SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

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 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 `(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)`.

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