SSC (Marathi Semi-English) 10th Standard Board Exam [इयत्ता १० वी] - Maharashtra State Board Important Questions for Geometry Mathematics 2

Prove that `1/("cosec"  theta - cot theta)` = 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°.

Find the value of sin 45° + cos 45° + tan 45°.

Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

What will be the value of sin 45° + `1/sqrt(2)`?

In the given figure, if sin θ = `7/13`, which angle will be θ?

Prove that: cot θ + tan θ = cosec θ·sec θ

Proof: L.H.S. = cot θ + tan θ

= `square/square + square/square`  ......`[∵ cot θ = square/square, tan θ = square/square]`

= `(square + square)/(square xx square)`  .....`[∵ square + square = 1]`

= `1/(square xx square)`

= `1/square xx 1/square`

= cosec θ·sec θ  ......`[∵ "cosec"  θ = 1/square, sec θ = 1/square]`

= R.H.S.

∴ L.H.S. = R.H.S.

∴ cot θ + tan θ = cosec·sec θ

If sec θ = `1/2`, what will be the value of cos θ?

Find will be the value of cos 90° + sin 90°.

If cot θ = `40/9`, find the values of cosec θ and sinθ,

We have, 1 + cot²θ = cosec²θ

1 + `square` = cosec²θ

1 + `square` = cosec²θ

`(square + square)/square` = cosec²θ

`square/square` = cosec²θ  ......[Taking root on the both side]

cosec θ = `41/9`

and sin θ = `1/("cosec"  θ)`

sin θ = `1/square`

∴ sin θ =  `9/41`

The value is cosec θ = `41/9`, and sin θ = `9/41`