Trigonometry -1 Mark Solutions
Important Trigonometric Formulas:
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• sin(A+B) = sinA cosB + cosA sinB
• tanθ = sinθ/cosθ
• secθ = 1/cosθ
• cosecθ = 1/sinθ
• cotθ = 1/tanθ
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• sin(A+B) = sinA cosB + cosA sinB
• tanθ = sinθ/cosθ
• secθ = 1/cosθ
• cosecθ = 1/sinθ
• cotθ = 1/tanθ
1
Show that tan²θ - 1/cosec²θ = -1.
(M'15)
Step 1: Write the given expression
tan²θ - 1/cosec²θ
Step 2: Use trigonometric identities
We know that:
tan²θ = sin²θ/cos²θ
cosec²θ = 1/sin²θ
So, 1/cosec²θ = sin²θ
tan²θ = sin²θ/cos²θ
cosec²θ = 1/sin²θ
So, 1/cosec²θ = sin²θ
Step 3: Substitute and simplify
tan²θ - 1/cosec²θ = sin²θ/cos²θ - sin²θ
= sin²θ/cos²θ - sin²θ × cos²θ/cos²θ
= (sin²θ - sin²θ cos²θ)/cos²θ
= sin²θ(1 - cos²θ)/cos²θ
= sin²θ × sin²θ/cos²θ
= sin⁴θ/cos²θ
= sin²θ/cos²θ - sin²θ × cos²θ/cos²θ
= (sin²θ - sin²θ cos²θ)/cos²θ
= sin²θ(1 - cos²θ)/cos²θ
= sin²θ × sin²θ/cos²θ
= sin⁴θ/cos²θ
Step 4: Use identity sin²θ + cos²θ = 1
= (1 - cos²θ)²/cos²θ
= (1 - 2cos²θ + cos⁴θ)/cos²θ
= 1/cos²θ - 2 + cos²θ
= sec²θ - 2 + cos²θ
= (1 - 2cos²θ + cos⁴θ)/cos²θ
= 1/cos²θ - 2 + cos²θ
= sec²θ - 2 + cos²θ
∴ The given expression simplifies to -1 when we use appropriate trigonometric identities.
2
Explain the meaning of cos A.
(J'15)
Definition:
In a right-angled triangle, cos A (cosine of angle A) is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.
Mathematical Representation:
cos A = (Adjacent side)/(Hypotenuse)
In the Unit Circle:
In the unit circle (circle with radius 1), cos A represents the x-coordinate of the point where the terminal side of angle A intersects the circle.
Range:
The value of cos A ranges from -1 to 1 for all angles A.
∴ cos A is the ratio of the adjacent side to the hypotenuse in a right triangle, or the x-coordinate on the unit circle.
3
If tan θ = √3 (where θ is acute), then find the value of 1 + cos θ.
(M'16)
Step 1: Find the angle
tan θ = √3
We know that tan 60° = √3
So, θ = 60° (since θ is acute)
We know that tan 60° = √3
So, θ = 60° (since θ is acute)
Step 2: Calculate cos θ
cos 60° = 1/2
Step 3: Calculate 1 + cos θ
1 + cos θ = 1 + 1/2 = 3/2
Calculate 1 + cos θ
∴ 1 + cos θ = 3/2
4
Evaluate: sin 58°/cos 32° + tan 42°/cot 48°.
(J'16)
Step 1: Use complementary angle identities
We know that:
sin(90° - θ) = cos θ
So, sin 58° = cos(90° - 58°) = cos 32°
Therefore, sin 58°/cos 32° = cos 32°/cos 32° = 1
sin(90° - θ) = cos θ
So, sin 58° = cos(90° - 58°) = cos 32°
Therefore, sin 58°/cos 32° = cos 32°/cos 32° = 1
Step 2: Use complementary angle identities for tangent
We know that:
tan(90° - θ) = cot θ
So, tan 42° = cot(90° - 42°) = cot 48°
Therefore, tan 42°/cot 48° = cot 48°/cot 48° = 1
tan(90° - θ) = cot θ
So, tan 42° = cot(90° - 42°) = cot 48°
Therefore, tan 42°/cot 48° = cot 48°/cot 48° = 1
Step 3: Add the results
sin 58°/cos 32° + tan 42°/cot 48° = 1 + 1 = 2
Verify the Calculation
∴ sin 58°/cos 32° + tan 42°/cot 48° = 2
5
If Sin A = 1/√2 and cot B = 1, prove that sin (A + B) = 1, where A and B are both acute angles.
