NCERT MCQ Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

In a right-angled triangle ABC, with C as the right angle:
A + B = 90° (since A + B + C = 180° and C = 90°).

Thus, cos(A + B) = cos(90°) = 0.

We are given the equation: 5 tanθ – 4 = 0.

Step 1: Solve for tanθ
Rearrange the equation to solve for tanθ:
5 tanθ = 4
tanθ = 4/5.

Step 2: Express sinθ and cosθ in terms of tanθ
Using the identity tanθ = sinθ / cosθ, we can write:
sinθ = 4k and cosθ = 5k,
where k is a positive constant such that sin²θ + cos²θ = 1 (Pythagorean identity).

Substitute sinθ = 4k and cosθ = 5k into the identity:
(4k)² + (5k)² = 1
16k² + 25k² = 1
41k² = 1
k² = 1/41
k = √(1/41).

Thus: sinθ = 4k = 4/√41,
cosθ = 5k = 5/√41.

Step 3: Simplify the given expression
We are tasked with finding the value of:
(5 sinθ – 4 cosθ) / (5 sinθ + 4 cosθ).

Substitute sinθ = 4/√41 and cosθ = 5/√41 into the expression:
Numerator:
5 sinθ – 4 cosθ = 5(4/√41) – 4(5/√41)
= (20/√41) – (20/√41)
= 0.

Denominator:
5 sinθ + 4 cosθ = 5(4/√41) + 4(5/√41)
= (20/√41) + (20/√41)
= 40/√41.

Thus, the entire expression becomes:
(5 sinθ – 4 cosθ) / (5 sinθ + 4 cosθ) = 0 / (40/√41) = 0.

Step 4: Final Answer
The value of the given expression is 0.
The correct answer is zero.

We are given that the angles A, B, and C of a triangle △ABC form an increasing arithmetic progression (AP). We need to find the value of sin B.

Step 1: Properties of angles in a triangle
The sum of the angles in any triangle is:
A + B + C = 180°.

Since A, B, and C form an increasing arithmetic progression, let the common difference of the AP be d. Then we can write:
A = B – d, B = B, C = B + d.

Substitute these into the angle sum property:
(B – d) + B + (B + d) = 180°.

Simplify: 3B = 180°.
Solve for B: B = 60°.

Step 2: Find sin B
Now that we know B = 60°, we use the standard trigonometric value:
sin 60° = √3/2.

Step 3: Final Answer
The value of sin B is: √3/2.
The correct answer is √3/2.

We are given the equations:
a cosθ + b sinθ = m — (1)
a sinθ – b cosθ = n — (2)
We need to find the value of a² + b².

Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = m²
Expand the left-hand side:
a²cos²θ + 2ab cosθ sinθ + b²sin²θ = m² — (3)

Square both sides of equation (2):
(a sinθ – b cosθ)² = n²
Expand the left-hand side:
a²sin²θ – 2ab sinθ cosθ + b²cos²θ = n² — (4)

Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a²cos²θ + 2ab cosθ sinθ + b²sin²θ) + (a²sin²θ – 2ab sinθ cosθ + b²cos²θ) = m² + n²

Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) = m² + n²

Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.

Substitute this into the equation:
a²(1) + b²(1) = m² + n²

Simplify: a² + b² = m² + n².
The value of a² + b² is m² + n².

We are given the equations:
a cotθ + b cosecθ = p — (1)
b cotθ + a cosecθ = q — (2)
We need to find the value of p² – q².
Recall the identity for p² – q²
The difference of squares formula states: p² – q² = (p + q)(p – q).
Thus, we first calculate p + q and p – q.

Add equations (1) and (2)
Add the two equations:
(a cotθ + b cosecθ) + (b cotθ + a cosecθ) = p + q.

Simplify: (a + b)cotθ + (a + b)cosecθ = p + q.
Factor out (a + b): (a + b)(cotθ + cosecθ) = p + q. — (3)

Subtract equation (2) from (1)
Subtract equation (2) from equation (1):
(a cotθ + b cosecθ) – (b cotθ + a cosecθ) = p – q.

