Updated on: 2026-03-31 | Author: Aarti Kulkarni

Geometry (Math 2) Chapter 6 Trigonometry Solutions

Q1. `(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?

`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = `(1 - (1)^2)/(1 + (1)^2)` ......[∵ tan 45° = 1]

= `(1 - 1)/(1 + 1)`

= `0/2`

= 0

Q2. `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`

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

= `5 (1/(sin^2theta) - cot^2theta)`

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

= 5(1)

= 5.

Q3. Choose the correct alternative: 1 + cot 2 θ = ?

1 + cot2θ = cosec2θ

Q4. Choose the correct alternative: `(1 + cot^2"A")/(1 + tan^2"A")` = ?

cot2A

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

= `("cosec"^2"A")/("sec"^2"A")`

= `(1/("sin"^2"A"))/(1/("cos"^2"A"))`

= `("cos"^2"A")/("sin"^2"A")`

= cot2A

Q5. Choose the correct alternative: cos 45° = ?

sin 45°

cos 45° = `1/sqrt2`, sin 45° = `1/sqrt2`

∴ cos 45° = sin 45°.

Q6. Choose the correct alternative: cos θ. sec θ = ?

1

cos θ. sec θ = cos θ. `1/"cos θ"` = 1.

Q7. Choose the correct alternative: cot θ . tan θ = ?

1

cot θ. tan θ = `1/"tan θ"`. tan θ = 1.

Q8. Choose the correct alternative: If ∠A = 30°, then tan 2A = ?

`sqrt(3)`

tan 2A = tan 60° = `sqrt(3)`.

Q9. Choose the correct alternative: If sin θ = `3/5`, then cos θ = ?

`4/5`

cos θ = `sqrt(1 - sin^2 θ)`

∴ cos θ = `sqrt(1 - (3/5)^2)`
∴ cos θ = `sqrt(1 - (9/25))`
∴ cos θ = `sqrt(16/25)`
∴ cos θ = `4/5`.

Q10. Choose the correct alternative: sec 60° = ?

2

Q11. Choose the correct alternative: sec 2 θ – tan 2 θ =?

1

1 + tan2θ = sec2θ

∵ sec2θ – tan2θ = 1.

Q12. Choose the correct alternative: sin θ = `1/2`, then θ = ?

30°

sin θ = `1/2`

∴ θ = 30° ...[sin 30° = `1/2`]

Q13. Choose the correct alternative: tan (90 – θ) = ?

cot θ

Q14. Choose the correct alternative: Which is not correct formula?

1 + sec2θ = tan2θ

Q15. If 1 – cos 2 θ = `1/4`, then θ = ?

1 – cos2θ = `1/4` ......[Given]

∴ sin2θ = `1/4` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
∴ sin θ = `1/2` ......[Taking square root of both sides]
∴ θ = 30° ......`[because sin 30^circ = 1/2]`

Q16. If 2 sin θ = 3 cos θ, then tan θ = ?

2sin θ = 3cos θ ......[Given]

∴ `sintheta/costheta = 3/2`
∴ tan θ = `3/2`

Q17. If 3 sin θ = 4 cos θ, then sec θ = ?

3 sin θ = 4cos θ .....[Given]

∴ `(sintheta)/(costheta) = 4/3`
∴ tan θ = `4/3`

We know that,

1 + tan2θ = sec2θ

∴ `1 + (4/3)^2` = sec2θ
∴ `1 + 16/9` = sec2θ
∴ sec2θ = `(9 + 16)/9`
∴ sec2θ = `25/9`
∴ sec θ = `5/3` ......[Taking square root of both sides]

Q18. If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3

3 sin A + 5 cos A = 5 ....[Given]
∴ (3 sin A + 5 cos A)2 = 25 ......[Squaring both the sides]
∴ 9 sin2A + 30 sin A cos A + 25 cos2A = 25
∴ 9(1 – cos2A) + 30 sin A cos A + 25(1 – sin2A) = 25
∴ 9 – 9 cos2A + 30 sin A cos A + 25 – 25 sin2A = 25
∴ 25 sin2A – 30 sin A cos A + 9 cos2A = 9
∴ (5 sin A – 3 cos A)2 = 9 ......[∵ a2 – 2ab + b2 = (a – b)2]
∴ 5 sin A – 3 cos A = ± 3 .....[Taking square root of both sides]

Q19. If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ

5 sec θ – 12 cosec θ = 0 ......[Given]

∴ 5 sec θ = 12 cosec θ
∴ `5/costheta = 12/sintheta` ......`[because sectheta = 1/costheta, "cosec" theta = 1/sintheta]`
∴ `sintheta/costheta = 12/5`
∴ tan θ = `12/5`

