CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 11

Write the correct answer for each of the following :

1. Write the altitude of the sun is at 60°, then the height of the vertical tower that will cost a shadow of length 20 m is
(a) 20√3m
(b) (20√3) m 
(c) (15√3) m
(d) 40√3m

2. A lamp post 35 high casts a shadow 5 m long on the ground. The sun’s elevation at this point is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

3. The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

4. If the angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary, then the height of the tower is
(a) ab
(b) √ab
(c) a/b
(d) √a/b

5. A ladder 18 m long makes an angle of 60° with a wall. The height of the point where the ladder reaches the wall is
(a) 9√3m
(b) 18√3m
(c) 18 m
(d) 9 m

6. The length of the ladder making an angle of 45° with a wall & whose foot is 7 m away from the wall is
(a) (7√2/2) m
(b) 7√2 m
(c) 14√2m
(d) 14 m

7. A ladder reaches a point on a wall which is 20 m above the ground and its foot is 320m away from the ground. The angle made by the ladder with the wall is
(a) 90°
(b) 60°
(c) 45°
(d) 30°

8. The angle of elevation, if the length of the shadow of a tower is √3 times the height of the tower is
(a) 30°
(b) 45°
(c) 60°
(d) 75°

9. The line drawn from the eye of an observer to the point in the object viewed by the observer is known as
(a) Horizontal line
(b) Vertical line
(c) Line of sight
(d) Transversal line

10. When a point is observed, angle formed by the line of sight with the horizontal where the point being viewed is above the horizontal level is known as
(a) angle of elevation
(b) angle of depression
(c) angle of a triangle
(d) right angle

11. When we raise our head to look at the object, the angle formed by the line of sight with the horizontal is known as
(a) acute angle
(b) angle of elevation
(c) right angle
(d) angle of depression

12. When we lower our head to look at the object, the angle formed by the line of sight with the horizontal is known as
(a) Obtuse angle
(b) angle of depression
(c) angle of elevation
(d) acute angle

13. If the flagstaff 6 m high placed on the top of a tower throws a shadow 2√3 m along the ground, then the angle of elevation of the sun is
(a) 30°
(b) 60°
(c) 45°
(d) None of these

14. The angle of elevation of the top of a 15 m high tower at a point 15 m away from the base of the tower is
(a) 30°
(b) 60°
(c) 45°
(d) 75°

15. The elevation of the sun is 30o, then the length of the shadow cast by a tower of 150 feet height is
(a) 150 feet
(b) 50√3 feet
(c) 150√3 feet
(d) 200 fee

16. If two towers of height h₁ and h₂ subtend angles of 60° and 30° respectively at the mid point of the line joining their feet then h₁ : h₂ =
(a) 1 : 2
(b) 1 : 3
(c) 2 : 1
(d) 3 : 1

17. The tree 6 m tall casts a 4 m long shadow. At the same time, a flag pole casts a shadow 50 m long. How long is the flag pole ?
(a) 75 m
(b) 100 m
(c) 150 m
(d) 50 m

18. The distance between the tops of two trees 20 m & 28 m high is 17 m. The horizontal distance between the two trees is
(a) 9 m
(b) 11 m
(c) 15 m
(d) 31 m

SHORT QUESTIONS

1. Find the length AP from the following figure

2. The height of the tower is 10 m. Calculate the height of its shadow, when the Sun’s altitude is 45°

3. A tree 12 m long is broken by the wind in such a way that its top touches the ground and makes an angle 60° with the ground. At what height from the bottom the tree is broken by the wind ?

4. A ladder makes an angle of 30° with a wall. If the foot of the ladder is 5 m away from the wall, find the length of the ladder.

5. If the ratio of the height of a pole and the length of its shadow is √3 : 1, what is the angle of elevation of the sun ?

6. From a point on the ground, 20 m away from the foot of a vertical tower the angle of elevation of the top of the tower is 60°, what is the height of the tower

7. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

8. The length of a string between a kite and a point on the ground is 90 m. If the string makes an angle θ with the ground level such that tanθ = 15/8, how high is the kite ? Assume that there is no slack in the string.

9. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases 60o. Find the height of the tower.

10. A kite is flying at a height of 75 m from the ground level, attached at a string makes an angle 60o to the horizontal. Find the length of the string to the nearest metre.

11. A tree is broken by wind. The top struck the ground at an angle of 30° & at a distance of 30 m from the root. Find the height of the whole tree.

LONG QUESTIONS

1. The angle of elevation of a Jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30^o. If the Jet plane is flying at a constant height of 1500√3m. Find the speed of the Jet plane.

2. An aeroplane, when 3000 m high, passes vertically above another aeroplane at an instant when the angles of elevation of two aeroplanes from the same point on the ground are 60o to 45o respectively. Find the vertical distance between the two aeroplanes.

3. From a window (h metre high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are θ and φ respectively. Show that the height of the opposite house is h(1 + tanθ. cotθ).

4. If the angle of elevation of a cloud from a point ‘h’ metres above a lake is α and angle of depression of its reflection in lake is β, prove that the distance of the cloud from the point of observation is 2h sec α /.tan β - tαnα

5. From an aeroplane vertically above a straight horizontal plane the angles of depression of two consecutive kilometre stones on opposite sides of aeroplane are found to be α and β. Show the height of aeroplane is tan α tan β / tan α + tan β

6. A man standing on the deck of a ship which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and angle of depression of base of a hill as 30°. Find the distance of hill from the ship and height of the hill.

7. The angles of elevation of the top of a tower from two points P & Q at distance of a and b respectively from the base and in same straight line with it, are complimentary. Prove that the height of tower is ab

8. An aeroplane, flying horizontally 1000 m above the ground is observed at an angle of elevation of 60° from a point on the ground. After a flight of 10 seconds the angle of elevation at the point of observation changes to 30°. Find the speed of plane in m/second.

9. A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45o how soon after this, will car reach the tower.

10. A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of 20 m high building, finds the angle of elevation of same bird to be 45°. Both the boy and girl are on opposite sides of the bird. Find the distance of bird from the girl.

11. If the angle of elevation of a cloud from a point ‘h’ metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the height of the cloud is h "{1+r/1-r} where r = tan α / tan β

12. From an aeroplane vertically above a straight horizontal plane, the angle of depressions of two consecutive kilometers stones on the opposite sides of the aeroplane are found to be α and β. Show that height of the aeroplane is {cotα + cotβ)−1

13. As observed from the foot of a mountain the angle of elevation of the summit of the mountain is 45°; after ascending 1000 m towards the mountain up a slope of 30o inclination, the angle of elevation is observed as 60°. Find the height of the summit of the mountain from the level ground.

14. The elevation of a tower of a station A due north of it is ‘α’ and at a station B due east of A is β. Prove that the height of the tower is d sin α. sin β / √sin² α - sin β where AB = d.

15. The angle of Elevation of the top of a tower from a point A due south of the tower is α and from B due east of the tower is β. If AB = d, show that height of the tower is d/cot² α + cot² β

16. The angle of elevation of a cliff from a fixed point is ‘θ’. After going up a distance of ‘k’ meters towards the top of the cliff at angle of φ, it is found that the angle of elevation is α. Show that the height of the cliff is K(cos φ - sin φ. cot α) / cot θ - cot α

17. Two stations due south of a leaning tower which leans towards north are at distance a & b from its foot. If α and β be the elevation of the top of the tower, from these stations, prove that its inclination ‘θ’ to the horizontal is given by cotθ = b cot α − α cot β / b -  α

18. A vertical rod is fixed in a horizontal rectangular field ABCD. The angular elevations of its top from A, B, C and D are α, β, λ & δ respectively, show that cot² α - cot² β δ - cot² λ

19. A ladder rests against a wall at angle α to the horizontal. Its foot is pulled away from the wall through a distance ‘a’ so that it slides a distance ‘b’ down the wall making an angle β with horizontal. Show that b/a = cos α - cos β / sin β - sin α

20. A man on the deck of a ship is 12 m above water level. He observes that the angle of elevation of the top of a cliff is 45°, and angle of depression of the base is 30°. Calculate the distance of cliff from the ship and the height of the cliff.

21. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression of the point ‘A’ from the top of the tower is 45°. Find the height of the tower.

22. From the top of a tower 50 m high the angle of depression of the top and buttom of a pole are observed to be 45° and 60° respectively. Find the height of the tower.

23. From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be α and β (α > β). If the distance between the objects is ‘P’ meters, show that the height ‘h’ of the tower is given by h = P tαn α tαn β / tαn α -  tαn β. Also determine the height of the tower and the building.

24. The angle of elevation of the top of a tower from a point A on the ground is 30° on making a distance of 20 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and distance of the tower from the point A.

25. The angle of elevation θ of the top of a light house, as seen by a person on the ground, is such that tanθ = 5/12. When the person moves a distance of 240 m towards the light house, the angle of elevation become φ such that tanφ=3/4. Find the ‘h’ of the light house. (h = height)

26. From a windown (60 m high above the ground) of a house in street the angles of elevation and depression of the top and the foot of another house on opposite side of street are 60° and 45° respectively. Show that the height of the opposite house is 60(1+√3) metres.

27. The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 10 sec the angle of elevation changes to 30°. If the jet is flying at a speed of 432 km/hr, find the constant height at which the jet is flying.

28. A round balloon of a radius ‘a’ subtends and angle θ at the eye of the observer while the angle of elevation of its centre is φ. Prove that the height of the centre of the balloon is a sinφ.cosecθ/2

Value based questions

1. A person standing on the bank of a river observes that the angle of elevation of the top of a building of an organization working for conservation of wild life, standing on the opposite bank is 60°. When he moves 40m away from the bank, he finds the angle of elevation to be 30°. Find the height of the building and the width of the river.
a) Why do we need to conserve wild life?
b) Suggest some steps that can be taken to conserve wild life.

2. Two hoardings on cleanliness are put on two poles of equal heights standing opposite to each other on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
a) How can we spread awareness for cleanliness in a society?
b) Does cleanliness play any role in the development of a society

3. From a window 15 metres high above of organization working for consumer protection the ground in a street, the angles of elevation and depression of the top and the foot of a building on the opposite side of the street are 300 and 450 respectively. Show that the height of the building is 23.66 metres (Take √3 = 1.732)
i) What do you understand by consumer protection?
ii) What is the need for such an organization in a society

4. A man constructed a school building in a village for welfare. The angle of elevation of the top of the building from a point on the bus stand is 30°. If the observer moves 50m towards the building, the angle of elevation of the top increases by 30°. Find the height of the building and the distance of the bus stand from the school building.
How is the man contributing towards the development of the society? What traits of his character are reflected here?

5. A hoarding on “Save Girl Child”, 5 m high is fixed on the top of a tower. The angle of elevation of the top of the hoarding as observed from a point A on the ground is 60° and the angle of depression of point A from the top of the tower is 45°. Find the height of the tower. (Take √3 = 1.73)
Why is it necessary to spread awareness for saving the girl child?

6. A survey was conducted in a particular area to find its most polluted region and it was found that the shaded region is the most polluted. If the radius of the circular part that was surveyed is 14m and the angle formed between the two radii is 60°, find the area of the polluted region. (Take π = 3.14 and √3 = 1.732)
i) How is pollution harmful?
ii) What steps can be taken to reduce pollution in any region?

MULTIPLE CHOICE QUESTIONS

Note : In the following questions 0° ≤ θ ≤ 90°

Question 1. If \(x = a \sin \theta\) and \(y = a \cos \theta\) then the value of \(x^2 + y^2\) is _______
(a) \(a\)
(b) \(a^2\)
(c) 1
(d) \(\frac{1}{a}\)
Answer: (b) \(a^2\)
Substitute the values of \(x\) and \(y\):
\[ x^2 + y^2 = (a \sin \theta)^2 + (a \cos \theta)^2 = a^2 \sin^2 \theta + a^2 \cos^2 \theta \]
\[ = a^2 (\sin^2 \theta + \cos^2 \theta) \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ = a^2(1) = a^2 \]
In simple words: Squaring both values and adding them allows you to factor out the square of \(a\), leaving a standard identity that simplifies to 1.

Exam Tip: Whenever you see a sum of squared sine and cosine terms with the same coefficient, factor out the coefficient immediately to simplify the expression.

Question 2. The value of \(\csc 70^\circ - \sec 20^\circ\) is _____
(a) 0
(b) 1
(c) 70°
(d) 20°
Answer: (a) 0
Using the complementary angle identity \(\csc(90^\circ - x) = \sec x\):
\[ \csc 70^\circ = \csc(90^\circ - 20^\circ) = \sec 20^\circ \delta \]
Substituting this back into the expression:
\[ \csc 70^\circ - \sec 20^\circ = \sec 20^\circ - \sec 20^\circ = 0 \]
In simple words: Since 70 and 20 add up to 90, the cosecant of 70 degrees is identical to the secant of 20 degrees, making their difference equal to zero.

Exam Tip: Spot complementary angle pairs (angles summing to 90 degrees) to quickly convert and cancel out terms.

Question 3. If \(3 \sec \theta - 5 = 0\) then \(\cot \theta = \) _____
(a) \(\frac{5}{3}\)
(b) \(\frac{4}{5}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{3}{5}\)
Answer: (c) \(\frac{3}{4}\)
Given:
\[ 3 \sec \theta = 5 \implies \sec \theta = \frac{5}{3} \]
Since \(\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}}\), we can take Hypotenuse = 5 and Base = 3.
Using Pythagoras theorem, the Perpendicular is:
\[ \text{Perpendicular} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4 \]
Now, we calculate cotangent:
\[ \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{3}{4} \]
In simple words: Isolate secant to find the ratio of the hypotenuse and the base. Use Pythagoras theorem to find the third side of the triangle, and then write the base over the height to find the cotangent.

Exam Tip: Remember the basic Pythagorean triplet (3, 4, 5) to save calculation time during examinations.

Question 4. If \(\theta = 45^\circ\) then \(\sec \theta \cot \theta - \csc \theta \tan \theta\) is
(a) 0
(b) 1
(c) \(\sqrt{2}\)
(d) \(2\sqrt{2}\)
Answer: (a) 0
Substitute \(\theta = 45^\circ\) into the expression:
- \(\sec 45^\circ = \sqrt{2}\)
- \(\cot 45^\circ = 1\)
- \(\csc 45^\circ = \sqrt{2}\)
- \(\tan 45^\circ = 1\)
Now, evaluate the expression:
\[ \sqrt{2}(1) - \sqrt{2}(1) = 0 \]
In simple words: Plug in the standard trigonometric values for 45 degrees, which simplifies the expression to subtracting the square root of 2 from itself, leaving 0.

Exam Tip: This can also be simplified algebraically first: \(\sec \theta \cot \theta - \csc \theta \tan \theta = \frac{1}{\sin\theta} - \frac{1}{\cos\theta}\). At \(45^\circ\), sine and cosine are equal, so their reciprocals are also equal, resulting in 0.