Simplify: (a – b)cotθ + (b – a)cosecθ = p – q.

Factor out (a – b): (a – b)(cotθ – cosecθ) = p – q. — (4)
Substitute into p² – q²
Using the identity p² – q² = (p + q)(p – q), substitute the expressions for p + q and p – q from equations (3) and (4):
p² – q² = [(a + b)(cotθ + cosecθ)] × [(a – b)(cotθ – cosecθ)].
Simplify: p² – q² = (a + b)(a – b) × (cotθ + cosecθ)(cotθ – cosecθ).

Simplify further using the difference of squares
The term (cotθ + cosecθ)(cotθ – cosecθ) is a difference of squares:
(cotθ + cosecθ)(cotθ – cosecθ) = cot²θ – cosec²θ.

From the trigonometric identity: cosec²θ – cot²θ = 1,
we can write: cot²θ – cosec²θ = -1.
Substitute this into the expression for p² – q²:
p² – q² = (a + b)(a – b) × (-1).
Simplify: p² – q² = -(a + b)(a – b).

Expand (a + b)(a – b) using the difference of squares:
(a + b)(a – b) = a² – b².
Thus: p² – q² = -(a² – b²).

We are given the equations:
a cosθ + b sinθ = 4 — (1)
a sinθ – b cosθ = 3 — (2)

We need to find the value of a² + b².
Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = 4²
Expand the left-hand side:
a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3)

Square both sides of equation (2):
(a sinθ – b cosθ)² = 3²
Expand the left-hand side:
a² sin²θ – 2ab sinθ cosθ + b² cos²θ = 9 — (4)

Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a² cos²θ + 2ab cosθ sinθ + b² sin²θ) + (a² sin²θ – 2ab sinθ cosθ + b² cos²θ) = 16 + 9

Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = 25

Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.

Substitute this into the equation:
a² (1) + b² (1) = 25

Simplify: a² + b² = 25.
The value of a² + b² is 25.

We know the trigonometric identity:
sec²A – tan²A = 1.
Thus 9(sec²A – tan²A) = 9(1) = 9.

We are given:
sinθ = x and secθ = y.
Recall the trigonometric identities
1. The definition of secθ is secθ = 1/cosθ.
Therefore, cosθ = 1/secθ = 1/y.
2. The definition of tanθ is tanθ = sinθ / cosθ.

Substitute the values of sinθ and cosθ
From the problem, sinθ = x and cosθ = 1/y. Substituting these into the formula for tanθ:
tanθ = sinθ / cosθ = x / (1/y)

Simplify the expression
Dividing by 1/y is equivalent to multiplying by y:

tanθ = x * y = xy
Thus, tanθ is equal to xy.

We are given the equation:
cos A + cos²A = 1.
We need to find the value of sin²A + sin⁴A.

Express cos²A in terms of cos A
Rearrange the given equation: cos²A = 1 – cos A.

Use the Pythagorean identity
From the Pythagorean identity, we know: sin²A + cos²A = 1.
Substitute cos²A = 1 – cos A into the identity:
sin²A + (1 – cos A) = 1.
Simplify: sin²A = cos A.

Express sin⁴A in terms of sin²A
Since sin²A = cos A, we can write:
sin⁴A = (sin²A)² = (cos A)² = cos²A.

Substitute into sin²A + sin⁴A
Now substitute sin²A = cos A and sin⁴A = cos²A into the expression sin²A + sin⁴A:
sin²A + sin⁴A = cos A + cos²A.

From the given equation, we know: cos A + cos²A = 1.
Thus: sin²A + sin⁴A = 1.

We are given the equation:
x tan 45° cos 60° = sin 60° cot 60°.
Substitute the trigonometric values
Using standard trigonometric values:
– tan 45° = 1,
– cos 60° = 1/2,
– sin 60° = √3/2,
– cot 60° = 1/√3.

Substitute these values into the equation:
x (1) (1/2) = (√3/2) (1/√3).

Simplify the left-hand side:
x (1/2) = x/2.