We know that,

1 + tan2θ = sec2θ

∴ `1 + (12/5)^2` = sec2θ
∴ `1 + 144/25` = sec2θ
∴ `(25 + 144)/25` = sec2θ
∴ sec2θ = `169/25`
∴ secθ = `13/5` ......[Taking square root of both sides]

Now, cos θ = `1/sectheta`

= `1/((13/5))`

∴ cos θ = `5/13`

We know that,

sin2θ + cos2θ = 1

∴ `sin^2theta + (5/13)^2` = 1
∴ `sin^2theta + 25/169` = 1
∴ sec2θ = `1 - 25/169`
∴ sec2θ = `(169 - 25)/169`
∴ sec2θ = `144/169`
∴ sin θ = `12/13` ......[Taking square root of both sides]
∴ sin θ = `12/13`, sec θ = `13/5`.

Q20. If cos θ = `24/25`, then sin θ = ?

cos θ = `24/25` ......[Given]

We know that,

sin2θ + cos2θ = 1

∴ `sin^2theta + (24/25)^2` = 1
∴ `sin^2theta + 576/625` = 1
∴ sin2θ = `1 - 576/625`
∴ sin2θ = `(625 - 576)/625`
∴ sin2θ = `49/625`
∴ sin θ = `7/25` ......[Taking square root of both sides]

Q21. If cos(45° + x) = sin 30°, then x = ?

cos(45° + x) = sin 30° .....[Given]
∴ cos(45° + x) = sin(90° – 60°)
∴ cos(45° + x) = cos 60° .....[∵ sin (90° – θ) = cos θ]
∴ 45° + x = 60°
∴ x = 60° – 45°
∴ x = 15°

Q22. If cos A = `(2sqrt("m"))/("m" + 1)`, then prove that cosec A = `("m" + 1)/("m" - 1)`

cos A = `(2sqrt("m"))/("m" + 1)` ......[Given]

We know that,

sin2A + cos2A = 1
∴ `sin^2"A" + ((2sqrt("m"))/("m" + 1))^2` = 1
∴ `sin^2"A" + (4"m")/("m" + 1)^2` = 1
∴ sin2A = `1 - (4"m")/("m" + 1)^2`

= `(("m" + 1)^2 - 4"m")/("m" + 1)^2`

= `("m"^2 + 2"m" + 1 - 4"m")/("m" + 1)^2` ......[∵ (a + b)2 = a2 + 2ab + b2]

= `("m"^2 - 2"m" + 1)/("m" + 1)^2`

∴ sin2A = `("m" - 1)^2/("m" + 1)^2` ......[∵ a2 – 2ab + b2 = (a – b)2]
∴ sin A = `("m" - 1)/("m" + 1)` .....[Taking square root of both sides]

Now, cosec A = `1/"sin A"`

= `1/(("m" - 1)/("m" + 1))`

∴ cosec A = `("m" + 1)/("m" - 1)`

Q23. If cos A + cos 2 A = 1, then sin 2 A + sin4 A = ?

cos A + cos2A = 1 ......[Given]
∴ cos A = 1 – cos2A
∴ cos A = sin2A ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
∴ cos2A = sin4A .....[Squaring both the sides]
∴ 1 – sin2A = sin4A
∴ 1 = sin4A + sin2A
∴ sin2A + sin4A = 1

Q24. 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

cosec A – sin A = p ......[Given]
∴ `1/"sin A" - sin "A"` = p
∴ `(1 - sin^2"A")/"sin A"` = p
∴ `(cos^2"A")/"sin A"` = p ......`(i) [(because sin^2"A" + cos^2"A" = 1),(therefore 1 - sin^2"A" = cos^2"A")]`
sec A – cos A = q ......[Given]
∴ `1/"cos A" - cos "A"` = q
∴ `(1 - cos^2"A")/"cos A"` = q
∴ `(sin^2"A")/"cos A"` = q .....`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`

L.H.S = `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)`

= `[((cos^2"A")/(sin "A"))^2 ((sin^2"A")/(cos"A"))]^(2/3) + [((cos^2"A")/(sin "A"))((sin^2"A")/(cos"A"))^2]^(2/3)` ......[From (i) and (ii)]

= `((cos^4"A")/(sin^2"A") xx (sin^2"A")/(cos"A"))^(2/3) + ((cos^2"A")/(sin"A") xx (sin^4"A")/(cos^2"A"))^(2/3)`

= `(cos^3"A")^(2/3) + (sin^3"A")^(2/3)`

= cos2A + sin2A

= 1

= R.H.S

∴ `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1

Q25. If cot( 90 – A ) = 1, then ∠A = ?

