SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

Prove that sin⁶A + cos⁶A = 1 – 3sin²A . cos²A.

Prove that 2(sin⁶A + cos⁶A) – 3(sin⁴A + cos⁴A) + 1 = 0.

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

If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3.

If cos A + cos²A = 1, then sin²A + sin⁴A = ?

If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.

Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`.

If `sin θ + cos θ = sqrt(3)`, then show that tan θ + cot θ = 1.

If tan θ – sin²θ = cos²θ, then show that `sin^2θ = 1/2`.

Prove that (1 – cos²A) . sec²B + tan²B (1 – sin²A) = sin²A + tan²B.

In ΔABC, ∠ABC = 90°, ∠BAC = ∠BCA = 45°. If AC = `9sqrt(2)`, then find the value of AB.

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"`

From the information given in the figure, determine whether MP is the bisector of ∠KMN.

If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?

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`

Draw a line segment AB of length 10 cm and divide it internally in the ratio of 2:5 Justify the division of line segment AB.

A pizza has 8 slices all equally spaced. Suppose pizza is a flat circle of radius 28 cm, find the area covered between 3 slices of pizza.

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