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## Solutions for Chapter 6: Trigonometry

Below listed, you can find solutions for Chapter 6 of Maharashtra State Board SCERT Maharashtra Question Bank for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board.

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.1 (A)

#### MCQ [1 Mark]

**Choose the correct alternative:**

cos θ. sec θ = ?

1

0

`1/2`

`sqrt(2)`

**Choose the correct alternative:**

sec 60° = ?

`1/2`

2

`2/sqrt(3)`

`sqrt(3)`

**Choose the correct alternative:**

1 + cot^{2}θ = ?

tan

^{2}θsec

^{2}θcosec

^{2}θcos

^{2}θ

**Choose the correct alternative:**

cot θ . tan θ = ?

1

0

2

`sqrt(2)`

**Choose the correct alternative:**

sec^{2}θ – tan^{2}θ =?

0

1

2

`sqrt(2)`

sin^{2}θ + sin^{2}(90 – θ) = ?

0

1

2

`sqrt(2)`

**Choose the correct alternative:**

`(1 + cot^2"A")/(1 + tan^2"A")` = ?

tan

^{2}Asec

^{2}Acosec

^{2}Acot

^{2}A

**Choose the correct alternative:**

sin θ = `1/2`, then θ = ?

30°

45°

60°

90°

**Choose the correct alternative:**

tan (90 – θ) = ?

sin θ

cos θ

cot θ

tan θ

**Choose the correct alternative:**

cos 45° = ?

sin 45°

sec 45°

cot 45°

tan 45°

**Choose the correct alternative:**

If sin θ = `3/5`, then cos θ = ?

`5/3`

`3/5`

`4/5`

`5/4`

**Choose the correct alternative:**

Which is not correct formula?

1 + tan

^{2}θ = sec^{2}θ1 + sec

^{2}θ = tan^{2}θcosec

^{2}θ − cot^{2}θ = 1sin

^{2}θ + cos^{2}θ = 1

**Choose the correct alternative:**

If ∠A = 30°, then tan 2A = ?

1

0

`1/sqrt(3)`

`sqrt(3)`

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.1 (B)

#### Solve the following questions: [ 1 Mark]

`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?

If tan θ = `13/12`, then cot θ = ?

Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1

If tan θ = 1, then sin θ . cos θ = ?

If 2 sin θ = 3 cos θ, then tan θ = ?

If cot( 90 – A ) = 1, then ∠A = ?

If 1 – cos^{2}θ = `1/4`, then θ = ?

Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`

If tan θ × A = sin θ, then A = ?

(sec θ + tan θ) . (sec θ – tan θ) = ?

`(sin 75^circ)/(cos 15^circ)` = ?

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.2 (A)

#### Complete the following activities [2 Marks]

Prove that cos^{2}θ . (1 + tan^{2}θ) = 1. Complete the activity given below.

**Activity:**

L.H.S = `square`

= `cos^2theta xx square .....[1 + tan^2theta = square]`

= `(cos theta xx square)^2`

= 1^{2}

= 1

= R.H.S

`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.

**Activity:**

`5/(sin^2theta) - 5cot^2theta`

= `square (1/(sin^2theta) - cot^2theta)`

= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`

= 5(1)

= `square`

If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ

**Activity:**

`square` = 1 + tan^{2}θ ......[Fundamental trigonometric identity]

`square` – tan^{2}θ = 1

(sec θ + tan θ) . (sec θ – tan θ) = `square`

`sqrt(3)*(sectheta - tan theta)` = 1

(sec θ – tan θ) = `square`

If tan θ = `9/40`, complete the activity to find the value of sec θ.

**Activity:**

sec^{2}θ = 1 + `square` ......[Fundamental trigonometric identity]

sec^{2}θ = 1 + `square^2`

sec^{2}θ = 1 + `square`

sec θ = `square`

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.2 (B)

#### Solve the following questions [2 Marks]

If cos θ = `24/25`, then sin θ = ?

Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ

Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ

If cos(45° + x) = sin 30°, then x = ?

If tan θ + cot θ = 2, then tan^{2}θ + cot^{2}θ = ?

Prove that sec^{2}θ + cosec^{2}θ = sec^{2}θ × cosec^{2}θ

Prove that cot^{2}θ × sec^{2}θ = cot^{2}θ + 1

If 3 sin θ = 4 cos θ, then sec θ = ?