cot(90 – A) = 1 .....[Given]
∴ tan A = 1
∴ A = 45° .....[∵ tan 45° = 1]

Q26. If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ

sec θ = `41/40` ......[Given]

∴ cos θ = `1/sectheta = 1/(41/40)`
∴ cos θ = `40/41`

We know that,

sin2θ + cos2θ = 1

∴ `sin^2theta + (40/41)^2` = 1
∴ `sin^2theta + 1600/1681` = 1
∴ sin2θ = `1 - 1600/1681`
∴ sin2θ = `(1681- 1600)/1681`
∴ sin2θ = `81/1681`
∴ sin θ = `9/41` .......[Taking square root of both sides]

Now, cosec θ = `1/sintheta`

= `1/((9/41))`

= `41/9`

cot θ = `costheta/sintheta`

= `((40/41))/((9/41))`

= `40/9`

∴ sin θ = `9/41`, cot θ = `40/9`, cosec θ = `41/9`

Q27. If sec A = `x + 1/(4x)`, then show that sec A + tan A = 2x or `1/(2x)`

sec A = `x + 1/(4x)` .....[Given]

We know that,

1 + tan2A = sec2A
∴ tan2A = sec2A – 1

= `(x + 1/(4x))^2 - 1`

= `x^2 + 2 xx x xx 1/(4x) + (1/(4x))^2 - 1` ......[∵ (a + b)2 = a2 + 2ab + b2]

= `x^2 + 1/2 + 1/(16x^2) - 1`

= `x^2 - 1/2 + 1/(16x^2)`

∴ tan2A = `(x - 1/(4x))^2` ......[∵ a2 – 2ab + b2 = (a – b)2]
∴ tan A = `x - 1/(4x)` or tan A = `-(x - 1/(4x))`

When tan A = `x - 1/(4x)`,

sec A + tan A

= `x + 1/(4x) + x - 1/(4x)`

= 2x

When tan A = `-(x - 1/(4x))`,

sec A + tan A

= `x + 1/(4x) - (x - 1/(4x))`

= `x + 1/(4x) - x + 1/(4x)`

= `2/(4x)`

= `1/(2x)`

Q28. 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`

sec2θ = 1 + tan2θ ......[Fundamental trigonometric identity]

sec2θ – tan2θ = 1

(sec θ + tan θ) . (sec θ – tan θ) = 1

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

(sec θ – tan θ) = `1/sqrt(3)`

Q29. If sin 3A = cos 6A, then ∠A = ?

sin 3A = cos 6A .....[Given]
∴ sin 3A = sin(90° – 6A) .....[∵ cos θ = sin(90° – θ)]
∴ 3A = 90° – 6A
∴ 3A + 6A = 90°
∴ 9A = 90°
∴ A = `(90^circ)/9`
∴ A = 10°

Q30. If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A

Diagram: Refer textbook

Q31. If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1

sin θ + cos θ = `sqrt(3)` ......[Given]

∴ (sin θ + cos θ)2 = 3 ......[Squaring on both sides]
∴ sin2θ + 2sinθ cosθ + cos2θ = 3 ......[∵ (a + b)2 = a2 + 2ab + b2]
∴ (sin2θ + cos2θ) + 2sinθ cosθ = 3
∴ 1 + 2 sin θ cos θ = 3 ......[∵ sin2θ + cos2θ = 1]
∴ 2 sin θ cos θ = 2
∴ sin θ cos θ = 1 ......(i)

tan θ + cot θ = `sintheta/costheta + costheta/sintheta`

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

= `1/(sintheta costheta)` ......[∵ sin2θ + cos2θ = 1]

= `1/1` ......[From (i)]

= 1

Q32. If tan θ = 1, then sin θ . cos θ = ?

tan θ = 1 ......[Given]

∴ θ = 45° ......[∵ tan45° = 1]
∴ sin θ . cos θ = sin 45° cos 45°

= `1/sqrt(2)*1/sqrt(2)`

= `1/2`

Q33. If tan θ = `13/12`, then cot θ = ?

cot θ = `1/tantheta`

= `1/(13/12)`

∴ cot θ = `12/13`

Q34. 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]`

sec2θ = 1 + tan2θ ......[Fundamental tri. identity]

∴ sec2θ = 1 + `(7/24)^2`
∴ sec2θ = 1 + `49/576`
∴ sec2θ =`(576 + 49)/576`
∴ sec2θ = `625/576`
∴ sec θ = `25/24`
∴ cos θ = `24/25` .......`[cos theta = 1/sectheta]`

Q35. 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`

sec2θ = 1 + tan2θ ......[Fundamental trigonometric identity]