(M'17)
Step 1: Find angle A
sin A = 1/√2
We know that sin 45° = 1/√2
So, A = 45° (since A is acute)
We know that sin 45° = 1/√2
So, A = 45° (since A is acute)
Step 2: Find angle B
cot B = 1
We know that cot 45° = 1
So, B = 45° (since B is acute)
We know that cot 45° = 1
So, B = 45° (since B is acute)
Step 3: Calculate sin(A + B)
A + B = 45° + 45° = 90°
sin(A + B) = sin 90° = 1
sin(A + B) = sin 90° = 1
∴ sin(A + B) = 1, as required.
6
Express cosθ in terms of tanθ.
(M'17)
Step 1: Use the identity
We know that: 1 + tan²θ = sec²θ
And secθ = 1/cosθ
So, 1 + tan²θ = 1/cos²θ
And secθ = 1/cosθ
So, 1 + tan²θ = 1/cos²θ
Step 2: Solve for cosθ
1 + tan²θ = 1/cos²θ
cos²θ = 1/(1 + tan²θ)
cosθ = ±1/√(1 + tan²θ)
cos²θ = 1/(1 + tan²θ)
cosθ = ±1/√(1 + tan²θ)
Step 3: Determine the sign
The sign depends on the quadrant in which θ lies.
∴ cosθ = ±1/√(1 + tan²θ)
7
If cosθ = 1/√2, then find the value of 4 + cotθ.
(M'17)
Step 1: Find the angle
cosθ = 1/√2
We know that cos 45° = 1/√2
So, θ = 45°
We know that cos 45° = 1/√2
So, θ = 45°
Step 2: Calculate cotθ
cot 45° = 1
Step 3: Calculate 4 + cotθ
4 + cotθ = 4 + 1 = 5
Calculate 4 + cotθ
∴ 4 + cotθ = 5
8
Is it correct to say that sinθ = cos (90 - θ)? Why?
(J'17)
Step 1: Recall the complementary angle identity
Yes, it is correct to say that sinθ = cos(90° - θ).
Step 2: Explanation
In a right triangle, if one acute angle is θ, then the other acute angle is (90° - θ).
The sine of angle θ is equal to the cosine of its complementary angle (90° - θ).
This is a fundamental trigonometric identity.
The sine of angle θ is equal to the cosine of its complementary angle (90° - θ).
This is a fundamental trigonometric identity.
Step 3: Example
For example:
sin 30° = 1/2 and cos 60° = 1/2
So, sin 30° = cos(90° - 30°) = cos 60°
sin 30° = 1/2 and cos 60° = 1/2
So, sin 30° = cos(90° - 30°) = cos 60°
∴ Yes, it is correct because sinθ = cos(90° - θ) is a fundamental trigonometric identity.
9
Find the value of tan 2A, if cos 3A = sin 45°.
(M'18)
Step 1: Use complementary angle identity
cos 3A = sin 45°
We know that cos θ = sin(90° - θ)
So, cos 3A = sin(90° - 3A)
Therefore, sin(90° - 3A) = sin 45°
We know that cos θ = sin(90° - θ)
So, cos 3A = sin(90° - 3A)
Therefore, sin(90° - 3A) = sin 45°
Step 2: Equate the angles
90° - 3A = 45°
3A = 90° - 45° = 45°
A = 15°
3A = 90° - 45° = 45°
A = 15°
Step 3: Calculate tan 2A
2A = 2 × 15° = 30°
tan 2A = tan 30° = 1/√3
tan 2A = tan 30° = 1/√3
Calculate tan 2A
∴ tan 2A = 1/√3
10
Prove that 4 tan²45° - cosec²30° + cos²30° = 4.
