1.2 Why the ratios are well-defined
If two right triangles have the same acute angle A, they are similar by the AA criterion. Corresponding sides are proportional, so ratios such as opposite ÷ hypotenuse, adjacent ÷ hypotenuse, and opposite ÷ adjacent have the same values in both triangles. Therefore, each trigonometric ratio depends only on angle A, not on the size of the triangle.
1.3 Special angles and relationships
The standard values at 30°, 45°, and 60° can be obtained from an equilateral triangle split into two right triangles and from an isosceles right triangle. The values at 0° and 90° are defined using limiting or coordinate-based interpretations; reciprocal ratios are undefined whenever their denominator is zero.
Worked example
In a right triangle, relative to angle A, the opposite side is 6 cm and the adjacent side is 8 cm. The hypotenuse is √(6² + 8²) = 10 cm. Hence sin A = 6/10 = 3/5, cos A = 8/10 = 4/5, tan A = 6/8 = 3/4, cosec A = 5/3, sec A = 5/4, and cot A = 4/3.
Common mistakes
-
Using the side next to the angle as the adjacent side even when it is the hypotenuse.
-
Keeping the same opposite and adjacent sides after changing the reference angle.
-
Writing tan A = adjacent/opposite instead of opposite/adjacent.
-
Treating cosec 0°, tan 90°, or sec 90° as ordinary finite values.
Practice questions
-
In a right triangle, the opposite and hypotenuse relative to A are 5 cm and 13 cm. Find sin A and cos A.
-
Find all six ratios when tan A = 5/12 and A is acute.
-
State the values of sin 30°, cos 60°, tan 45°, and sec 60°.
-
Explain in two or three sentences why sin A is well-defined.
Trigonometric Identities
2.1 The fundamental identity
For an acute angle A in a right triangle, let the opposite side be p, the adjacent side be b, and the hypotenuse be h. By the Pythagorean theorem, p² + b² = h². Dividing every term by h² gives p²/h² + b²/h² = 1. Since p/h = sin A and b/h = cos A, we obtain the identity sin²A + cos²A = 1.
Here sin²A means (sin A)², not sin(A²). The identity is true for every angle for which these ratios are defined in the syllabus context.
2.2 Useful relationships
From sin²A + cos²A = 1, we can write sin²A = 1 − cos²A or cos²A = 1 − sin²A. The reciprocal relationships are cosec A = 1/sin A, sec A = 1/cos A, and cot A = 1/tan A. Also, tan A = sin A/cos A and cot A = cos A/sin A whenever the denominators are non-zero.
Simple identities may require replacing a ratio with its reciprocal, combining fractions, factoring, or using sin²A + cos²A = 1. Work line by line and keep both sides balanced.
Worked example
If cos A = 12/13 and A is acute, find sin A and tan A. Using sin²A + cos²A = 1, sin²A = 1 − (12/13)² = 1 − 144/169 = 25/169. Since A is acute, sin A is positive, so sin A = 5/13. Therefore tan A = sin A/cos A = (5/13)/(12/13) = 5/12.
Unit 3 — Heights and Distances
3.1 Reading the situation
A line of sight joins the observer’s eye to the object being viewed. The angle of elevation is measured upward from the horizontal when the object is above the observer. The angle of depression is measured downward from the horizontal when the object is below the observer. Draw the horizontal line, mark the right angle, and label the known lengths before choosing a ratio.
When the observer and the object are at different heights, use the vertical difference between their levels. If the horizontal lines are parallel, an angle of depression equals the corresponding angle of elevation because they are alternate interior angles.
Situation
Most useful ratio
Typical equation
Height and horizontal distance are involved
tan θ
tan θ = height ÷ horizontal distance
Height and line of sight are involved
sin θ
sin θ = height ÷ line of sight
Horizontal distance and line of sight are involved
cos θ
cos θ = horizontal distance ÷ line of sight
Worked example
From a point on level ground, the angle of elevation of the top of a tower is 45°. If the point is 20 m from the foot of the tower, find the tower’s height. Let the height be h. Since tan 45° = h/20 and tan 45° = 1, h/20 = 1, so h = 20 m. The tower is 20 m high, assuming the observer’s eye height is neglected.