∴ sec2θ = 1 + `(9/40)^2`
∴ sec2θ = 1 + `81/1600`
∴ sec2θ = `1681/1600`
∴ sec θ = `41/40`

Q36. If tan θ × A = sin θ, then A = ?

tan θ × A = sin θ .....[Given]
∴ `(sin theta)/(cos theta) xx "A"` = sin θ
∴`1/(cos theta) xx "A"` = 1
∴ A = cos θ

Q37. If tan θ + cot θ = 2, then tan 2 θ + cot 2 θ = ?

tan θ + cot θ = 2 ....[Given]

∴ (tan θ + cot θ)2 = 4 .....[Squaring both sides]
∴ tan2θ + 2tan θ.cot θ + cot2θ = 4 ......[∵ (a + b)2 = a2 + 2ab + b2]
∴ tan2θ + 2(1) + cot2θ = 4 ......[∵ tan θ ⋅ cot θ = 1]
∴ tan2θ + cot2θ = 4 – 2
∴ tan2θ + cot2θ = 2

Q38. If tan θ – sin 2 θ = cos 2 θ, then show that sin 2 θ = `1/2`.

tan θ – sin2θ = cos2θ ......[Given]

∴ tan θ = sin2θ + cos2θ
∴ tan θ = 1 ....[∵ sin2θ + cos2θ = 1]

But, tan 45° = 1

∴ tan θ = tan 45°
∴ θ = 45°

sin2θ = sin245°

= `(1/sqrt(2))^2`

= `1/2`

Q39. In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin 2 A + sin 2 B + sin 2 C = 2, then ∠A = ? , ∠B = ?, ∠C = ?

Diagram: Refer textbook

Q40. In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?

Diagram: Refer textbook

Q41. Prove that (1 – cos 2 A) . sec 2 B + tan 2 B(1 – sin 2 A) = sin 2 A + tan 2 B

L.H.S = (1 – cos2A) . sec2B + tan2B(1 – sin2A)
= `sin^2"A"* 1/(cos^2"B") + (sin^2"B")/(cos^2"B") (1 - sin^2"A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`

= `(sin^2"A")/(cos^2"B") + (sin^2"B")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B")`

= `(sin^2"A")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B") + (sin^2"B")/(cos^2"B")`

= `(sin^2"A")/(cos^2"B") (1 - sin^2"B") + tan^2"B"`

= `(sin^2"A")/(cos^2"B") (cos^2"B") + tan^2"B"`

= sin2A + tan2B

= R.H.S

∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B

Q42. Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A

L.H.S = `sqrt((1 + cos "A")/(1 - cos"A"))`

= `sqrt((1 + cos "A")/(1 - cos "A") xx (1 + cos "A")/(1 + cos "A"))` ......[On rationalising the denominator]

= `sqrt((1 + cos "A")^2/(1 - cos^2 "A"))`

= `sqrt((1 + cos "A")^2/(sin^2 "A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`

= `(1 + cos"A")/"sin A"`

= `1/"sin A" + "cos A"/"sin A"`

= cosec A + cot A

= R.H.S

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

Q43. Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`

L.H.S = `(1 + sec "A")/"sec A"`

= `1/"sec A" + "sec A"/"sec A"`

= cos A + 1

= `(1 + cos "A") xx (1 - cos"A")/(1 - cos"A")`

= `(1 - cos^2"A")/(1 - cos"A")`

= `(sin^2"A")/(1 - cos"A")` .......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`

= R.H.S

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

Q44. Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ) 2

L.H.S = `(1 + sintheta)/(1 - sin theta)`

= `((1 + sintheta)/(costheta))/((1 - sintheta)/(costheta))` ......[Dividing numerator and denominator by cos θ]

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

= `(sectheta + tantheta)/(sectheta - tantheta)`

= `(sectheta + tantheta)/(sectheta - tantheta) xx (sectheta + tantheta)/(sectheta + tantheta)` ......[On rationalising the denominator]

= `(sectheta + tantheta)^2/(sec^2theta - tan^2theta)`

= `(sectheta + tantheta)^2/1` ......`[(because 1 + tan^2theta = sec^2theta),(therefore sec^2theta - tan^2theta = 1)]`

= (sec θ + tan θ)2

= R.H.S

∴ `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2

Q45. Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B

L.H.S = `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")`

= `((1 +sin "B")^2 + cos^2"B")/(cos "B"(1 + sin "B"))`

= `(1 +2sin"B" + sin^2"B" + cos^2"B")/(cos"B"(1 + sin"B"))` ......[∵ (a + b)2 = a2 + 2ab + b2]
= `(1 + 2sin"B" + 1)/(cos"B"(1+ sin"B"))` .....[∵ sin2B + cos2B = 1]