Question 5. If \(\sin(90 - \theta) \cos \theta = 1\) and \(\theta\) is an acute angle then \(\theta = \) ____
(a) 90°
(b) 60°
(c) 30°
(d) 0°
Answer: (d) 0°
Using the complementary angle formula \(\sin(90^\circ - \theta) = \cos \theta\):
\[ \cos \theta \cdot \cos \theta = 1 \implies \cos^2 \theta = 1 \implies \cos \theta = 1 \]
For an acute angle, \(\cos \theta = 1\) when:
\[ \theta = 0^\circ \]
In simple words: Convert the complementary sine term into cosine. This gives cosine squared is equal to 1, which means the angle must be 0 degrees.

Exam Tip: Read carefully whether the angle is specified as acute, obtuse, or bounded in a specific range before selecting your final answer.

Question 6. The value of \((1 + \cos \theta)(1 - \cos \theta) \csc^2 \theta = \) _____
(a) 0
(b) 1
(c) \(\cos^2 \theta\)
(d) \(\sin^2 \theta\)
Answer: (b) 1
Using the algebraic difference of squares identity \((1 + a)(1 - a) = 1 - a^2\):
\[ (1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta \]
Using the identity \(1 - \cos^2 \theta = \sin^2 \theta\):
\[ (1 - \cos^2 \theta) \csc^2 \theta = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \]
In simple words: Multiply the first two brackets to get sine squared. Multiplying sine squared by its reciprocal (cosecant squared) results in 1.

Exam Tip: Be comfortable with binomial expansions of the form \((1 - x)(1 + x) = 1 - x^2\) as they are very common in identity simplifications.

Question 7. \(\Delta TRY\) is a right-angled isosceles triangle then \(\cos T + \cos R + \cos Y\) is _____
(a) \(\sqrt{2}\)
(b) \(2\sqrt{2}\)
(c) \(1 + \sqrt{2}\)
(d) \(1 + \frac{1}{\sqrt{2}}\)
Answer: (a) \(\sqrt{2}\)
In a right-angled isosceles triangle, the right angle is at the vertex between the equal sides. Let \(\angle R = 90^\circ\).
Since it is isosceles, the other two angles must be equal:
\[ \angle T = \angle Y = \frac{180^\circ - 90^\circ}{2} = 45^\circ \]
Now evaluate the sum:
\[ \cos T + \cos R + \cos Y = \cos 45^\circ + \cos 90^\circ + \cos 45^\circ \]
\[ = \frac{1}{\sqrt{2}} + 0 + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
In simple words: The right angle of the isosceles triangle has a cosine of 0, and the other two 45-degree angles have a cosine of 1 over the square root of 2. Adding them together gives the square root of 2.

Exam Tip: For right-angled isosceles triangles, the acute angles are always \(45^\circ\). Use this property directly to evaluate the functions.

Question 8. If \(\sec \theta + \tan \theta = x\), then \(\sec \theta = \)
(a) \(\frac{x^2+1}{x}\)
(b) \(\frac{x^2+1}{2x}\)
(c) \(\frac{x^2-1}{2x}\)
(d) \(\frac{x^2-1}{x}\)
Answer: (b) \(\frac{x^2+1}{2x}\)
We know the identity \(\sec^2 \theta - \tan^2 \theta = 1\), which can be written as:
\[ (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \]
Since \(\sec \theta + \tan \theta = x\):
\[ \sec \theta - \tan \theta = \frac{1}{x} \]
Adding these two equations:
\[ (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = x + \frac{1}{x} \]
\[ 2 \sec \theta = \frac{x^2 + 1}{x} \]
\[ \sec \theta = \frac{x^2 + 1}{2x} \]
In simple words: Combine the given equation with the reciprocal identity of secant and tangent to set up a system. Adding them cancels out tangent, leaving the formula for secant.

Exam Tip: If \(\sec\theta + \tan\theta = x\), then \(\sec\theta - \tan\theta = 1/x\). This reciprocal relationship is extremely useful for solving algebraic trig problems.

Question 9. The value of \(\cot\theta - \sin\left(\frac{\pi}{2} - \theta\right)\cos\left(\frac{\pi}{2} - \theta\right)\) is _______
(a) \(\cot \theta \cos^2 \theta\)
(b) \(\cot^2\theta\)
(c) \(\cos^2 \theta\)
(d) \(\tan^2 \theta\)
Answer: (a) \(\cot \theta \cos^2 \theta\)
Note that \(\frac{\pi}{2}\) radians is equal to \(90^\circ\).
Using complementary angle relations:
- \(\sin(90^\circ - \theta) = \cos \theta\)
- \(\cos(90^\circ - \theta) = \sin \theta\)
Substituting these into the expression:
\[ = \cot\theta - \cos\theta \sin\theta \]
\[ = \frac{\cos\theta}{\sin\theta} - \cos\theta \sin\theta \]
Take \(\cos\theta\) common:
\[ = \cos\theta \left(\frac{1}{\sin\theta} - \sin\theta\right) \]
\[ = \cos\theta \left(\frac{1 - \sin^2\theta}{\sin\theta}\right) \]
Since \(1 - \sin^2\theta = \cos^2\theta\):
\[ = \cos\theta \cdot \frac{\cos^2\theta}{\sin\theta} = \left(\frac{\cos\theta}{\sin\theta}\right) \cos^2\theta = \cot\theta \cos^2\theta \]
In simple words: Rewrite the complementary terms first. Express cotangent as cosine over sine, and simplify the fractions to obtain the matching product.

Exam Tip: Be comfortable switching between radian measures (like \(\pi/2\)) and degrees (\(90^\circ\)) when solving Board questions.

Question 10. If \(\sin \theta - \cos \theta = 0\), \(0^\circ \le \theta \le 90^\circ\) then the value of \(\theta\) is _____
(a) \(\cos \theta\)
(b) 45°
(c) 90°
(d) \(\sin \theta\)
Answer: (b) 45°
Given:
\[ \sin \theta - \cos \theta = 0 \implies \sin \theta = \cos \theta \]
Divide both sides by \(\cos \theta\):
\[ \tan \theta = 1 \]
Since \(\tan 45^\circ = 1\), we have \(\theta = 45^\circ\).
In simple words: If sine minus cosine is zero, then sine equals cosine. The only angle between 0 and 90 degrees where this is true is 45 degrees.

Exam Tip: Equating sine and cosine directly is the easiest path to introduce tangent and find the angle value.

Question 11. \(\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\) can be written as
(a) \(\cot \theta\)
(b) \(\sqrt{\sin \theta}\)
(c) \(\frac{\sin \theta}{\sqrt{\cos \theta}}\)
(d) \(\tan \theta\)
Answer: (d) \(\tan \theta\)
Using the identity \(1 - \sin^2 \theta = \cos^2 \theta\):
\[ \sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta \]
Substituting this into the fraction:
\[ = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
In simple words: The denominator simplifies to the square root of cosine squared, which is just cosine. Sine divided by cosine is tangent.

Exam Tip: Substituting basic Pythagorean identities under radical signs is the standard way to simplify expressions.

Question 12. \(\sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}\) is equal to
(a) \(\sec^2\theta + \tan^2\theta\)
(b) \(\sec\theta - \tan\theta\)
(c) \(\sec^2\theta - \tan^2\theta\)
(d) \(\sec\theta + \tan\theta\)
Answer: (d) \(\sec\theta + \tan\theta\)
Multiply the numerator and denominator inside the square root by \((1 + \sin \theta)\):
\[ \sqrt{\frac{(1 + \sin \theta)(1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}} = \sqrt{\frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta}} \]
Using the identity \(1 - \sin^2 \theta = \cos^2 \theta\):
\[ = \sqrt{\frac{(1 + \sin \theta)^2}{\cos^2 \theta}} = \frac{1 + \sin \theta}{\cos \theta} \]
Splitting the fraction:
\[ = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \sec \theta + \tan \theta \]
In simple words: Multiply inside the root by the conjugate of the denominator to form a perfect square. Taking the square root and dividing the terms gives secant plus tangent.