Common mistakes
-
Using the sloping line of sight as the horizontal distance.
-
Confusing an angle of depression with an angle measured from the vertical.
-
Forgetting to subtract the observer’s height when the eye is above ground level and that information is given.
-
Using a calculator approximation when an exact standard value is available.
-
Giving a height without a unit or without checking whether it is positive and reasonable.
Practice questions
-
A pole casts a 10 m horizontal shadow when the angle of elevation of the sun is 60°. Find the pole’s height in exact form.
-
From a point 15 m from a building, the angle of elevation of its top is 30°. Find the building’s height, ignoring eye height.
-
From the top of a 12 m building, the angle of depression of a point on the ground is 45°. Find the horizontal distance to the point.
-
Draw a labelled diagram for a problem involving two right triangles and identify the two horizontal distances before calculating.
8.1 exercise
Question 1
In triangle ABC, angle B is 90°.
Given:
AB = 24 cm
BC = 7 cmFirst, find AC.
AC is the hypotenuse.
Using Pythagoras theorem:
AC² = AB² + BC²
AC² = 24² + 7²
AC² = 576 + 49
AC² = 625
AC = 25 cm
For angle A
For angle A:
- Opposite side = BC = 7 cm
- Base = AB = 24 cm
- Hypotenuse = AC = 25 cm
Therefore:
sin A = Opposite side / Hypotenuse
sin A = 7/25
cos A = Base / Hypotenuse
cos A = 24/25
For angle C
For angle C:
- Opposite side = AB = 24 cm
- Base = BC = 7 cm
- Hypotenuse = AC = 25 cm
Therefore:
sin C = 24/25
cos C = 7/25
Answer
sin A = 7/25
cos A = 24/25sin C = 24/25
cos C = 7/25Question 2
In triangle PQR
QR² = PR² – PQ²
= (13)² -(12)²
= 169 – 144
= 25
QR = 5 cm
hence, tan P – cot R = QR/PQ- QR/PQ
=5/12 – 5/12 = 0
Question 3
Given:
tan A = 3/4
We know:
tan A = Opposite side / Base
Using the k-method:
Opposite side = 3k
Base = 4kNow find the hypotenuse.
Hypotenuse² = Opposite side² + Base²
Hypotenuse² = (3k)² + (4k)²
Hypotenuse² = 9k² + 16k²
Hypotenuse² = 25k²
Hypotenuse = 5k
Now,
sin A = Opposite side / Hypotenuse
sin A = 3k / 5k
sin A = 3/5
Also,
cos A = Base / Hypotenuse
cos A = 4k / 5k
cos A = 4/5
Answer
sin A = 3/5
cos A = 4/5
Question 4
Given:
15 cot A = 8
Divide both sides by 15:
cot A = 8/15
We know:
cot A = Base / Opposite side
Using the k-method:
Base = 8k
Opposite side = 15kNow find the hypotenuse.
Hypotenuse² = Base² + Opposite side²
Hypotenuse² = (8k)² + (15k)²
Hypotenuse² = 64k² + 225k²
Hypotenuse² = 289k²
Hypotenuse = 17k
Now,
sin A = Opposite side / Hypotenuse
sin A = 15k / 17k
sin A = 15/17
Also,
sec A = Hypotenuse / Base
sec A = 17k / 8k
sec A = 17/8
Answer
sin A = 15/17
sec A = 17/8
Question 5
Given:
sec θ = 13/12
We know:
sec θ = Hypotenuse / Base
Using the k-method:
Hypotenuse = 13k
Base = 12kNow find the opposite side.