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

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

= `2/"cos B"`

= 2 sec B

= R.H.S

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

Q46. Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ

L.H.S = `1/("cosec" theta - cot theta)`

= `1/("cosec" theta - cot theta) xx ("cosec"theta + cottheta)/("cosec"theta + cottheta)` ......[On rationalising the denominator]

= `("cosec"theta + cottheta)/("cosec"^2theta - cot^2theta)` ......[∵ (a – b)(a + b) = a2 – b2]
= `("cosec"theta cottheta)/1` ......`[(∵ 1 + cot^2θ = "cosec"^2θ),(∴ "cosec"^2θ - cot^2θ = 1)]`

= cosecθ + cotθ

= R.H.S

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

Q47. Prove that 2(sin 6 A + cos 6 A) – 3(sin 4 A + cos 4 A) + 1 = 0

sin6A + cos6A = (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= 1 – 3 cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ......[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]

= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A

= 1 – 3 cos2A sin2A

sin4A + cos4A = (sin2A)2 + (cos2A)2

= (1 – cos2A)2 + (cos2A)2

= 1 – 2 cos2A + (cos2A)2 + (cos2A)2 ......[∵ (a – b)2 = a2 – 2ab + b2]

= 1 – 2 cos2A + 2 cos4A

= 1 – 2 cos2A(1 – cos2A)

= 1 – 2 cos2A sin2A

L.H.S = 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1

= 2(1 – 3 cos2A sin2A) – 3(1 – 2 cos2A sin2A) + 1

= 2 – 6 cos2A sin2A – 3 + 6 cos2A sin2A + 1

= 0

= R.H.S

∴ 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0

Q48. Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`

L.H.S = `costheta/(1 + sintheta)`

= `costheta/(1 + sintheta) xx (1 - sintheta)/(1 - sintheta)` ......[On rationalising the denominator]

= `(costheta(1 - sintheta))/(1 - sin^2theta)`

= `(costheta(1 - sintheta))/(cos^2theta)` ......`[(because sin^2theta +cos^2theta = 1),(therefore 1 -sin^2theta = cos^2theta)]`

= `(1 - sintheta)/costheta`

= R.H.S

∴ `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`

Q49. Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`

L.H.S = `(cos(90 - "A"))/(sin "A")`

= `"sin A"/"sin A"`

= 1

R.H.S = `(sin(90 - "A"))/(cos "A")`

= `"cos A"/"cos A"`

= 1

∴ L.H.S = R.H.S

Q50. 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

L.H.S. = `cos^2theta*(1 + tan^2theta)`

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

= `(cos theta xx sectheta)^2`

= 12

= 1

= R.H.S

Q51. Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ

L.H.S = `(cos^2theta)/(sintheta) + sintheta`

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

= `1/sintheta` .......[∵ sin2θ + cos2θ = 1]

= cosec θ

= R.H.S

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

Q52. Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1

L.H.S = `"cosec" θ xx sqrt(1 - cos^2theta)`

= `"cosec" θ xx sqrt(sin^2theta)` ......`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`

= cosec θ × sin θ

= 1 ......[∵ sin θ × cosec θ = 1]

= R.H.S

Q53. Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`

L.H.S = cosec θ – cot θ

= `1/sintheta - costheta/sintheta`

= `(1 -costheta)/sintheta`

= `(1 - costheta)/sintheta xx (1 + costheta)/(1 +costheta)` .....[On rationalising the numerator]

= `(1 - cos^2theta)/(sintheta(1 +costheta))`

= `(sin^2theta)/(sintheta(1 + costheta))` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`

= `sintheta/(1 + costheta)`

= R.H.S

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

Q54. Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1

L.H.S = `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")`

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

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

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

= `1/(tan "A" - 1) + "tan A"/(1 - tan "A")` ......[∵ cot A tan A = 1]

= `- 1/(1 - tan "A") + "tan A"/(1 - tan "A")`

= `- (1/(1 -tan "A") - "tan A"/(1- tan "A"))`

= `-((1 - tan "A")/(1 - tan "A"))`

= – 1

= R.H.S

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

Q55. Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1

`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")`

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

= `((cos "A")/(sin "A"))/((cos "A" - sin "A")/(cos "A")) + ((sin "A")/(cos "A"))/((sin "A" - cos "A")/(sin "A"))`

= `"cos A"/"sin A" xx "cos A"/(cos "A" - sin "A") + "sin A"/"cos A" xx "sin A"/(sin "A" - cos "A")`

= `(cos^2"A")/(sin "A"(cos "A" - sin "A")) + (sin^2"A")/(cos"A"(sin"A" - cos"A"))`