If sin 3A = cos 6A, then ∠A = ?

Prove that sec^{2}θ − cos^{2}θ = tan^{2}θ + sin^{2}θ

Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`

Prove that `(sintheta + tantheta)/cos theta` = tan θ(1 + sec θ)

Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ

Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.3 (A)

#### Complete the following activities [3 Marks]

sin^{4}A – cos^{4}A = 1 – 2cos^{2}A. For proof of this complete the activity given below.

**Activity:**

L.H.S = `square`

= (sin^{2}A + cos^{2}A) `(square)`

= `1 (square)` .....`[sin^2"A" + square = 1]`

= `square` – cos^{2}A .....[sin^{2}A = 1 – cos^{2}A]

= `square`

= R.H.S

tan^{2}θ – sin^{2}θ = tan^{2}θ × sin^{2}θ. For proof of this complete the activity given below.

**Activity:**

L.H.S = `square`

= `square (1 - (sin^2theta)/(tan^2theta))`

= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`

= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`

= `tan^2theta (1 - square)`

= `tan^2theta xx square` .....[1 – cos^{2}θ = sin^{2}θ]

= R.H.S

If tan θ = `7/24`, then to find value of cos θ complete the activity given below.

**Activity:**

sec^{2}θ = 1 + `square` ......[Fundamental tri. identity]

sec^{2}θ = 1 + `square^2`

sec^{2}θ = 1 + `square/576`

sec^{2}θ = `square/576`

sec θ = `square`

cos θ = `square` .......`[cos theta = 1/sectheta]`

To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.

**Activity:**

L.H.S = `square`

= `square/sintheta + sintheta/costheta`

= `(cos^2theta + sin^2theta)/square`

= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`

= `1/sintheta xx 1/square`

= `square`

= R.H.S

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.3 (B)

#### Solve the following questions [3 Marks]

If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ

If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ

Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ

Prove that cot^{2}θ – tan^{2}θ = cosec^{2}θ – sec^{2}θ

Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)^{2}

Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ

Prove that `sec"A"/(tan "A" + cot "A")` = sin A

Prove that `(sintheta + "cosec" theta)/sin theta` = 2 + cot^{2}θ

Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1

Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A

Prove that sin^{4}A – cos^{4}A = 1 – 2cos^{2}A

Prove that sec^{2}θ – cos^{2}θ = tan^{2}θ + sin^{2}θ

Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`

In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?

Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`

If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A

Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.4

#### Solve the following questions: [Challenging questions, 4 marks]

Prove that

sin^{2}A . tan A + cos^{2}A . cot A + 2 sin A . cos A = tan A + cot A

Prove that

sec^{2}A – cosec^{2}A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

Prove that

`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`

Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ

If cos A = `(2sqrt("m"))/("m" + 1)`, then prove that cosec A = `("m" + 1)/("m" - 1)`

If sec A = `x + 1/(4x)`, then show that sec A + tan A = 2x or `1/(2x)`

In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin^{2}A + sin^{2}B + sin^{2}C = 2, then ∠A = ? , ∠B = ?, ∠C = ?

Prove that sin^{6}A + cos^{6}A = 1 – 3sin^{2}A . cos^{2}A

Prove that 2(sin^{6}A + cos^{6}A) – 3(sin^{4}A + cos^{4}A) + 1 = 0

Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1

### SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board Chapter 6 Trigonometry Q.5

#### Solve the following questions: [Creative questions, 3 Marks]

If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3

If cos A + cos^{2}A = 1, then sin^{2}A + sin4 A = ?

If cosec A – sin A = p and sec A – cos A = q, then prove that `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1

Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`

If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1

If tan θ – sin^{2}θ = cos^{2}θ, then show that sin^{2 }θ = `1/2`.

Prove that (1 – cos^{2}A) . sec^{2}B + tan^{2}B(1 – sin^{2}A) = sin^{2}A + tan^{2}B

## Solutions for Chapter 6: Trigonometry

## SCERT Maharashtra Question Bank solutions for 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board chapter 6 - Trigonometry

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Concepts covered in 10th Standard SSC Mathematics 2 Geometry Maharashtra State Board chapter 6 Trigonometry are Trigonometry Ratio of Zero Degree and Negative Angles, Trigonometric Ratios in Terms of Coordinates of Point, Angles in Standard Position, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Trigonometric Table, Heights and Distances, Trigonometric Ratios, Application of Trigonometry.

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