Opposite side² = Hypotenuse² − Base²
Opposite side² = (13k)² − (12k)²
Opposite side² = 169k² − 144k²
Opposite side² = 25k²
Opposite side = 5k
Now find all the ratios.
sin θ = Opposite side / Hypotenuse
sin θ = 5k / 13k
sin θ = 5/13
cos θ = Base / Hypotenuse
cos θ = 12k / 13k
cos θ = 12/13
tan θ = Opposite side / Base
tan θ = 5k / 12k
tan θ = 5/12
cosec θ = Hypotenuse / Opposite side
cosec θ = 13k / 5k
cosec θ = 13/5
cot θ = Base / Opposite side
cot θ = 12k / 5k
cot θ = 12/5
Answer
sin θ = 5/13
cos θ = 12/13
tan θ = 5/12
cosec θ = 13/5
cot θ = 12/5
Question 6
Given:
cos A = cos B
We have to prove:
A = B
Similarity method
Make two right-angled triangles:
- In the first triangle, one acute angle is A.
- In the second triangle, one acute angle is B.
- Take the hypotenuse of both triangles equal to 1 unit.
For the first triangle:
cos A = Base / Hypotenuse
Since the hypotenuse is 1:
cos A = Base / 1
Therefore:
Base of the first triangle = cos A
For the second triangle:
cos B = Base / Hypotenuse
Again, the hypotenuse is 1:
cos B = Base / 1
Therefore:
Base of the second triangle = cos B
Given:
cos A = cos B
Therefore, the bases of the two triangles are equal.
So, in the two triangles:
- Hypotenuses are equal.
- Bases are equal.
- Both triangles have a right angle.
Therefore, the two triangles are similar.
In similar triangles, corresponding angles are equal.
Hence:
A = B
Answer
A = B
Question 7
Given:
cot θ = 7/8
Part (i)
We have to find:
(1 + sin θ)(1 − sin θ)
divided by
(1 + cos θ)(1 − cos θ)First simplify the numerator:
(1 + sin θ)(1 − sin θ)
Using:
(1 + x)(1 − x) = 1 − x²
Numerator = 1 − sin² θ
We know:
1 − sin² θ = cos² θ
So, numerator = cos² θ
Now simplify the denominator:
(1 + cos θ)(1 − cos θ)
Denominator = 1 − cos² θ
We know:
1 − cos² θ = sin² θ
So, denominator = sin² θ
Therefore, the expression becomes:
cos² θ / sin² θ
cos² θ / sin² θ = cot² θ
Given:
cot θ = 7/8
Therefore:
cot² θ = (7/8)²
cot² θ = 49/64
Answer for part (i)
49/64
Part (ii)
Given:
cot θ = 7/8
Squaring both sides:
cot² θ = (7/8)²
cot² θ = 49/64
Answer for part (ii)
49/64
Question 8
Given:
3 cot A = 4
Divide both sides by 3:
cot A = 4/3
Therefore:
tan A = 3/4
Using the k-method:
Opposite side = 3k
Base = 4kThe hypotenuse is:
Hypotenuse² = (3k)² + (4k)²
Hypotenuse² = 9k² + 16k²
Hypotenuse² = 25k²
Hypotenuse = 5k
Therefore:
sin A = 3/5
cos A = 4/5
tan A = 3/4
We have to check whether:
(1 − tan² A) / (1 + tan² A)
is equal to:
cos² A − sin² A
Left-hand side
LHS = (1 − tan² A) / (1 + tan² A)
LHS = (1 − (3/4)²) / (1 + (3/4)²)
LHS = (1 − 9/16) / (1 + 9/16)
LHS = (7/16) / (25/16)
LHS = 7/25
Right-hand side
RHS = cos² A − sin² A
RHS = (4/5)² − (3/5)²
RHS = 16/25 − 9/25
RHS = 7/25
Therefore:
LHS = RHS = 7/25
Answer
The given statement is correct.
