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

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.  

Choose the correct alternative answer for the following question.

The curved surface area of a cylinder is 440 cm 2 and its radius is 5 cm. Find its height.

Choose the correct alternative answer for the following question.

Choose the correct alternative answer for the following question.

The area of a sector of a circle of 6 cm radius is 15 π sq. cm. Find the measure of the arc and the length of the arc corresponding to the sector.

In the given figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion. ( \[\pi\] = 3.14, \[\sqrt{3}\]= 1.73 )

In Δ ABC, if ∠ A = 65° ; ∠ B = 40° then find the measure of ∠ C.