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

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`

Show that, cotθ + tanθ = cosecθ × secθ

Solution :

L.H.S. = cotθ + tanθ

= `cosθ/sinθ + sinθ/cosθ`

= `(square + square)/(sinθ xx cosθ)`

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

= `1/sinθ xx 1/square`

= cosecθ × secθ

L.H.S. = R.H.S

∴ cotθ + tanθ = cosecθ × secθ

Eliminate θ if x = r cosθ and y = r sinθ.

`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.

Find the value of sin²θ  + cos²θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB² + BC² = `square`   .....(Pythagoras theorem)

Divide both sides by AC²

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 

The radius of a circle is 7 cm. find the circumference of the circle.

In figure, ΔABC is an isosceles triangle with perimeter 44 cm. The base BC is of length 12 cm. Side AB and side AC are congruent. A circle touches the three sides as shown in the figure below. Find the length of the tangent segment from A to the circle.

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its volume \[\pi\] = 3.14) 

The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.( \[\pi = \frac{22}{7}\]) 

In the given figure,

\[\square\] PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y, and z.