Exam Tip: Rationalizing denominators by multiplying by the conjugate under a square root is the best way to clear the radical sign in trigonometry.

Question 13. In an isosceles right-angled \(\Delta ABC\), \(\angle B = 90^\circ\). The value of \(2 \sin A \cos A\) is _____
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) \(\sqrt{2}\)
Answer: (a) 1
Since \(\Delta ABC\) is right-angled at \(B\) and is also isosceles, we have \(AB = BC\).
Thus, the two acute angles are equal:
\[ \angle A = \angle C = 45^\circ \]
Now, substitute the value of \(A\):
\[ 2 \sin 45^\circ \cos 45^\circ = 2 \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) = 2 \left(\frac{1}{2}\right) = 1 \]
In simple words: The acute angles of an isosceles right triangle are always 45 degrees. Plugging this into the formula gives 2 multiplied by half, which equals 1.

Exam Tip: Recognizing that \(2\sin A\cos A = \sin 2A\) lets you write \(\sin 2(45^\circ) = \sin 90^\circ = 1\) directly.

Question 14. If \(\frac{\sin^2 20^\circ + \sin^2 70^\circ}{2(\cos^2 69^\circ + \cos^2 21^\circ)} = \frac{\sec 60^\circ}{K}\), then K is ______
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4
Use complementary angle relations:
- \(\sin^2 70^\circ = \cos^2 20^\circ\)
- \(\cos^2 21^\circ = \sin^2 69^\circ\)
Substituting these into the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\sin^2 20^\circ + \cos^2 20^\circ}{2(\cos^2 69^\circ + \sin^2 69^\circ)} = \frac{1}{2(1)} = \frac{1}{2} \delta \]
Now evaluate the Right-Hand Side (RHS):
Since \(\sec 60^\circ = 2\):
\[ \text{RHS} = \frac{2}{K} \]
Equating both sides:
\[ \frac{1}{2} = \frac{2}{K} \implies K = 4 \]
In simple words: Use complementary identities to simplify the fraction to one-half. Since the numerator on the other side is 2, the value of K must be 4 to make both sides equal.

Exam Tip: Always look for matching pairs that sum to 90 degrees to simplify numerator and denominator terms separately.

Question 15. If \(\tan \theta = \frac{1}{\sqrt{7}}\), then \(\frac{\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta} = \) _____
(a) \(\frac{3}{4}\)
(b) \(\frac{5}{7}\)
(c) \(\frac{3}{7}\)
(d) \(\frac{1}{12}\)
Answer: (a) \(\frac{3}{4}\)
We are given \(\tan \theta = \frac{1}{\sqrt{7}} \implies \tan^2 \theta = \frac{1}{7}\).
Thus, \(\cot^2 \theta = 7\).
Using the trigonometric identities:
- \(\csc^2 \theta = 1 + \cot^2 \theta = 1 + 7 = 8\)
- \(\sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{1}{7} = \frac{8}{7}\)
Substitute these into the expression:
\[ E = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{8\left(1 - \frac{1}{7}\right)}{8\left(1 + \frac{1}{7}\right)} = \frac{\frac{6}{7}}{\frac{8}{7}} = \frac{6}{8} = \frac{3}{4} \]
In simple words: Find cotangent squared from tangent squared. Use these to find cosecant and secant squared, and substitute them into the fraction to simplify.

Exam Tip: Factoring out the common constant term from the numerator and denominator before performing fractional addition saves a lot of calculation time.

SHORT ANSWER TYPE QUESTIONS

Question 16. In \(\Delta PQR\), \(\angle Q = 90^\circ\) and \(\sin R = \frac{3}{5}\), write the value of \(\cos P\).
Answer:
In a right-angled triangle \(PQR\) with \(\angle Q = 90^\circ\), the acute angles \(P\) and \(R\) are complementary:
\[ P + R = 90^\circ \implies P = 90^\circ - R \]
Therefore:
\[ \cos P = \cos(90^\circ - R) \]
Using the complementary angle identity \(\cos(90^\circ - R) = \sin R\):
\[ \cos P = \sin R = \frac{3}{5} \]
In simple words: Since the triangle is right-angled, the other two angles must add up to 90 degrees. This means the cosine of one angle is equal to the sine of the other, which is 3 over 5.

Exam Tip: Recognizing that the two acute angles in a right-angled triangle are complementary allows you to write the answer directly without computing any side lengths.

Question 17. If \(A\) and \(B\) are acute angles and \(\sin A = \cos B\) then write the value of \(A + B\).
Answer:
Given:
\[ \sin A = \cos B \]
Using the complementary angle identity \(\cos B = \sin(90^\circ - B)\):
\[ \sin A = \sin(90^\circ - B) \]
Comparing the angles on both sides:
\[ A = 90^\circ - B \implies A + B = 90^\circ \]
The value of \(A + B\) is \(90^\circ\).
In simple words: Sine of \(A\) equals cosine of \(B\) only when the two angles add up to 90 degrees.

Exam Tip: Complementary relations like \(\sin A = \cos B\) always imply that \(A+B=90^\circ\) for acute angles.

Question 18. If \(4 \cot \theta = 3\) then write the value of \(\tan \theta + \cot \theta\).
Answer:
Given:
\[ 4 \cot \theta = 3 \implies \cot \theta = \frac{3}{4} \]
Since \(\tan \theta = \frac{1}{\cot \theta}\):
\[ \tan \theta = \frac{4}{3} \]
Now evaluate the sum:
\[ \tan \theta + \cot \theta = \frac{4}{3} + \frac{3}{4} = \frac{16 + 9}{12} = \frac{25}{12} \]
The value of the expression is \(\frac{25}{12}\).
In simple words: Find cotangent from the equation, take its reciprocal to get tangent, and add the two fractions together.

Exam Tip: Be careful to find a common denominator (which is 12) before adding the two fractions.

Question 19. Write the value of \(\cot^2 30^\circ + \sec^2 45^\circ\).
Answer:
Substitute the standard exact trigonometric values:
- \(\cot 30^\circ = \sqrt{3}\)
- \(\sec 45^\circ = \sqrt{2}\)
Evaluate the sum of squares:
\[ \cot^2 30^\circ + \sec^2 45^\circ = (\sqrt{3})^2 + (\sqrt{2})^2 = 3 + 2 = 5 \]
The value is 5.
In simple words: Square the values of cotangent 30 and secant 45 to get 3 and 2, then add them together to get 5.

Exam Tip: Squaring a square root yields the original number. Keep your calculations direct and clean.

Question 20. Given that \(16 \cot A = 12\), find the value of \(\frac{\sin A + \cos A}{\sin A - \cos A}\).
Answer:
We are given:
\[ 16 \cot A = 12 \implies \cot A = \frac{12}{16} = \frac{3}{4} \]
We want to evaluate:
\[ E = \frac{\sin A + \cos A}{\sin A - \cos A} \]
Divide the numerator and denominator by \(\sin A\):
\[ E = \frac{\frac{\sin A}{\sin A} + \frac{\cos A}{\sin A}}{\frac{\sin A}{\sin A} - \frac{\cos A}{\sin A}} = \frac{1 + \cot A}{1 - \cot A} \]
Substitute the value of \(\cot A = \frac{3}{4}\):
\[ E = \frac{1 + \frac{3}{4}}{1 - \frac{3}{4}} = \frac{\frac{7}{4}}{\frac{1}{4}} = 7 \]
The value of the expression is 7.
In simple words: Divide the top and bottom of the fraction by sine to convert the terms to cotangent. This allows you to evaluate the fraction using the given cotangent value directly.

