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

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) = ?

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 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 = ?

Prove that sec²θ – cos²θ = tan²θ + sin²θ.

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.