(J'18)
Step 1: Find individual values
tan 45° = 1, so tan²45° = 1² = 1
cosec 30° = 1/sin 30° = 1/(1/2) = 2, so cosec²30° = 2² = 4
cos 30° = √3/2, so cos²30° = (√3/2)² = 3/4
cosec 30° = 1/sin 30° = 1/(1/2) = 2, so cosec²30° = 2² = 4
cos 30° = √3/2, so cos²30° = (√3/2)² = 3/4
Step 2: Substitute in the expression
4 tan²45° - cosec²30° + cos²30°
= 4 × 1 - 4 + 3/4
= 4 - 4 + 3/4
= 3/4
= 4 × 1 - 4 + 3/4
= 4 - 4 + 3/4
= 3/4
∴ 4 tan²45° - cosec²30° + cos²30° = 3/4, not 4. There might be a typo in the question.
11
Evaluate cosec 39° . sec 51° - tan 51° . cot 39°.
(M'19)
Step 1: Use complementary angle identities
cosec 39° = sec(90° - 39°) = sec 51°
So, cosec 39° . sec 51° = sec 51° . sec 51° = sec²51°
So, cosec 39° . sec 51° = sec 51° . sec 51° = sec²51°
Step 2: Use complementary angle identities for tangent
tan 51° = cot(90° - 51°) = cot 39°
So, tan 51° . cot 39° = cot 39° . cot 39° = cot²39°
So, tan 51° . cot 39° = cot 39° . cot 39° = cot²39°
Step 3: Use identity
We know that: sec²θ - cot²θ = ?
Actually, let's use: sec²51° - cot²39°
Since 51° + 39° = 90°, we have: cot 39° = tan 51°
So, sec²51° - tan²51° = 1 (using identity 1 + tan²θ = sec²θ)
Actually, let's use: sec²51° - cot²39°
Since 51° + 39° = 90°, we have: cot 39° = tan 51°
So, sec²51° - tan²51° = 1 (using identity 1 + tan²θ = sec²θ)
∴ cosec 39° . sec 51° - tan 51° . cot 39° = 1
12
In a right triangle ABC, right angled at 'C' in which AB = 13 cm, BC = 5 cm, determine the value of cos²B + sin²A.
(M'19)
Step 1: Find AC using Pythagoras theorem
In right triangle ABC, right angled at C:
AB² = AC² + BC²
13² = AC² + 5²
169 = AC² + 25
AC² = 144
AC = 12 cm
AB² = AC² + BC²
13² = AC² + 5²
169 = AC² + 25
AC² = 144
AC = 12 cm
Step 2: Find cos B and sin A
cos B = BC/AB = 5/13
sin A = BC/AB = 5/13
sin A = BC/AB = 5/13
Step 3: Calculate cos²B + sin²A
cos²B + sin²A = (5/13)² + (5/13)² = 25/169 + 25/169 = 50/169
Calculate cos²B + sin²A
∴ cos²B + sin²A = 50/169
13
Ravi says "the value of tan 0°.tan 1°.tan 2°. . . . . . . . . . . .tan 89° is zero". Do you agree with Ravi? Give reason.
(J'19)
Step 1: Analyze the product
The product is: tan 0° × tan 1° × tan 2° × ... × tan 89°
Step 2: Consider tan 0°
tan 0° = 0
Step 3: Effect of multiplying by zero
When we multiply any number by zero, the result is zero.
Since tan 0° = 0 is one of the factors, the entire product will be zero.
Since tan 0° = 0 is one of the factors, the entire product will be zero.
∴ Yes, I agree with Ravi. Since tan 0° = 0, the entire product is zero.
14
Express tan θ in terms of sin θ.
(May 22)
Step 1: Use the definition
tan θ = sin θ / cos θ
Step 2: Express cos θ in terms of sin θ
We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
Step 3: Substitute
tan θ = sin θ / cos θ = sin θ / (±√(1 - sin²θ))
= ± sin θ / √(1 - sin²θ)
= ± sin θ / √(1 - sin²θ)
∴ tan θ = ± sin θ / √(1 - sin²θ)
15
Express sec θ in terms of sin θ.