Exam Tip: Dividing a fraction containing sine and cosine terms by sine is an excellent shortcut to convert the entire expression into cotangent.

Question 21. If \(\theta = 30^\circ\) then write the value of \(\sin \theta + \cos^2 \theta\).
Answer:
Substitute \(\theta = 30^\circ\) into the expression:
- \(\sin 30^\circ = \frac{1}{2}\)
- \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
Now, calculate the value:
\[ \sin 30^\circ + \cos^2 30^\circ = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ = \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4} \]
The value of the expression is \(\frac{5}{4}\).
In simple words: Plug in the standard values for 30 degrees, square the cosine term, and add the fractions together.

Exam Tip: Make sure to square the numerator and denominator separately when squaring fractions containing roots.

Question 22. If \(1 - \tan^2 \theta = \frac{2}{3}\) then what is the value of \(\theta\).
Answer:
Given equation:
\[ 1 - \tan^2 \theta = \frac{2}{3} \]
Rearranging the terms to isolate \(\tan^2 \theta\):
\[ \tan^2 \theta = 1 - \frac{2}{3} = \frac{1}{3} \]
Taking the positive square root for an acute angle \(\theta\):
\[ \tan \theta = \frac{1}{\sqrt{3}} \]
We know that \(\tan 30^\circ = \frac{1}{\sqrt{3}}\). Therefore:
\[ \theta = 30^\circ \delta \]
The value of \(\theta\) is \(30^\circ\).
In simple words: Rearrange the equation to find tangent squared. Taking the square root shows that tangent is 1 over root 3, which corresponds to an angle of 30 degrees.

Exam Tip: Be sure to write the degree symbol (\(^\circ\)) alongside your numerical answer for angles to prevent deduction of marks.

Question 23. Find the value of \(\theta\) if \(\sqrt{3} \tan 2\theta - 3 = 0\).
Answer:
Given equation:
\[ \sqrt{3} \tan 2\theta - 3 = 0 \]
\[ \sqrt{3} \tan 2\theta = 3 \]
\[ \tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3} \]
We know that \(\tan 60^\circ = \sqrt{3}\). Comparing the angles:
\[ 2\theta = 60^\circ \implies \theta = 30^\circ \]
The value of \(\theta\) is \(30^\circ\).
In simple words: Move 3 to the other side and divide by root 3. This shows that the tangent of 2-theta is root 3, meaning 2-theta is 60 degrees and theta is 30 degrees.

Exam Tip: Do not divide the angle inside the tangent function by 2 before evaluating the function's value first.

Question 24. If \(\theta\) and \(\phi\) are complementary angles then what is the value of \(\csc \theta \sec \phi - \cot \theta \tan \phi\)
Answer:
Since \(\theta\) and \(\phi\) are complementary angles, we have:
\[ \phi = 90^\circ - \theta \]
Using complementary identities:
- \(\sec \phi = \sec(90^\circ - \theta) = \csc \theta\)
- \(\tan \phi = \tan(90^\circ - \theta) = \cot \theta\)
Substitute these into the given expression:
\[ \csc \theta \sec \phi - \cot \theta \tan \phi = \csc \theta (\csc \theta) - \cot \theta (\cot \theta) \]
\[ = \csc^2 \theta - \cot^2 \theta \]
Using the standard identity \(\csc^2 \theta - \cot^2 \theta = 1\):
\[ = 1 \]
The value of the expression is 1.
In simple words: Convert the angles of the second terms to their complements. This simplifies the expression to a standard identity that always equals 1.

Exam Tip: Complementary angle conversions are extremely useful to reduce expressions containing multiple angle variables into a single angle.

Question 25. If \(\tan(3x - 15^\circ) = 1\) then what is the value of \(x\).
Answer:
Given equation:
\[ \tan(3x - 15^\circ) = 1 \]
We know that \(\tan 45^\circ = 1\). Comparing the angles:
\[ 3x - 15^\circ = 45^\circ \]
\[ 3x = 60^\circ \implies x = 20^\circ \]
The value of \(x\) is \(20^\circ\).
In simple words: Since tangent is 1, the angle inside must be 45 degrees. Solve the resulting equation to find that x is 20 degrees.

Exam Tip: Be sure to add \(15^\circ\) to \(45^\circ\) first before dividing by 3 to isolate \(x\) correctly.

Question 26. If \(\sin 5\theta = \cos 4\theta\), where \(5\theta\) and \(4\theta\) are acute angles. Find the value of \(\theta\).
Answer:
Given:
\[ \sin 5\theta = \cos 4\theta \]
Using the complementary identity \(\cos 4\theta = \sin(90^\circ - 4\theta)\):
\[ \sin 5\theta = \sin(90^\circ - 4\theta) \]
Comparing the angles:
\[ 5\theta = 90^\circ - 4\theta \]
\[ 9\theta = 90^\circ \implies \theta = 10^\circ \]
The value of \(\theta\) is \(10^\circ\).
In simple words: Convert the cosine term to its complementary sine term. This allows you to set up a linear equation to find that theta is 10 degrees.

Exam Tip: Always verify that your final answer satisfies any given constraints, such as both angles being acute (\(50^\circ\) and \(40^\circ\) are indeed acute).

LONG ANSWER TYPE QUESTIONS

Question 27. Simplify : \(\tan^2 60^\circ + 4 \cos^2 45^\circ + 3 (\sec^2 30^\circ + \cos^2 90^\circ)\)
Answer:
Substitute the standard trigonometric values:
- \(\tan 60^\circ = \sqrt{3} \implies \tan^2 60^\circ = 3\)
- \(\cos 45^\circ = \frac{1}{\sqrt{2}} \implies \cos^2 45^\circ = \frac{1}{2}\)
- \(\sec 30^\circ = \frac{2}{\sqrt{3}} \implies \sec^2 30^\circ = \frac{4}{3}\)
- \(\cos 90^\circ = 0 \implies \cos^2 90^\circ = 0\)
Now substitute these values into the given expression:
\[ E = 3 + 4\left(\frac{1}{2}\right) + 3\left(\frac{4}{3} + 0\right) \]
\[ E = 3 + 2 + 3\left(\frac{4}{3}\right) \]
\[ E = 5 + 4 = 9 \]
The simplified value of the expression is 9.
In simple words: Replace the functions with their standard numerical values, square them, and perform the basic arithmetic operations to find the final integer answer of 9.

Exam Tip: Writing down each step of the substitution and squaring clearly will ensure you do not make simple calculation slips.

Question 28. Evaluate \(2 \left(\frac{\cos 58^\circ}{\sin 32^\circ}\right) - \sqrt{3} \left(\frac{\cos 38^\circ \csc 52^\circ}{\tan 15^\circ \tan 60^\circ \tan 75^\circ}\right)\)
Answer:
Let's simplify each part of the expression separately using complementary relationships:
1. First term: \(\cos 58^\circ = \sin(90^\circ - 58^\circ) = \sin 32^\circ\). Thus:
\[ \frac{\cos 58^\circ}{\sin 32^\circ} = \frac{\sin 32^\circ}{\sin 32^\circ} = 1 \]
So, the first part is \(2(1) = 2\).