= `1/(sin "A" - cos "A") ((-cos^3"A" + sin^3"A")/(sin"A" cos"A"))`

= `1/(sin"A" - cos"A")((sin^3"A" - cos^3"A")/(sin"A" cos"A"))`

= `1/(sin"A" - cos"A")xx ((sin"A" - cos"A")(sin^2"A" + sin"A" cos"A" + cos^2"A"))/(sin"A" cos"A")` ......[∵ a3 – b3 = (a – b)(a2 + ab + b2)]

= `(sin^2"A" +sin"A" cos"A" + cos^2"A")/(sin"A" cos"A"` ......(i)

= `(1 + sin"A" cos"A")/(sin"A" cos"A")` .....[∵ sin2A + cos2A = 1]

= `1/(sin"A" cos"A") + (sin"A" cos"A")/(sin"A" cos"A")`

= cosec A sec A + 1 .....(ii)

`"cot A"/(1 - tan "A") + "tan A"/(1 - cot "A")`

= `(sin^2"A" + sin"A" cos"A" + cos^2"A")/(sin"A" cos"A")` ......[From (i)]

= `(sin^2"A")/(sin"A" cos"A") + "sin A cos A"/"sin A cos A" + (cos^2"A")/"sin A cos A"`

= `"sin A"/"cos A" + 1 + "cos A"/"sin A"`

= tan A + 1 + cot A ......(iii)

From (ii) and (iii), we get

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

Q56. Prove that cot 2 θ × sec 2 θ = cot 2 θ + 1

L.H.S = cot2θ × sec2θ

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

= `1/(sin^2theta)`

= cosec2θ

= 1 + cot2θ ......[∵ 1 + cot2θ = cosec2θ]

= R.H.S

∴ cot2θ × sec2θ = cot2θ + 1

Q57. Prove that cot 2 θ – tan 2 θ = cosec 2 θ – sec 2 θ

L.H.S = cot2θ – tan2θ
= (cosec2θ − 1) − (sec2θ − 1) ......`[(because tan^2theta = sec^2theta - 1),(cot^2theta = "cosec"^2 theta - 1)]`

= cosec2θ − 1 − sec2θ + 1

= cosec2θ − sec2θ

= R.H.S

∴ cot2θ – tan2θ = cosec2θ – sec2θ

Q58. Prove that `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`

L.H.S = `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1)`

= `(cot"A" + "cosec A" - ("cosec"^2"A" - cot^2"A"))/(cot"A" - "cosec A" + 1)` .....`[(because 1 + cot^2"A" = "cosec"^2"A"),(therefore "cosec"^2"A" - cot^2"A" = 1)]`
= `(cot"A" + "cosec A" - ("cosec A" + cot"A")("cosec A" - cot"A"))/(cot"A" - "cosec A" + 1)` .....[∵ a2 – b2 = (a + b) (a – b)]

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

= cot A + cosec A

= `"cos A"/"sin A" + 1/"sin A"`

= `(cos "A" + 1)/"sin A"`

= R.H.S

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

Q59. Prove that sec 2 θ − cos 2 θ = tan 2 θ + sin 2 θ

L.H.S = sec2θ − cos2θ

= 1 + tan2θ – cos2θ .......[∵ 1 + tan2θ = sec2θ]

= tan2θ + (1 – cos2θ)

= tan2θ + sin2θ ......`[(because sin^2theta +cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`

= R.H.S

∴ sec2θ − cos2θ = tan2θ + sin2θ

Q60. Prove that sec 2 θ – cos 2 θ = tan 2 θ + sin 2 θ

L.H.S = sec2θ – cos2θ
= sec2θ – (1 – sin2θ) ......`[(because sin^2theta + cos^2theta = 1),(therefore 1 - sin^2theta = cos^2theta)]`

= sec2θ – 1 + sin2θ

= tan2θ + sin2θ ......`[(because 1 + tan^2theta = sec^2theta),(therefore tan^2theta = sec^2theta - 1)]`

= R.H.S

∴ sec2θ – cos2θ = tan2θ + sin2θ

Q61. Prove that sec 2 θ + cosec 2 θ = sec 2 θ × cosec 2 θ

L.H.S = sec2θ + cosec2θ

= `1/(cos^2theta) + 1/(sin^2theta)`

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

= `1/(cos^2theta*sin^2theta)` ......[∵ sin2θ + cos2θ = 1]

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

= sec2θ × cosec2θ

= R.H.S

∴ sec2θ + cosec2θ = sec2θ × cosec2θ

Q62. Prove that sec 2 A – cosec 2 A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

L.H.S = sec2A – cosec2A

= `1/(cos^2"A") - 1/(sin^2"A")`

= `(sin^2"A" - cos^2"A")/(cos^2"A"*sin^2"A")`

= `(sin^2"A" - (1 - sin^2"A"))/(sin^2"A"*cos^2"A")` .....`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - sin^2"A" = cos^2"A")]`