Simplify the right-hand side:
(√3/2) (1/√3) = (√3 / √3) / 2 = 1/2.
Thus, the equation becomes x/2 = 1/2.

Solve for x
Multiply through by 2 to isolate x:
x = 1.
The correct answer is 1.

To determine which of the given trigonometric expressions is not defined, we analyze each option based on the definitions and properties of trigonometric functions.

Analyze each option
a) cos 0°
The cosine function is defined for all real numbers. The value of cos 0° is:
cos 0° = 1.
Thus, cos 0° is defined.

b) tan 45°
The tangent function is defined as tan θ = sin θ / cos θ. For θ = 45°:
tan 45° = sin 45° / cos 45° = (1/√2) / (1/√2) = 1.
Thus, tan 45° is defined.

c) sec 90°
The secant function is defined as sec θ = 1 / cos θ. For θ = 90°:
cos 90° = 0.
Since division by zero is undefined, sec 90° = 1 / cos 90° = 1 / 0 is not defined.
Thus, sec 90° is not defined.

d) sin 90°
The sine function is defined for all real numbers. The value of sin 90° is:
sin 90° = 1.
Thus, sin 90° is defined.
The expression that is not defined is sec 90°.

Given: sinθ – cosθ = 0.
sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45°.
sinθ = cosθ = 1/√2.
Calculate sin⁴θ + cos⁴θ:
sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)²
= (1/2)² + (1/2)²
= 1/4 + 1/4
= 1/2.

We are tasked with finding the maximum value of 1/secθ for 0° ≤ θ < 90°. Recall the definition of secant The secant function is defined as secθ = 1/cosθ. Thus, the reciprocal of secant is 1/secθ = cosθ. Analyze the behavior of cosθ in the given range For 0° ≤ θ < 90°: - The cosine function decreases from cos 0° = 1 to cos 90° = 0 (but does not actually reach 0 since θ < 90°). - Therefore, the maximum value of cosθ occurs at θ = 0°. At θ = 0°: cos 0° = 1. The maximum value of 1/secθ is equal to the maximum value of cosθ, which is 1. The correct answer is 1.

sin(A – B) = 0
⇒ A – B = 0°
⇒ A = B.

2cos(A + B) – 1 = 0
⇒ cos(A + B) = 1/2
⇒ A + B = 60°.

Substitute A = B into A + B = 60°:
2A = 60°
⇒ A = 30°.

We are given:
8 tan x = 15.

Rearrange the equation to solve for tan x:
tan x = 15/8.
Using the identity tan x = sin x / cos x, we can write:
sin x = 15k and cos x = 8k,
where k is a positive constant such that sin²x + cos²x = 1 (Pythagorean identity).

Substitute sin x = 15k and cos x = 8k into the identity:
(15k)² + (8k)² = 1
225k² + 64k² = 1
289k² = 1
k² = 1/289
k = √(1/289)
k = 1/17.

Thus:
sin x = 15k = 15/17,
cos x = 8k = 8/17.

Find sin x – cos x
Now, calculate sin x – cos x:
sin x – cos x = (15/17) – (8/17)
= (15 – 8)/17
= 7/17.
The value of sin x – cos x is 7/17.

We are given:
16 cot x = 12.
Rearrange the equation to solve for cot x:
cot x = 12/16 = 3/4.

Using the identity cot x = 1/tan x, we can write:
tan x = 1/cot x = 1/(3/4) = 4/3.

Using the identity tan x = sin x / cos x, we can write:
sin x = 4k and cos x = 3k,
where k is a positive constant such that sin²x + cos²x = 1 (Pythagorean identity).

Substitute sin x = 4k and cos x = 3k into the identity:
(4k)² + (3k)² = 1
16k² + 9k² = 1
25k² = 1
k² = 1/25
k = √(1/25)
k = 1/5.

Thus:
sin x = 4k = 4/5,
cos x = 3k = 3/5.
Substitute sin x = 4/5 and cos x = 3/5 into the expression:

Numerator:
sin x – cos x = (4/5) – (3/5)
= (4 – 3)/5
= 1/5.