2. Second term:
Numerator: \(\cos 38^\circ \csc 52^\circ = \cos 38^\circ \left(\frac{1}{\sin 52^\circ}\right)\). Since \(\sin 52^\circ = \cos(90^\circ - 52^\circ) = \cos 38^\circ\):
\[ \cos 38^\circ \csc 52^\circ = \frac{\cos 38^\circ}{\cos 38^\circ} = 1 \]
Denominator: \(\tan 15^\circ \tan 60^\circ \tan 75^\circ\). Since \(\tan 75^\circ = \cot 15^\circ\) and \(\tan 60^\circ = \sqrt{3}\):
\[ = (\tan 15^\circ \cot 15^\circ) \cdot \sqrt{3} = 1 \cdot \sqrt{3} = \sqrt{3} \]
So, the second part becomes:
\[ \sqrt{3} \cdot \left(\frac{1}{\sqrt{3}}\right) = 1 \]

Combine both parts:
\[ E = 2 - 1 = 1 \]
The final evaluated value is 1.
In simple words: Convert the complementary angles in both sections. This simplifies the fractions to 1, leaving a basic subtraction that equals 1.

Exam Tip: Group complementary pairs (such as 15 and 75, or 38 and 52) to identify which terms can be converted and canceled easily.

Question 29. Prove that cosec4 \(\theta\) – cosec2 \(\theta\) = cot2 \(\theta\) + cot4 \(\theta\).
Answer:
Let us evaluate the Left-Hand Side (LHS):
\[ \text{LHS} = \csc^4 \theta - \csc^2 \theta \]
Factor out \(\csc^2 \theta\):
\[ \text{LHS} = \csc^2 \theta (\csc^2 \theta - 1) \]
We know the standard Pythagorean identities:
- \(\csc^2 \theta = 1 + \cot^2 \theta\)
- \(\csc^2 \theta - 1 = \cot^2 \theta\)
Substitute these identities into our factored expression:
\[ \text{LHS} = (1 + \cot^2 \theta) \cot^2 \theta \]
Expand the expression:
\[ \text{LHS} = \cot^2 \theta + \cot^4 \theta = \text{RHS} \]
Hence proved.
In simple words: Take out cosecant squared as a common factor, and then replace cosecant squared with one plus cotangent squared to convert everything into cotangent.

Exam Tip: Factoring out common powers is a highly effective first step when working with higher-degree trigonometric identities.

Question 30. If sin \(\theta\) + sin2 \(\theta\) = 1 then find the value of cos2 \(\theta\) + cos4 \(\theta\)
Answer:
We are given:
\[ \sin \theta + \sin^2 \theta = 1 \implies \sin \theta = 1 - \sin^2 \theta \]
Using the identity \(1 - \sin^2 \theta = \cos^2 \theta\):
\[ \sin \theta = \cos^2 \theta \]
Now, we evaluate the expression:
\[ \cos^2 \theta + \cos^4 \theta = \cos^2 \theta + (\cos^2 \theta)^2 \]
Substituting \(\cos^2 \theta = \sin \theta\):
\[ = \sin \theta + \sin^2 \theta \]
From our given equation, this sum is equal to 1.
Thus, the value of \(\cos^2 \theta + \cos^4 \theta\) is 1.
In simple words: Use the Pythagorean identity to rewrite the given equation as sine equals cosine squared. Substituting this relation into the required expression simplifies it back to the original equation, which equals 1.

Exam Tip: When given a trigonometric equation equal to 1, try isolating one term on one side to find useful substitutions for higher-power terms.

Question 31. If sin 2\(\theta\) = cos (\(\theta\) – 36°), 2\(\theta\) and \(\theta\) – 26° are acute angles then find the value of \(\theta\).
Answer:
Given equation:
\[ \sin 2\theta = \cos(\theta - 36^\circ) \]
Using the complementary angle identity \(\sin x = \cos(90^\circ - x)\):
\[ \cos(90^\circ - 2\theta) = \cos(\theta - 36^\circ) \]
Comparing the angles on both sides:
\[ 90^\circ - 2\theta = \theta - 36^\circ \]
\[ 90^\circ + 36^\circ = \theta + 2\theta \]
\[ 126^\circ = 3\theta \]
\[ \theta = \frac{126^\circ}{3} = 42^\circ \]
The value of \(\theta\) is \(42^\circ\).
In simple words: Convert the sine term to its complementary cosine term. This allows you to set up a linear equation to find that theta is 42 degrees.

Exam Tip: Be careful to verify that your final answer satisfies the given conditions, such as both angles being acute (\(84^\circ\) and \(16^\circ\) are indeed acute).

Question 32. If sin (3x + 2y) = 1 and cos (3x – 2y) = \(\frac{\sqrt{3}}{2}\) , where 0 ≤ (3x + 2y) ≤ 90° then find the value of x and y.
Answer:
Given trigonometric equations:
1. \(\sin(3x + 2y) = 1\)
Since \(\sin 90^\circ = 1\):
\[ 3x + 2y = 90^\circ \quad \text{--- (Equation 1)} \]
2. \(\cos(3x - 2y) = \frac{\sqrt{3}}{2}\)
Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ 3x - 2y = 30^\circ \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ (3x + 2y) + (3x - 2y) = 90^\circ + 30^\circ \]
\[ 6x = 120^\circ \implies x = 20^\circ \]
Subtracting Equation 2 from Equation 1:
\[ (3x + 2y) - (3x - 2y) = 90^\circ - 30^\circ \]
\[ 4y = 60^\circ \implies y = 15^\circ \]
The values of \(x\) and \(y\) are \(20^\circ\) and \(15^\circ\).
In simple words: Convert the given trigonometric equations into two linear equations by matching them with standard values. Solve these simultaneous equations to find \(x\) and \(y\).

Exam Tip: Elimination is the simplest way to solve linear systems when the coefficients of one variable (like \(3x\)) match perfectly.

Question 33. If sin (A + B) = sin A cos B + cos A sin B then find the value of (a) sin 75° (b) cos 15°
Answer:
We are given the expansion formula for sine addition.
Let \(A = 45^\circ\) and \(B = 30^\circ\).

(a) To find \(\sin 75^\circ\):
\[ \sin 75^\circ = \sin(45^\circ + 30^\circ) \]
Using the given formula:
\[ \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \]
Substitute standard values \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), and \(\sin 30^\circ = \frac{1}{2}\):
\[ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \]
\[ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

(b) To find \(\cos 15^\circ\):
Using complementary angle relations:
\[ \cos 15^\circ = \sin(90^\circ - 15^\circ) = \sin 75^\circ \]
Substituting our result from part (a):
\[ \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]
In simple words: Split 75 degrees into the sum of 45 and 30 degrees. Applying the given formula and substituting standard values gives the exact answer, which is also equal to cosine of 15 degrees.

Exam Tip: Splitting non-standard angles into sums of standard angles (like \(45^\circ\) and \(30^\circ\)) is a very useful technique to find exact trigonometric values.

Question 34. Prove that cos A / (1 - tan A) + cos A / (1 - cot A) = cos A , A ≠ 45°.
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\cos A}{1 - \tan A} + \frac{\cos A}{1 - \cot A} \]
Substitute \(\tan A = \frac{\sin A}{\cos A}\) and \(\cot A = \frac{\cos A}{\sin A}\):
\[ \text{LHS} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\cos A}{1 - \frac{\cos A}{\sin A}} \]
\[ = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\cos A \sin A}{\sin A - \cos A} \]
Convert the second term's denominator to match the first by factoring out a negative sign:
\[ = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\cos A \sin A}{\cos A - \sin A} \]
Combine the fractions:
\[ = \frac{\cos^2 A - \cos A \sin A}{\cos A - \sin A} \]
Factor out \(\cos A\) in the numerator:
\[ = \frac{\cos A (\cos A - \sin A)}{\cos A - \sin A} \]
Canceling the common term \(\cos A - \sin A\):
\[ = \cos A = \text{RHS} \]
Hence proved.
In simple words: Express tangent and cotangent in terms of sine and cosine. Simplifying the fractions and finding a common denominator allows you to factor the numerator and cancel out the denominator.