Question 9
In triangle ABC, angle B is 90°.
Given:
tan A = 1/√3
We know:
tan A = Opposite side / Base
Using the k-method:
Opposite side = k
Base = √3kNow find the hypotenuse.
Hypotenuse² = k² + (√3k)²
Hypotenuse² = k² + 3k²
Hypotenuse² = 4k²
Hypotenuse = 2k
Now:
sin A = Opposite side / Hypotenuse
sin A = k / 2k
sin A = 1/2
cos A = Base / Hypotenuse
cos A = √3k / 2k
cos A = √3/2
Since angle B is 90°:
A + C = 90°
Therefore:
sin C = cos A = √3/2
cos C = sin A = 1/2
Part (i)
sin A cos C + cos A sin C
= (1/2 × 1/2) + (√3/2 × √3/2)
= 1/4 + 3/4
= 1
Answer
1
Part (ii)
cos A cos C − sin A sin C
= (√3/2 × 1/2) − (1/2 × √3/2)
= √3/4 − √3/4
= 0
Answer
0
Question 10
In triangle PQR, angle Q is 90°.
Given:
PR + QR = 25 cm
PQ = 5 cm
Let:
QR = x cm
Then:
PR = 25 − x cm
Using Pythagoras theorem:
PR² = PQ² + QR²
Therefore:
(25 − x)² = 5² + x²
Expand the left side:
625 − 50x + x² = 25 + x²
Cancel x² from both sides:
625 − 50x = 25
625 − 25 = 50x
600 = 50x
x = 12
Therefore:
QR = 12 cm
Now:
PR = 25 − 12
PR = 13 cm
For angle P:
sin P = QR / PR
sin P = 12/13
cos P = PQ / PR
cos P = 5/13
tan P = QR / PQ
tan P = 12/5
Answer
sin P = 12/13
cos P = 5/13
tan P = 12/5
Question 11
State whether the following statements are true or false.
Part (i)
Statement:
tan A is always less than 1.
This is false because:
tan 60° = √3
And √3 is greater than 1.
So, tan A is not always less than 1.
Answer
False
Part (ii)
Statement:
sec A = 12/5 is possible for some angle A.
This is true because sec A can be greater than 1.
Also:
12/5 is greater than 1.
Answer
True
Part (iii)
Statement:
cos A is an abbreviation for cosecant of angle A.
This is false.
cos A means cosine of angle A.
Cosecant is written as cosec A.
Answer
False
Part (iv)
Statement:
cot A means cot multiplied by A.
This is false.
cot A means cotangent of angle A.
It does not mean cot multiplied by A.
Answer
False
Part (v)
Statement:
sin θ = 4/3 is possible for some angle θ.
This is false because the value of sine always lies between −1 and 1.
But:
4/3 is greater than 1.
Therefore, sin θ cannot be 4/3.
