SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] Standard 10 Maharashtra State Board chapter 6 - Trigonometry [Latest edition]

cos θ . sec θ = ?

sec 60° = ?

1 + cot²θ = ? 

cot θ . tan θ = ?

sec²θ – tan²θ = ?

sin²θ + sin²(90 – θ) = ?

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

`sin θ = 1/2`, then θ = ?

tan (90 – θ) = ?

cos 45° = ?

If `sin θ = 3/5`, then cos θ = ?

Which is not correct formula?

If ∠A = 30°, then tan 2A = ?

`(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^2θ) = 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^2θ = 1/4`, then θ = ?

Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - 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)` = ?

Prove that cos²θ . (1 + tan²θ) = 1. Complete the activity given below.

Activity:

L.H.S. = `square`

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

= `(cos θ xx square)^2`

= 1²

= 1

= R.H.S.

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

Activity:

`5/(sin^2θ) - 5cot^2θ`

= `square (1/(sin^2θ) - cot^2θ)`

= `5(square - cot^2θ)   ...[1/(sin^2θ) = 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^2θ`   ...[Fundamental trigonometric identity]

`square - tan^2θ = 1`

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

`sqrt(3)  . (sec θ - tan θ) = 1`

`(sec θ - tan θ) = square`

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

Activity:

sec²θ = 1 + `square`   ...[Fundamental trigonometric identity]

sec²θ = 1 + `square^2`

sec²θ = 1 + `square` 

sec θ = `square` 

If `cos θ = 24/25`, then sin θ = ?

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

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

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

If tan θ + cot θ = 2, then tan²θ + cot²θ = ?

Prove that sec²θ + cosec²θ = sec²θ × cosec²θ.

Prove that cot²θ × sec²θ = cot²θ + 1.

If 3 sin θ = 4 cos θ, then sec θ = ?

If sin 3A = cos 6A, then ∠A = ?

Prove that sec²θ – cos²θ = tan²θ + sin²θ.

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

Prove that `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`.

Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec"  θ`.

Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.

sin⁴A – cos⁴A = 1 – 2cos²A. For proof of this complete the activity given below.

Activity:

L.H.S. = `square`

 = (sin²A + cos²A) `(square)`

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

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

= `square`

= R.H.S.

tan²θ – sin²θ = tan²θ × sin²θ. For proof of this complete the activity given below.

Activity:

L.H.S. = `square`

= `square (1 - (sin^2θ)/(tan^2θ))`

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

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

= `tan^2θ (1 - square)`

= `tan^2θ xx square`   ...[1 – cos²θ = sin²θ]

= R.H.S.

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

Activity:

sec²θ = 1 + `square`   ...[Fundamental tri. identity]

sec²θ = 1 + `square^2`

sec²θ = 1 + `square/576`

sec²θ = `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/(sinθ) + (sinθ)/(cosθ)`

= `(cos^2θ + sin^2θ)/square`

= `1/(sinθ.cosθ)`   ...`[cos^2θ + sin^2θ = square]`

= `1/(sinθ) xx 1/square`

= `square`

= R.H.S.

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 - θ) + cot(90 - θ))/("cosec"  θ) = sec θ`.

Prove that cot²θ – tan²θ = cosec²θ – sec²θ.

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

Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.

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

Prove that `(sin θ + "cosec"  θ)/(sin θ) = 2 + cot^2θ`.

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 sin⁴A – cos⁴A = 1 – 2 cos²A.

Prove that sec²θ – cos²θ = tan²θ + sin²θ.

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

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

Prove that `(1 + sec A)/(sec A) = (sin^2A)/(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`.

Prove that sin²A . tan A + cos²A . cot A + 2 sin A . cos A = tan A + cot A.

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

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, sin²A + sin²B + sin²C = 2, then ∠A = ?, ∠B = ?, ∠C = ?

Prove that sin⁶A + cos⁶A = 1 – 3sin²A . cos²A.

Prove that 2(sin⁶A + cos⁶A) – 3(sin⁴A + cos⁴A) + 1 = 0.

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

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

If cos A + cos²A = 1, then sin²A + sin⁴A = ?

If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(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 – sin²θ = cos²θ, then show that `sin^2θ = 1/2`.

Prove that (1 – cos²A) . sec²B + tan²B (1 – sin²A) = sin²A + tan²B.