Denominator:
sin x + cos x = (4/5) + (3/5)
= (4 + 3)/5
= 7/5.

Thus, the entire expression becomes:
(sin x – cos x) / (sin x + cos x) = (1/5) / (7/5).

Simplify: (1/5) / (7/5) = 1/7.
The value of (sin x – cos x) / (sin x + cos x) is 1/7.
The correct answer is 1/7.

We are given the equation:
sin 2A = 2 sin A.

Recall the double-angle identity for sine
The double-angle identity for sine is:
sin 2A = 2 sin A cos A.
Substitute this into the given equation:
2 sin A cos A = 2 sin A.

Simplify the equation
Divide both sides of the equation by 2 (assuming sin A ≠ 0):
sin A cos A = sin A.
Rearrange the terms: sin A cos A – sin A = 0.
Factor out sin A: sin A (cos A – 1) = 0.

Solve for A
This equation is satisfied if either:
sin A = 0 or cos A – 1 = 0.

Case 1: sin A = 0
The sine function is zero when A = 0°, 180°, etc. Among the given options, A = 0° satisfies this condition.
Case 2: cos A – 1 = 0
Solve for cos A:
cos A = 1.
The cosine function equals 1 when A = 0°, 360°, etc. Again, among the given options, A = 0° satisfies this condition.

Verify the solution
Substitute A = 0° into the original equation:
sin 2(0°) = 2 sin(0°).

The left-hand side: sin 2(0°) = sin 0° = 0.
The right-hand side: 2 sin(0°) = 2(0) = 0.
Since both sides are equal, A = 0° satisfies the equation.
The correct answer is 0°.

We are given that 0° ≤ A, B ≤ 90°, sin A = 1/2, and cos B = 1/2. We need to find the value of A + B.
Solve for A using sin A = 1/2
The sine function is defined as:
sin A = opposite/hypotenuse.

From trigonometric values, we know:
sin 30° = 1/2.
Since 0° ≤ A ≤ 90°, the only possible value for A is:
A = 30°.

Solve for B using cos B = 1/2
The cosine function is defined as:
cos B = adjacent/hypotenuse.
From trigonometric values, we know:
cos 60° = 1/2.
Since 0° ≤ B ≤ 90°, the only possible value for B is:
B = 60°.

Calculate A + B
Now, add the values of A and B:
A + B = 30° + 60° = 90°.
The value of A + B is 90°.

We are given the equation: sin θ + sin² θ = 1.
We need to find the value of cos² θ + cos⁴ θ.
Express sin² θ in terms of sin θ
Rearrange the given equation:
sin² θ = 1 – sin θ.

Use the Pythagorean identity
From the Pythagorean identity, we know:
sin² θ + cos² θ = 1.
Substitute sin² θ = 1 – sin θ into the identity:
(1 – sin θ) + cos² θ = 1.
Simplify: cos² θ = sin θ.

Express cos⁴ θ in terms of cos² θ
Since cos² θ = sin θ, we can write:
cos⁴ θ = (cos² θ)² = (sin θ)² = sin² θ.

Substitute into cos² θ + cos⁴ θ
Now substitute cos² θ = sin θ and cos⁴ θ = sin² θ into the expression cos² θ + cos⁴ θ:
cos² θ + cos⁴ θ = sin θ + sin² θ.
From the given equation, we know: sin θ + sin² θ = 1.

Thus: cos² θ + cos⁴ θ = 1.
The value of cos² θ + cos⁴ θ is 1.

Given: (1 + cotθ – cosecθ)(1 + tanθ + secθ).

Rewrite in terms of sinθ and cosθ:
(1 + cosθ/sinθ – 1/sinθ)(1 + sinθ/cosθ + 1/cosθ).

Simplify each factor:
First factor: (sinθ + cosθ – 1)/sinθ.
Second factor: (cosθ + sinθ + 1)/cosθ.

Multiply and simplify:
Numerator: (sinθ + cosθ)² – 1 = 2sinθcosθ.
Denominator: sinθcosθ.

Result: 2sinθcosθ / sinθcosθ = 2.