SSC (English Medium) 10th Standard - Maharashtra State Board Important Questions

Choose the correct alternative:

1 + tan² θ = ?

If sec θ = `25/7`, then find the value of tan θ.

If tan θ = 2, where θ is an acute angle, find the value of cos θ. 

If sin θ = `1/2`, then find the value of θ. 

Choose the correct alternative: 
sinθ × cosecθ =?

If cosθ = `5/13`, then find sinθ. 

Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle. 

If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.

Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`

In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.

ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.

If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan² θ = sec² θ

∴ 1 + tan² θ = `(25/7)^square`

∴ tan² θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)

If sin θ + sin² θ = 1 show that: cos² θ + cos⁴ θ = 1

sin²θ + sin²(90 – θ) = ?

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

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` 

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

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]`

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

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