SSC (Marathi Semi-English) १० वीं कक्षा - Maharashtra State Board Question Bank Solutions

sin²θ + sin²(90 – θ) = ?

Choose the correct alternative:

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

Choose the correct alternative:

sin θ = `1/2`, then θ = ?

Choose the correct alternative:

tan (90 – θ) = ?

Choose the correct alternative:

cos 45° = ?

Choose the correct alternative:

Which is not correct formula?

`(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 1 – cos²θ = `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) = ?

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

Activity:

L.H.S = `square`

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

= `(cos theta xx square)^2`

= 1²

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

`square` – tan²θ = 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:

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

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