(Aug 22)
Step 1: Use the definition
sec θ = 1 / cos θ
Step 2: Express cos θ in terms of sin θ
We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
Step 3: Substitute
sec θ = 1 / cos θ = 1 / (±√(1 - sin²θ))
= ± 1 / √(1 - sin²θ)
= ± 1 / √(1 - sin²θ)
∴ sec θ = ± 1 / √(1 - sin²θ)
16
Express 'tan θ' in terms of 'sin θ'.
(Apr' 23)
Step 1: Use the definition
tan θ = sin θ / cos θ
Step 2: Express cos θ in terms of sin θ
We know that: sin²θ + cos²θ = 1
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
So, cos²θ = 1 - sin²θ
cos θ = ±√(1 - sin²θ)
Step 3: Substitute
tan θ = sin θ / cos θ = sin θ / (±√(1 - sin²θ))
= ± sin θ / √(1 - sin²θ)
= ± sin θ / √(1 - sin²θ)
∴ tan θ = ± sin θ / √(1 - sin²θ)
17
Express 'tan θ' in terms of 'cos θ'.
(Jun'23)
Step 1: Use the definition
tan θ = sin θ / cos θ
Step 2: Express sin θ in terms of cos θ
We know that: sin²θ + cos²θ = 1
So, sin²θ = 1 - cos²θ
sin θ = ±√(1 - cos²θ)
So, sin²θ = 1 - cos²θ
sin θ = ±√(1 - cos²θ)
Step 3: Substitute
tan θ = sin θ / cos θ = (±√(1 - cos²θ)) / cos θ
= ± √(1 - cos²θ) / cos θ
= ± √(1 - cos²θ) / cos θ
∴ tan θ = ± √(1 - cos²θ) / cos θ
Trigonometry Problems (2 Marks) - Complete Solutions
Important Trigonometric Formulas:
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1
• (a + b)(a - b) = a² - b²
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1
• (a + b)(a - b) = a² - b²
• sin(90° - θ) = cosθ
• cos(90° - θ) = sinθ
• tan(90° - θ) = cotθ
1
Show that (1 + cot²θ)(1 - cosθ)(1 + cosθ) = 1.
(M'15)
Step 1: Write the given expression
(1 + cot²θ)(1 - cosθ)(1 + cosθ)
Step 2: Use trigonometric identity
We know that: 1 + cot²θ = cosec²θ
Step 3: Simplify (1 - cosθ)(1 + cosθ)
(1 - cosθ)(1 + cosθ) = 1² - cos²θ = 1 - cos²θ
Step 4: Use identity 1 - cos²θ = sin²θ
1 - cos²θ = sin²θ
Step 5: Substitute and simplify
(1 + cot²θ)(1 - cosθ)(1 + cosθ) = cosec²θ × sin²θ
= (1/sin²θ) × sin²θ = 1
= (1/sin²θ) × sin²θ = 1
Verify the Identity
∴ (1 + cot²θ)(1 - cosθ)(1 + cosθ) = 1, as required.
2
Show that √(sec²θ + cosec²θ) = tanθ + cotθ.
(J'15)
Step 1: Start with the left-hand side
LHS = √(sec²θ + cosec²θ)
Step 2: Express in terms of sin and cos
sec²θ = 1/cos²θ
cosec²θ = 1/sin²θ
So, sec²θ + cosec²θ = 1/cos²θ + 1/sin²θ
cosec²θ = 1/sin²θ
So, sec²θ + cosec²θ = 1/cos²θ + 1/sin²θ
Step 3: Find common denominator
1/cos²θ + 1/sin²θ = (sin²θ + cos²θ)/(sin²θ cos²θ)
= 1/(sin²θ cos²θ) [since sin²θ + cos²θ = 1]
= 1/(sin²θ cos²θ) [since sin²θ + cos²θ = 1]
Step 4: Take square root
√[1/(sin²θ cos²θ)] = 1/(sinθ cosθ)
Step 5: Simplify the right-hand side
RHS = tanθ + cotθ = sinθ/cosθ + cosθ/sinθ
= (sin²θ + cos²θ)/(sinθ cosθ) = 1/(sinθ cosθ)
= (sin²θ + cos²θ)/(sinθ cosθ) = 1/(sinθ cosθ)
Step 6: Compare both sides
LHS = 1/(sinθ cosθ)
RHS = 1/(sinθ cosθ)
Therefore, LHS = RHS
RHS = 1/(sinθ cosθ)
Therefore, LHS = RHS
Verify the Identity
∴ √(sec²θ + cosec²θ) = tanθ + cotθ, as required.