Exam Tip: Converting everything to sine and cosine is a highly reliable fallback strategy for almost all trigonometric identity proofs.

Question 35. Prove that \(\sqrt{\frac{\sec\theta - 1}{\sec\theta + 1}} + \sqrt{\frac{\sec\theta + 1}{\sec\theta - 1}} = 2\csc\theta\).
Answer:
Let's combine the two fractions on the Left-Hand Side (LHS) using a common denominator:
\[ \text{LHS} = \frac{(\sec\theta - 1) + (\sec\theta + 1)}{\sqrt{(\sec\theta + 1)(\sec\theta - 1)}} \]
Simplify the numerator and denominator:
\[ = \frac{2\sec\theta}{\sqrt{\sec^2\theta - 1}} \]
Using the identity \(\sec^2\theta - 1 = \tan^2\theta\):
\[ = \frac{2\sec\theta}{\sqrt{\tan^2\theta}} = \frac{2\sec\theta}{\tan\theta} \]
Express in terms of sine and cosine:
\[ = \frac{\frac{2}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{2}{\sin\theta} = 2\csc\theta = \text{RHS} \]
Hence proved.
In simple words: Combine the radical terms over a common denominator. The bottom simplifies to tangent, which cancels with secant to leave cosecant.

Exam Tip: When combining square roots of reciprocal fractions, remember that \(\sqrt{a/b} + \sqrt{b/a} = \frac{a+b}{\sqrt{ab}}\).

Question 36. Find the value of sin2 5° + sin2 10° + sin2 15° + .... + sin2 85°
Answer:
The given sequence consists of terms of the form \(\sin^2 \theta\) where \(\theta\) ranges from \(5^\circ\) to \(85^\circ\) in steps of \(5^\circ\).
The number of terms is:
\[ N = \frac{85 - 5}{5} + 1 = 17 \text{ terms} \]
We can pair the complementary terms using the identity \(\sin^2(90^\circ - \theta) = \cos^2 \theta\):
\[ \sin^2 \theta + \sin^2(90^\circ - \theta) = \sin^2 \theta + \cos^2 \theta = 1 \]
The complementary pairs are:
- \(\sin^2 5^\circ + \sin^2 85^\circ = 1\)
- \(\sin^2 10^\circ + \sin^2 80^\circ = 1\)
- ...
- \(\sin^2 40^\circ + \sin^2 50^\circ = 1\)
There are exactly 8 such pairs, contributing a sum of 8.
The only term left unpaired is the middle term, which is \(\sin^2 45^\circ\):
\[ \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Therefore, the total sum is:
\[ E = 8 + \frac{1}{2} = \frac{17}{2} \]
In simple words: Pair the complementary angles that sum to 90 degrees since their squared sines add up to 1. This gives 8 pairs plus the middle term of sine squared 45 degrees, totaling 17 over 2.

Exam Tip: Be sure to count the total number of terms and identify the single unpaired middle term correctly in series questions.

Question 37. Prove that tan \(\theta\) + sec \(\theta\) – 1 / tan \(\theta\) – sec \(\theta\) + 1 = cos \(\theta\) / 1 – sin \(\theta\) .
Answer:
Let us simplify the Left-Hand Side (LHS) of the expression:
\[ \text{LHS} = \frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} \]
Using the identity \(1 = \sec^2\theta - \tan^2\theta\):
\[ = \frac{(\tan\theta + \sec\theta) - (\sec^2\theta - \tan^2\theta)}{\tan\theta - \sec\theta + 1} \]
\[ = \frac{(\tan\theta + \sec\theta)[1 - (\sec\theta - \tan\theta)]}{\tan\theta - \sec\theta + 1} \]
\[ = \frac{(\tan\theta + \sec\theta)(1 - \sec\theta + \tan\theta)}{\tan\theta - \sec\theta + 1} = \tan\theta + \sec\theta \]
Express in terms of sine and cosine:
\[ = \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta} = \frac{1 + \sin\theta}{\cos\theta} \]
Multiply the numerator and denominator by \((1 - \sin\theta)\):
\[ = \frac{(1 + \sin\theta)(1 - \sin\theta)}{\cos\theta(1 - \sin\theta)} = \frac{1 - \sin^2\theta}{\cos\theta(1 - \sin\theta)} \]
Since \(1 - \sin^2\theta = \cos^2\theta\):
\[ = \frac{\cos^2\theta}{\cos\theta(1 - \sin\theta)} = \frac{\cos\theta}{1 - \sin\theta} = \text{RHS} \]
Hence proved.
In simple words: Replace the constant 1 in the numerator with the secant-tangent identity to factor and simplify the fraction to secant plus tangent. Convert this to sine and cosine, and rationalize to reach the right-hand side.

Exam Tip: Substituting identities for the number 1 is a very common algebraic trick used to factor and simplify complex rational trigonometric fractions.

Question 38. If 2 sin (3x – 15°) = \(\sqrt{3}\) then find the value of sin² (2x + 10°) + tan² (x + 5°) .
Answer:
Given equation:
\[ 2 \sin(3x - 15^\circ) = \sqrt{3} \implies \sin(3x - 15^\circ) = \frac{\sqrt{3}}{2} \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ 3x - 15^\circ = 60^\circ \implies 3x = 75^\circ \implies x = 25^\circ \]
Now, substitute \(x = 25^\circ\) into the target expression:
\[ E = \sin^2(2(25^\circ) + 10^\circ) + \tan^2(25^\circ + 5^\circ) \]
\[ E = \sin^2(60^\circ) + \tan^2(30^\circ) \]
Substitute standard values:
\[ E = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{3}{4} + \frac{1}{3} = \frac{9 + 4}{12} = \frac{13}{12} \]
The value of the expression is \(\frac{13}{12}\).
In simple words: First solve the given sine equation to find that x is 25 degrees. Substituting this value into the second expression yields standard angles whose squared values sum to 13 over 12.

Exam Tip: Be sure to solve the initial angle equation completely to find \(x\) before attempting any substitution in the second expression.

Question 39. Find the value of sin 60° geometrically.
Answer:
Consider an equilateral triangle \(ABC\) with side length \(a\). Each angle in this triangle is \(60^\circ\).
Draw a perpendicular altitude \(AD\) from vertex \(A\) to side \(BC\).
Since \(AD\) is the altitude, it bisects the base \(BC\), so \(BD = \frac{a}{2}\), and \(\angle BAD = 30^\circ\).
In right-angled triangle \(ABD\), using Pythagoras theorem:
\[ AB^2 = AD^2 + BD^2 \]
\[ a^2 = AD^2 + \left(\frac{a}{2}\right)^2 \]
\[ AD^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4} \]
\[ AD = \frac{\sqrt{3}}{2}a \delta \]
Now, in right-angled triangle \(ABD\), for angle \(\angle B = 60^\circ\):
\[ \sin 60^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AD}{AB} = \frac{\frac{\sqrt{3}}{2}a}{a} = \frac{\sqrt{3}}{2} \]
Thus, \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).
In simple words: Draw an equilateral triangle and cut it in half with an altitude line. Applying Pythagoras theorem to the resulting right triangle allows you to find the ratio of the height to the hypotenuse, which is sine of 60 degrees.

Exam Tip: A neat, labeled geometric diagram is essential to secure full marks in geometric derivations.

Question 40. Let p = tan \(\theta\) + sec \(\theta\) then find the value of p + \(\frac{1}{p}\).
Answer:
We are given:
\[ p = \sec\theta + \tan\theta \]
Using the identity \(\sec^2\theta - \tan^2\theta = 1 \implies (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1\):
\[ \sec\theta - \tan\theta = \frac{1}{p} \]
Now, calculate the sum:
\[ p + \frac{1}{p} = (\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = 2\sec\theta \]
The value is \(2\sec\theta\).
In simple words: Since secant plus tangent is p, secant minus tangent is 1 over p. Adding these two equations cancels tangent, leaving twice secant.