= `(sin^2"A" - 1 + sin^2"A")/(sin^2"A"*cos^2"A")`

= `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

= R.H.S

∴ sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

Q63. Prove that `sec"A"/(tan "A" + cot "A")` = sin A

L.H.S = `sec"A"/(tan "A" + cot "A")`

= `sec"A"/((sin"A")/(cos"A") + (cos"A")/(sin"A"))`

= `sec"A"/((sin^2"A" + cos^2"A")/(cos"A" sin"A"))`

= `sec"A"/(1/(cos"A" sin"A"))` ......[∵ sin2A + cos2A = 1]

= sec A cos A sin A

= `1/cos"A" xx cos "A" sin "A"`

= sin A

= R.H.S.

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

Q64. Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ

L.H.S = sin θ (1 – tan θ) – cos θ (1 – cot θ)

= `sintheta (1 - (sintheta)/(costheta)) - costheta (1 - (costheta)/(sintheta))`

= `sintheta - (sin^2theta)/costheta - costheta + (cos^2theta)/sintheta`

= `sintheta + (cos^2theta)/sintheta - (sin^2theta)/costheta - costheta`

= `(sin^2theta + cos^2theta)/sintheta - ((sin^2theta + cos^2theta)/costheta)`

= `1/sintheta - 1/costheta` ......[∵ sin2θ + cos2θ = 1]

= cosec θ – sec θ

= R.H.S

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

Q65. Prove that `(sintheta + "cosec" theta)/sin theta` = 2 + cot 2 θ

L.H.S = `(sintheta + "cosec" theta)/sin theta`

= `sintheta/sintheta + ("cosec"theta)/sintheta`

= 1 + cosec θ × cosec θ ......`[∵ "cosec" theta = 1/sin theta]`

= 1 + cosec2θ

= 1 + 1 + cot2θ .......[∵ 1 + cot2θ = cosec2θ]

= 2 + cot2θ

= R.H.S

∴ `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ

Q66. Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ

L.H.S = `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)`

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

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

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

= `sin theta costheta (1 /(1 + costheta) + 1/(1 - costheta))`

= `sintheta costheta [(1 - costheta + 1 + costheta)/((1 + costheta)(1 - costheta))]`

= `sintheta costheta (2/(1 - cos^2theta))` ......[∵ (a + b)(a – b) = a2 – b2]
= `sintheta costheta xx 2/(sin^2theta)` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`

= `2 xx (costheta)/(sintheta)`

= 2cot θ

= R.H.S

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

Q67. Prove that `(sintheta + tantheta)/cos theta` = tan θ(1 + sec θ)

L.H.S = `(sintheta + tantheta)/cos theta`

= `sintheta/costheta + tantheta/costheta`

= tan θ + tan θ sec θ

= tan θ(1 + sec θ)

= R.H.S

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

Q68. Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ

L.H.S = `(sin^2theta)/(cos theta) + cos theta`

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

= `1/costheta` ......[∵ sin2θ + cos2θ = 1]

= sec θ

= R.H.S

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

Q69. Prove that sin 2 A . tan A + cos 2 A . cot A + 2 sin A . cos A = tan A + cot A

L.H.S = sin2A . tan A + cos2A . cot A + 2 sin A . cos A

= `sin^2"A"* (sin "A")/(cos "A") + cos^2"A"* (cos"A")/(sin"A") + 2sin"A" *cos"A"`

= `(sin^3"A")/"cosA" + (cos^3"A")/"sinA" + 2sin"A"*cos"A"`

= `(sin^4"A" + cos^4"A" + 2sin^2"A"cos^2"A")/(sin"A"cos"A")`

= `(sin^2"A" + cos^2"A")^2/(sin"A"cos"A")` .....[∵ a2 + b2 + 2ab = (a + b)2]
= `1^2/(sin"A"cos"A")` ......[∵ sin2A + cos2A = 1]

= `1/(sin"A"cos"A")`

= `(sin^2"A"+ cos^2"A")/(sin"A"cos"A")` ......[∵ 1 = sin2A + cos2A]

= `(sin^2"A")/(sin"A"cos"A") + (cos^2"A")/(sin"A"cos"A")`

= `"sin A"/"cos A" + "cos A"/"sin A"`

= tan A + cot A

= R.H.S

∴ sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A

Q70. Prove that sin 4 A – cos 4 A = 1 – 2cos 2 A

L.H.S = sin4A – cos4A

= (sin2A)2 – (cos2A)2

= (sin2A + cos2A)(sin2A – cos2A) .....[∵ a2 – b2 = (a + b)(a – b)]
= (1)(sin2A – cos2A) ......[∵ sin2A + cos2A = 1]