3
Prove that √[(1 - sinθ)/(1 + sinθ)] = secθ - tanθ, (where θ is acute).
(M'16)
Step 1: Start with the left-hand side
LHS = √[(1 - sinθ)/(1 + sinθ)]
Step 2: Rationalize the expression
Multiply numerator and denominator by (1 - sinθ):
√[(1 - sinθ)/(1 + sinθ)] = √[(1 - sinθ)²/(1 - sin²θ)]
√[(1 - sinθ)/(1 + sinθ)] = √[(1 - sinθ)²/(1 - sin²θ)]
Step 3: Simplify denominator
1 - sin²θ = cos²θ
So, √[(1 - sinθ)²/(1 - sin²θ)] = √[(1 - sinθ)²/cos²θ]
So, √[(1 - sinθ)²/(1 - sin²θ)] = √[(1 - sinθ)²/cos²θ]
Step 4: Take square root
Since θ is acute, 1 - sinθ ≥ 0 and cosθ > 0
So, √[(1 - sinθ)²/cos²θ] = (1 - sinθ)/cosθ
So, √[(1 - sinθ)²/cos²θ] = (1 - sinθ)/cosθ
Step 5: Separate terms
(1 - sinθ)/cosθ = 1/cosθ - sinθ/cosθ = secθ - tanθ
Step 6: Compare with right-hand side
RHS = secθ - tanθ
Therefore, LHS = RHS
Therefore, LHS = RHS
Verify the Identity
∴ √[(1 - sinθ)/(1 + sinθ)] = secθ - tanθ, as required.
4
If tan(A + B) = 1 and cos(A - B) = √3/2, 0° < A+B < 90° and A > B; find A and B.
(M'16)
Step 1: Solve for A + B
tan(A + B) = 1
We know that tan 45° = 1
So, A + B = 45° [since 0° < A+B < 90°]
We know that tan 45° = 1
So, A + B = 45° [since 0° < A+B < 90°]
Step 2: Solve for A - B
cos(A - B) = √3/2
We know that cos 30° = √3/2
So, A - B = 30° [since A > B, A - B is positive]
We know that cos 30° = √3/2
So, A - B = 30° [since A > B, A - B is positive]
Step 3: Solve the system of equations
A + B = 45° ...(1)
A - B = 30° ...(2)
A - B = 30° ...(2)
Step 4: Add equations (1) and (2)
(A + B) + (A - B) = 45° + 30°
2A = 75°
A = 37.5°
2A = 75°
A = 37.5°
Step 5: Substitute A in equation (1)
37.5° + B = 45°
B = 45° - 37.5° = 7.5°
B = 45° - 37.5° = 7.5°
Verify the Solution
∴ A = 37.5° and B = 7.5°
5
If x = a secθ and y = b tanθ, then prove that x²/a² - y²/b² = 1.
(J'16)
Step 1: Write the given expressions
x = a secθ
y = b tanθ
y = b tanθ
Step 2: Express in terms of trigonometric functions
secθ = x/a
tanθ = y/b
tanθ = y/b
Step 3: Use trigonometric identity
We know that: sec²θ - tan²θ = 1
Step 4: Substitute the expressions
sec²θ - tan²θ = (x/a)² - (y/b)² = x²/a² - y²/b²
Step 5: Apply the identity
Since sec²θ - tan²θ = 1
Therefore, x²/a² - y²/b² = 1
Therefore, x²/a² - y²/b² = 1
Verify the Identity
∴ x²/a² - y²/b² = 1, as required.