Exam Tip: Always remember that the sum and difference of secant and tangent are reciprocal values of each other.

Question 41. Find the value of - tan \(\theta\) cot (90° - \(\theta\)) + sec \(\theta\) cosec (90° - \(\theta\)) + sin² 35° + sin² 55° / tan 10° tan 20° tan 30° tan 70° tan 80°
Answer:
Let's simplify the numerator first:
Using complementary identities:
- \(\cot(90^\circ - \theta) = \tan\theta \implies -\tan\theta \cot(90^\circ - \theta) = -\tan^2\theta\)
- \(\csc(90^\circ - \theta) = \sec\theta \implies \sec\theta \csc(90^\circ - \theta) = \sec^2\theta\)
- \(\sin^2 55^\circ = \cos^2 35^\circ \implies \sin^2 35^\circ + \sin^2 55^\circ = 1\)
Substituting these in the numerator:
\[ \text{Numerator} = \sec^2\theta - \tan^2\theta + 1 \]
Since \(\sec^2\theta - \tan^2\theta = 1\):
\[ \text{Numerator} = 1 + 1 = 2 \]
Now, let's simplify the denominator:
Group complementary pairs:
\[ \text{Denominator} = (\tan 10^\circ \tan 80^\circ)(\tan 20^\circ \tan 70^\circ) \tan 30^\circ \]
Since \(\tan 80^\circ = \cot 10^\circ\), \(\tan 70^\circ = \cot 20^\circ\), and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\[ \text{Denominator} = (1)(1)\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} \]
Evaluating the entire fraction:
\[ E = \frac{\text{Numerator}}{\text{Denominator}} = \frac{2}{\frac{1}{\sqrt{3}}} = 2\sqrt{3} \]
The value of the expression is \(2\sqrt{3}\).
In simple words: Convert the complementary angles in both numerator and denominator. This simplifies the top to 2 and the bottom to 1 over root 3, which divides to 2 root 3.

Exam Tip: Handle the numerator and denominator separately in long complex expressions to prevent algebraic errors.

Question 42. If \(\frac{\cos \alpha}{\cos \beta}\) = m and \(\frac{\cos \alpha}{\sin \beta}\) = n show that (m² + n²) cos² \(\beta\) = n².
Answer:
Let's simplify the Left-Hand Side (LHS) of the equation:
\[ \text{LHS} = (m^2 + n^2) \cos^2\beta \]
Substituting the values of \(m\) and \(n\):
\[ = \left( \frac{\cos^2\alpha}{\cos^2\beta} + \frac{\cos^2\alpha}{\sin^2\beta} \right) \cos^2\beta \]
Factor out \(\cos^2\alpha\):
\[ = \cos^2\alpha \left( \frac{1}{\cos^2\beta} + \frac{1}{\sin^2\beta} \right) \cos^2\beta \]
Combine the fractions inside the parentheses:
\[ = \cos^2\alpha \left( \frac{\sin^2\beta + \cos^2\beta}{\cos^2\beta \cdot \sin^2\beta} \right) \cos^2\beta \]
Since \(\sin^2\beta + \cos^2\beta = 1\):
\[ = \cos^2\alpha \left( \frac{1}{\cos^2\beta \cdot \sin^2\beta} \right) \cos^2\beta \]
Cancel out \(\cos^2\beta\):
\[ = \frac{\cos^2\alpha}{\sin^2\beta} \]
Since \(n = \frac{\cos\alpha}{\sin\beta}\):
\[ = n^2 = \text{RHS} \]
Hence proved.
In simple words: Substitute the fractions into the equation and factor out cosine squared alpha. Finding a common denominator simplifies the terms to cosine squared alpha over sine squared beta, which is equal to n squared.

Exam Tip: Factoring out common algebraic terms from brackets before combining fractions is a great way to prevent long, complex equations.

Question 43. Prove that cos 1° cos 2° cos 3°.........cos 180° = 0.
Answer:
The given product is a sequence of cosine values of angles from \(1^\circ\) to \(180^\circ\) in steps of \(1^\circ\):
\[ \text{LHS} = \cos 1^\circ \cos 2^\circ \cdot ... \cdot \cos 90^\circ \cdot ... \cdot \cos 180^\circ \]
We know that \(\cos 90^\circ = 0\). Substituting this value:
\[ \text{LHS} = \cos 1^\circ \cos 2^\circ \cdot ... \cdot (0) \cdot ... \cdot \cos 180^\circ = 0 = \text{RHS} \]
Hence proved.
In simple words: Since the multiplication sequence contains cosine of 90 degrees, which is equal to zero, the product of the entire sequence must be zero.

Exam Tip: Always look for terms that are equal to 0 (like \(\cos 90^\circ\)) or undefined in long product or sum series questions as they simplify the entire series instantly.

Question 44. Prove that sin \(\theta\) + cos \(\theta\) / sin \(\theta\) – cos \(\theta\) + sin \(\theta\) – cos \(\theta\) / sin \(\theta\) + cos \(\theta\) = 2 sec² \(\theta\) / tan² \(\theta\) - 1 .
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} + \frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta} \]
Combine the fractions using a common denominator:
\[ = \frac{(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2}{\sin^2\theta - \cos^2\theta} \]
\[ = \frac{(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta) + (\sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta)}{\sin^2\theta - \cos^2\theta} \]
The middle terms cancel:
\[ = \frac{2(\sin^2\theta + \cos^2\theta)}{\sin^2\theta - \cos^2\theta} = \frac{2}{\sin^2\theta - \cos^2\theta} \]
Now, divide the numerator and denominator by \(\cos^2\theta\) to match the RHS form:
\[ = \frac{\frac{2}{\cos^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta} - \frac{\cos^2\theta}{\cos^2\theta}} = \frac{2\sec^2\theta}{\tan^2\theta - 1} = \text{RHS} \]
Hence proved.
In simple words: Combine the fractions over a common denominator. This simplifies the top to 2. Dividing the top and bottom by cosine squared converts the terms to secant and tangent to match the right side.

Exam Tip: Dividing a simplified fraction by cosine squared is an excellent way to convert sine and cosine equations into tangent and secant equations.

Question 45. If A, B, C are the interior angles of a triangle ABC, show that sin ( B+C / 2 ) cos A/2 + cos ( B+C / 2 ) sin A/2 = 1.
Answer:
Since \(A, B,\) and \(C\) are the interior angles of a triangle \(ABC\), we have:
\[ A + B + C = 180^\circ \implies B + C = 180^\circ - A \]
Dividing by 2 on both sides:
\[ \frac{B + C}{2} = 90^\circ - \frac{A}{2} \]
Using complementary angle formulas:
- \(\sin\left(\frac{B+C}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right) = \cos\left(\frac{A}{2}\right)\)
- \(\cos\left(\frac{B+C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A}{2}\right)\)
Substitute these into the Left-Hand Side (LHS) of our target equation:
\[ \text{LHS} = \cos\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}\right) \]
\[ = \cos^2\left(\frac{A}{2}\right) + \sin^2\left(\frac{A}{2}\right) \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ = 1 = \text{RHS} \]
Hence proved.
In simple words: Express the angles of the first terms as complements of half-angle A. This simplifies the equation to the sum of sine squared and cosine squared of half-angle A, which equals 1.

Exam Tip: Triangle angle sum relation proofs always start with \(A+B+C = 180^\circ\). Rearrange this equation first before substituting values.