Answer
False
NCERT Class 10 Maths
Chapter 8: Introduction to Trigonometry
Exercise 8.2 Solutions
I am using the standard values of trigonometric ratios:
Angle
sin θ
cos θ
tan θ
0°
0
1
0
30°
1/2
√3/2
1/√3
45°
1/√2
1/√2
1
60°
√3/2
1/2
√3
90°
1
0
Not defined
Also,
- sec θ = 1/cos θ
- cosec θ = 1/sin θ
- cot θ = 1/tan θ
Question 1: Evaluate
(i) sin 60° cos 30° + sin 30° cos 60°
Using standard values:
sin 60° = √3/2
cos 30° = √3/2
sin 30° = 1/2
cos 60° = 1/2Therefore,
sin 60° cos 30° + sin 30° cos 60°
= (√3/2 × √3/2) + (1/2 × 1/2)
= 3/4 + 1/4
= 4/4
Answer = 1
(ii) 2 tan² 45° + cos² 30° − sin² 60°
We know:
tan 45° = 1
cos 30° = √3/2
sin 60° = √3/2Therefore,
2 tan² 45° + cos² 30° − sin² 60°
= 2(1)² + (√3/2)² − (√3/2)²
= 2 + 3/4 − 3/4
= 2
Answer = 2
(iii) cos 45° / (sec 30° + cosec 30°)
We know:
cos 45° = 1/√2
sec 30° = 1/cos 30°
= 1/(√3/2)
= 2/√3cosec 30° = 1/sin 30°
= 1/(1/2)
= 2Therefore,
cos 45° / (sec 30° + cosec 30°)
= (1/√2) / (2/√3 + 2)
Taking the denominator:
2/√3 + 2 = (2 + 2√3)/√3
= 2(1 + √3)/√3So,
= 1/√2 × √3/[2(1 + √3)]
= √3/[2√2(1 + √3)]
After rationalising and simplifying:
Answer = (3√2 − √6)/8
(iv)
(sin 30° + tan 45° − cosec 60°)
/
(sec 30° + cos 60° + cot 45°)We know:
sin 30° = 1/2
tan 45° = 1
cosec 60° = 1/sin 60° = 2/√3sec 30° = 2/√3
cos 60° = 1/2
cot 45° = 1Substituting these values:
= (1/2 + 1 − 2/√3) / (2/√3 + 1/2 + 1)
= (3/2 − 2/√3) / (3/2 + 2/√3)
Multiplying numerator and denominator by 2√3:
= (3√3 − 4)/(3√3 + 4)
Rationalising:
= [(3√3 − 4)(3√3 − 4)] / [(3√3 + 4)(3√3 − 4)]
= (27 − 24√3 + 16)/(27 − 16)
= (43 − 24√3)/11
Answer = (43 − 24√3)/11
(v)
[5 cos² 60° + 4 sec² 30° − tan² 45°]
/
[sin² 30° + cos² 30°]We know:
cos 60° = 1/2
sec 30° = 2/√3
tan 45° = 1
sin 30° = 1/2
cos 30° = √3/2Substituting:
= [5(1/2)² + 4(2/√3)² − (1)²]
/
[(1/2)² + (√3/2)²]= [5/4 + 4(4/3) − 1] / [1/4 + 3/4]
= [5/4 + 16/3 − 1] / 1
Now,
5/4 − 1 = 1/4
Therefore,
= 1/4 + 16/3
Taking LCM 12:
= 3/12 + 64/12
= 67/12
Answer = 67/12
Question 2: Choose the correct answer
(i)
2 tan 30° / (1 + tan² 30°)
We know:
tan 30° = 1/√3
Therefore,
= [2/√3] / [1 + (1/√3)²]
= [2/√3] / [1 + 1/3]
= [2/√3] / [4/3]
= 2/√3 × 3/4
= 3/(2√3)
= √3/2
We know:
sin 60° = √3/2
Therefore,
Answer = sin 60°
Correct option: A
(ii)
(1 − tan² 45°) / (1 + tan² 45°)
We know:
tan 45° = 1
Therefore,
= (1 − 1²)/(1 + 1²)
= (1 − 1)/(1 + 1)
= 0/2
= 0
Answer = 0
Correct option: D
(iii)
Given:
sin 2A = 2 sin A
We know:
sin 2A = 2 sin A cos A
Therefore,
2 sin A cos A = 2 sin A
Taking everything on one side:
2 sin A cos A − 2 sin A = 0
2 sin A(cos A − 1) = 0
This is true when:
sin A = 0, or cos A = 1
For angles between 0° and 90°:
A = 0°
Answer = 0°
Correct option: A
(iv)
2 tan 30° / (1 − tan² 30°)
We know:
tan 30° = 1/√3
Therefore,
= [2/√3] / [1 − (1/√3)²]
= [2/√3] / [1 − 1/3]
= [2/√3] / [2/3]
= 2/√3 × 3/2
= 3/√3
= √3
We know:
tan 60° = √3
Therefore,
Answer = tan 60°
Correct option: C
Question 3
If
tan(A + B) = √3
and
tan(A − B) = 1/√3,
find A and B.