= sin2A – cos2A

= (1 – cos2A) – cos2A ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"" = sin^2"A")]`

= 1 – 2cos2A

= R.H.S

∴ sin4A – cos4A = 1 – 2cos2A

Q71. Prove that sin 6 A + cos 6 A = 1 – 3sin 2 A . cos 2 A

L.H.S = sin6A + cos6A

= (sin2A)3 + (cos2A)3

= (1 – cos2A)3 + (cos2A)3 ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2A")]`
= 1 – 3cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ......[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]

= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A

= 1 – 3 cos2A sin2A

= R.H.S

∴ sin6A + cos6A = 1 – 3sin2A . cos2A

Q72. Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ

L.H.S = `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)`

= `1/("cosec" theta)(cottheta + tantheta)` .....`[(because tan(90 - theta) = cot theta),(cot(90 - theta) = tantheta)]`

= sin θ (cot θ + tan θ)

= `sintheta ((costheta)/(sintheta) + (sintheta)/(costheta))`

= `sintheta ((cos^2theta + sin^2theta)/(sintheta costheta))`

= `sintheta (1/(sintheta costheta))` ......[∵ sin2θ + cos2θ = 1]

= `1/costheta`

= sec θ

= R.H.S

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

Q73. Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`

R.H.S = `(sec^2"A")/("cosec"^2"A")`

= `(1 + tan^2"A")/(1 + cot^2"A")` .....`[(because 1 + tan^2"A" = sec^2"A"),(1 + cot^2"A" = "cosec"^2"A")]`

= `(1 + (sin^2"A")/(cos^2"A"))/(1 + (cos^2"A")/(sin^2"A"))`

= `((cos^2"A" + sin^2"A")/(cos^2"A"))/((sin^2"A" + cos^2"A")/(sin^2"A"))`

= `(1/(cos^2"A"))/(1/(sin^2"A"))` .......[∵ sin2A + cos2A = 1]

= `(sin^2"A")/(cos^2"A")`

= tan2A

= tan A . tan A

= `"tan A"/"cot A"`

= L.H.S

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

Q74. (sec θ + tan θ) . (sec θ – tan θ) = ?

(sec θ + tan θ)(sec θ – tan θ)

= sec2θ – tan2θ ......[∵ (a + b)(a – b) = a2 – b2]
= 1 ......`[(because 1 + tan^2theta = sec^2theta),(therefore sec^2theta - tan^2theta = 1)]`

Q75. Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`

L.H.S = tan 7° × tan 23° × tan 60° × tan 67° × tan 83°

= tan 7° × tan 23° × `sqrt(3)` × tan(90° – 23°) × tan(90° – 7°)

= `sqrt(3)` × [tan 7° × tan(90° – 7°)] × [tan 23° × tan(90° – 23°)]

= `sqrt(3) xx 1 xx 1` ......[∵ tan θ × tan(90° – θ) = 1]

= `sqrt(3)`

= R.H.S

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

Q76. sin 2 θ + sin 2 (90 – θ) = ?

1

Explanation:

(sin (90 – θ))2 = (cosθ)2
sin2 (90 – θ) = cos2θ ...(1)

sin2θ + cos2θ = 1

∴ sin2θ + sin2(90 – θ) = 1 ...From (1)

Q77. 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

L.H.S = sin4A – cos4A

= (sin2A)2 – (cos2A)2

= (sin2A + cos2A) (sin2A – cos2A) .....[∵ a2 – b2 = (a + b)(a – b)]
= 1(sin2A – cos2A) .....[∵ sin2A + cos2A = 1]

= sin2A – cos2A

= 1 – cos2A – cos2A .....[sin2A = 1 – cos2A]

= 1 – 2cos2A

= R.H.S

Q78. `(sin 75^circ)/(cos 15^circ)` = ?

`(sin 75^circ)/(cos 15^circ)` = `(sin(90^circ - 15^circ))/(cos 15^circ)`

= `(cos 15^circ)/(cos 15^circ)` .....[∵ sin(90° – θ) = cos θ]

= 1

Q79. 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

L.H.S = tan2θ – sin2θ

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

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

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

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

= tan2θ × sin2θ .....[1 – cos2θ = sin2θ]

= R.H.S

Q80. 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

L.H.S = cot θ + tan θ

= `costheta/sintheta + sintheta/costheta`

= `(cos^2theta + sin^2theta)/(sintheta*costheta)`

= `1/(sintheta*costheta)` ......[cos2θ + sin2θ = 1]

= `1/sintheta xx 1/costheta`

= cosecθ × secθ

= R.H.S.

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