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Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Concept: Trigonometric Identities (Square Relations)
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
Concept: Trigonometric Identities (Square Relations)
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Concept: Trigonometric Identities (Square Relations)
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
Concept: Trigonometric Identities (Square Relations)
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Concept: Trigonometric Identities (Square Relations)
The radii of the ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its curved surface area.
Concept: Frustum of a Cone
The radius of a circle is 10 cm. The measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (\[\pi\]= 3.14 )
Concept: Length of an Arc

In the given figure, square ABCD is inscribed in the sector A - PCQ. The radius of sector C - BXD is 20 cm. Complete the following activity to find the area of shaded region
Concept: Length of an Arc
The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.
Concept: Ratios of Areas of Two Triangles
The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
Concept: Ratios of Areas of Two Triangles
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Concept: Ratios of Areas of Two Triangles
In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`

Concept: Ratios of Areas of Two Triangles
In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.

- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
Concept: Ratios of Areas of Two Triangles
In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Concept: Property of Three Parallel Lines and Their Transversals
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.

Concept: Property of an Angle Bisector of a Triangle
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, Find BQ.

Concept: Property of Three Parallel Lines and Their Transversals
Find QP using given information in the figure.

Concept: Property of an Angle Bisector of a Triangle
In the given figure, if AB || CD || FE then find x and AE.

Concept: Property of an Angle Bisector of a Triangle
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Concept: Ratios of Areas of Two Triangles
In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.

`"A(∆ ABD)"/"A(∆ ADC)"`
Concept: Ratios of Areas of Two Triangles