We know:
tan 60° = √3
Therefore,
A + B = 60°
Also,
tan 30° = 1/√3
Therefore,
A − B = 30°
Now we have:
A + B = 60° ……… (1)
A − B = 30° ……… (2)
Adding equations (1) and (2):
A + B + A − B = 60° + 30°
2A = 90°
A = 45°
Substituting A = 45° in equation (1):
45° + B = 60°
B = 60° − 45°
B = 15°
Therefore,
A = 45° and B = 15°
Question 4: State whether the following statements are true or false
(i) sin(A + B) = sin A + sin B
This statement is false.
For example, take:
A = 30° and B = 60°
Left side:
sin(A + B) = sin 90° = 1
Right side:
sin 30° + sin 60°
= 1/2 + √3/2
These two values are not equal.
Therefore,
The statement is false.
The correct formula is:
sin(A + B) = sin A cos B + cos A sin B
(ii) The value of sin θ increases as θ increases from 0° to 90°
We know:
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2
sin 90° = 1The value keeps increasing.
Therefore,
The statement is true.
(iii) The value of cos θ increases as θ increases from 0° to 90°
We know:
cos 0° = 1
cos 30° = √3/2
cos 45° = 1/√2
cos 60° = 1/2
cos 90° = 0The value of cos θ decreases as θ increases.
Therefore,
The statement is false.
(iv) The value of sin θ is always less than 1
For angles from 0° to 90°, the value of sin θ is at most 1.
At θ = 90°:
sin 90° = 1
So, if the statement says that sin θ is always less than 1, it is not correct because it can be equal to 1.
Therefore,
The statement is false.
The correct statement is:
The value of sin θ is less than or equal to 1.
(v) cot 0° is not defined
We know:
cot θ = cos θ/sin θ
Therefore,
cot 0° = cos 0°/sin 0°
= 1/0
Division by zero is not defined.
Therefore,
The statement is true.
Final Answers at a Glance
Question 1
- (i) 1
- (ii) 2
- (iii) (3√2 − √6)/8
- (iv) (43 − 24√3)/11
- (v) 67/12
Question 2
- (i) sin 60°, Option A
- (ii) 0, Option D
- (iii) 0°, Option A
- (iv) tan 60°, Option C
Question 3
A = 45°, B = 15°
Question 4
- False
- True
- False
- False
- True
Exercise 8.3: Prove the Identities
Below are the proof questions from Question 4 of Exercise 8.3, according to the latest NCERT syllabus/reprint.
Important identities used
- sin² A + cos² A = 1
- 1 + tan² A = sec² A
- 1 + cot² A = cosec² A
- sec A = 1/cos A
- cosec A = 1/sin A
- cot A = cos A/sin A
Assume that the angles are acute and all expressions are defined.
(i) Prove that
(cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ)
Proof
LHS = (cosec θ − cot θ)²
= (1/sin θ − cos θ/sin θ)²
= ((1 − cos θ)/sin θ)²
= (1 − cos θ)²/sin² θ
We know:
sin² θ = 1 − cos² θ
Therefore,
= (1 − cos θ)²/(1 − cos² θ)
= (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)]
Cancel (1 − cos θ):
= (1 − cos θ)/(1 + cos θ)
= RHS
Hence proved.
(ii) Prove that
cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
Proof
LHS = cos A/(1 + sin A) + (1 + sin A)/cos A
Rationalising the first term:
cos A/(1 + sin A)
= cos A(1 − sin A)/[(1 + sin A)(1 − sin A)]
= cos A(1 − sin A)/(1 − sin² A)
We know:
1 − sin² A = cos² A
Therefore,
cos A/(1 + sin A)
= cos A(1 − sin A)/cos² A
= (1 − sin A)/cos A
Now,
LHS = (1 − sin A)/cos A + (1 + sin A)/cos A
= [1 − sin A + 1 + sin A]/cos A
= 2/cos A
= 2 sec A
= RHS
Hence proved.
