angle of elevation changes to 30°. If the airplane is flying at a constant height of 3000 meters, find the speed of the airplane.
9. Two ships are sailing in the sea on either side of a lighthouse. The angle of elevation of the top of the lighthouse observed from the ships are 30° and 45° respectively. If the lighthouse is 100 meters high, find the distance between the ships.
10. A hot air balloon is rising vertically upward at a constant speed. An observer on the ground spots the balloon at an angle of elevation of 30° when it is 200 meters above the ground. After 10 minutes, the angle of elevation becomes 60°. Find the speed of the balloon.
Summary
Exercise 8.4 focuses on applying trigonometric ratios to solve problems involving heights and distances. It introduces students to practical applications of trigonometry in real-life scenarios. The problems typically involve calculating the height of tall structures or distances between objects using given angles of elevation or depression and known measurements. Students learn to identify the appropriate trigonometric ratio (sine, cosine, or tangent) to use based on the information provided and the unknown quantity they need to find.