SSC (Marathi Semi-English) इयत्ता १० वी - Maharashtra State Board Important Questions

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 cos (45° + x) = sin 30°, then x = ?

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`.

The value of 2tan45° – 2sin30° is ______.

Complete the following activity to prove:

cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= `cosθ/sinθ + square/cosθ`

= `(square + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ `square`

= `1/sinθ xx 1/cosθ`

= `square xx secθ`

∴ L.H.S. = R.H.S.

If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.

Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

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

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`

Find the value of sin 0° + cos 0° + tan 0° + sec 0°.