(iii) Prove that
tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ
Proof
Let tan θ = t.
Therefore,
cot θ = 1/t
LHS = t/(1 − 1/t) + (1/t)/(1 − t)
First term:
t/(1 − 1/t)
= t/[(t − 1)/t]
= t²/(t − 1)
Second term:
(1/t)/(1 − t)
= 1/[t(1 − t)]
= −1/[t(t − 1)]
Therefore,
LHS = t²/(t − 1) − 1/[t(t − 1)]
Taking the common denominator:
= [t³ − 1]/[t(t − 1)]
We know:
t³ − 1 = (t − 1)(t² + t + 1)
Therefore,
LHS = [(t − 1)(t² + t + 1)]/[t(t − 1)]
= (t² + t + 1)/t
= t + 1 + 1/t
= 1 + tan θ + cot θ
Now,
tan θ + cot θ
= sin θ/cos θ + cos θ/sin θ
= (sin² θ + cos² θ)/(sin θ cos θ)
= 1/(sin θ cos θ)
= sec θ cosec θ
Therefore,
LHS = 1 + sec θ cosec θ
= RHS
Hence proved.
(iv) Prove that
(1 + sec A)/sec A = sin² A/(1 − cos A)
Proof
LHS = (1 + sec A)/sec A
= 1/sec A + sec A/sec A
= cos A + 1
= 1 + cos A
Now consider the RHS:
RHS = sin² A/(1 − cos A)
Using sin² A = 1 − cos² A:
= (1 − cos² A)/(1 − cos A)
= [(1 − cos A)(1 + cos A)]/(1 − cos A)
Cancel (1 − cos A):
= 1 + cos A
Therefore,
LHS = RHS
Hence proved.
(v) Prove that
(cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A
Proof
LHS = (cos A − sin A + 1)/(cos A + sin A − 1)
Rearranging:
= (1 + cos A − sin A)/(cos A + sin A − 1)
Multiply numerator and denominator by (1 + cos A + sin A):
= [(1 + cos A − sin A)(1 + cos A + sin A)]
/ [(cos A + sin A − 1)(cos A + sin A + 1)]Using (x − y)(x + y) = x² − y²:
Numerator:
= (1 + cos A)² − sin² A
= 1 + 2 cos A + cos² A − sin² A
Since sin² A + cos² A = 1:
= 2 cos A + 2 cos² A
= 2 cos A(1 + cos A)
Denominator:
= (cos A + sin A)² − 1
= cos² A + 2 sin A cos A + sin² A − 1
= 1 + 2 sin A cos A − 1
= 2 sin A cos A
Therefore,
LHS = [2 cos A(1 + cos A)]/[2 sin A cos A]
Cancel 2 cos A:
= (1 + cos A)/sin A
= 1/sin A + cos A/sin A
= cosec A + cot A
= RHS
Hence proved.
(vi) Prove that
√[(1 + sin A)/(1 − sin A)] = sec A + tan A
Proof
LHS = √[(1 + sin A)/(1 − sin A)]
Multiply numerator and denominator inside the square root by (1 + sin A):
= √[(1 + sin A)²/(1 − sin² A)]
We know:
1 − sin² A = cos² A
Therefore,
= √[(1 + sin A)²/cos² A]
= (1 + sin A)/cos A
Since A is acute, cos A is positive.
= 1/cos A + sin A/cos A
= sec A + tan A
= RHS
Hence proved.
(vii) Prove that
(sin θ − 2 sin³ θ)/(2 cos³ θ − cos θ) = tan θ
Proof
LHS = (sin θ − 2 sin³ θ)/(2 cos³ θ − cos θ)
Taking sin θ common from the numerator:
= sin θ(1 − 2 sin² θ)
/ [2 cos³ θ − cos θ]Taking cos θ common from the denominator:
= sin θ(1 − 2 sin² θ)
/ [cos θ(2 cos² θ − 1)]Now we know:
sin² θ + cos² θ = 1
So,
1 − 2 sin² θ
= 1 − 2(1 − cos² θ)
= 1 − 2 + 2 cos² θ
= 2 cos² θ − 1
Therefore,
LHS = sin θ(2 cos² θ − 1)
/ [cos θ(2 cos² θ − 1)]Cancel (2 cos² θ − 1):
= sin θ/cos θ
= tan θ
= RHS
Hence proved.
(viii) Prove that
(sin A + cosec A)² + (cos A + sec A)²
= 7 + tan² A + cot² AProof
LHS = (sin A + cosec A)² + (cos A + sec A)²
Using (x + y)² = x² + 2xy + y²:
= sin² A + 2 sin A cosec A + cosec² A
- cos² A + 2 cos A sec A + sec² A
We know:
sin A cosec A = 1
cos A sec A = 1Therefore,
= sin² A + 2 + cosec² A
- cos² A + 2 + sec² A
= sin² A + cos² A + 4 + cosec² A + sec² A
Since sin² A + cos² A = 1:
= 1 + 4 + cosec² A + sec² A
= 5 + cosec² A + sec² A
Now,
cosec² A = 1 + cot² A
and
sec² A = 1 + tan² A
Therefore,
= 5 + 1 + cot² A + 1 + tan² A
= 7 + tan² A + cot² A
= RHS
Hence proved.
(ix) Prove that
(cosec A − sin A)(sec A − cos A)
= 1/(tan A + cot A)Proof
LHS = (cosec A − sin A)(sec A − cos A)
First bracket:
cosec A − sin A
= 1/sin A − sin A
= (1 − sin² A)/sin A
= cos² A/sin A
Second bracket:
sec A − cos A
= 1/cos A − cos A
= (1 − cos² A)/cos A
= sin² A/cos A
Therefore,
LHS = (cos² A/sin A)(sin² A/cos A)
Cancel common terms:
= sin A cos A
Now consider the RHS:
RHS = 1/(tan A + cot A)
= 1/[sin A/cos A + cos A/sin A]
Taking the common denominator:
= 1/[(sin² A + cos² A)/(sin A cos A)]
Since sin² A + cos² A = 1:
= 1/[1/(sin A cos A)]
= sin A cos A
Therefore,
LHS = RHS
Hence proved.
(x) Prove that
(1 + tan² A)/(1 + cot² A)
= [(1 − tan A)/(1 − cot A)]²
= tan² AFirst part
Using the identities:
1 + tan² A = sec² A
and
1 + cot² A = cosec² A
Therefore,
(1 + tan² A)/(1 + cot² A)
= sec² A/cosec² A
= (1/cos² A)/(1/sin² A)
= sin² A/cos² A
= tan² A
Thus,
(1 + tan² A)/(1 + cot² A) = tan² A
Second part
Consider:
[(1 − tan A)/(1 − cot A)]²
Since cot A = 1/tan A:
= [(1 − tan A)/(1 − 1/tan A)]²
Simplifying the denominator:
1 − 1/tan A
= (tan A − 1)/tan A
Therefore,
= [(1 − tan A)/((tan A − 1)/tan A)]²
= [(1 − tan A) × tan A/(tan A − 1)]²
Since:
1 − tan A = −(tan A − 1)
Therefore,
= [−tan A]²
= tan² A
Hence,
[(1 − tan A)/(1 − cot A)]² = tan² A
Therefore,
(1 + tan² A)/(1 + cot² A)
= [(1 − tan A)/(1 − cot A)]²
= tan² AHence proved.
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