If 6 tan A - 5 = 0 , find the value of 3 sin A - cos A 5 cos A + 9 sin A

Given: 6 tan A - 5 = 0 ⇒ 6 tan A = 5 ⇒ tan A = 5 6 Given, 3 sin A - cos A 5 cos A + 9 sin A Divide both numerator and the denominator by cos A . ⇒ 3 sin A - cos A cos A 5 cos A + 9 sin A cos A ⇒ 3 sin A cos A - cos A cos A 5 cos A cos A + 9 sin A cos A ⇒ 3 tan A - 1 5 + 9 tan A Substitute the value of tan A in the above equation: ⇒ 3 5 6 - 1 5 + 9 5 6 ⇒ 15 - 6 6 30 + 45 6 ⇒ 9 75 ⇒ 3 25 Hence, the value of 3 sin A - cos A 5 cos A + 9 sin A is 3 25 .

If cot θ = x y , find the value of x cos θ + y sin θ y sin θ - x cos θ

Given: cot θ = x y Given, x cos θ + y sin θ y sin θ - x cos θ Divide both numerator and the denominator by sin A . ⇒ x cos θ + y sin θ sin θ y sin θ - x cos θ sin θ ⇒ x cos θ sin θ + y sin θ sin θ y sin θ sin θ - x cos θ sin θ ⇒ x cot θ + y y - x cot θ Substitute the value of cot θ in the above equation: ⇒ x × x y + y y - x × x y ⇒ x 2 + y 2 y x 2 - y 2 y = x 2 + y 2 x 2 - y 2 Hence, the value of x cos θ + y sin θ y sin θ - x cos θ is x 2 + y 2 x 2 - y 2 .

In △ ABC , ∠ B = 90 ° and sin A = 8 11 , show that sec 2 A - tan 2 A = 1 .

Given: sin A = 8 11 We know that, sin A = opposite side H y p o t e n u s e = B C A C = 8 11 In △ ABC , by Pythagoras theorem: ⇒ AC 2 = AB 2 + BC 2 ⇒ 11 2 = A B 2 + 8 2 ⇒ A B 2 = 121 - 64 = 57 A B = 57 From the above triangle, tan A = Opposite side Adjacent side = BC A B = 8 57 Sec A = Hypotenuse Adjacent side = AC AB = 11 57 Consider L.H.S: sec 2 A - tan 2 A = 11 57 2 - 8 57 2 = 121 57 - 64 57 = 121 - 64 57 = 57 57 = 1 Hence, proved.

In △ ABC , ∠ B = 90 ° and sin A = 8 11 , show that cot 2 A - cosec 2 A + 1 = 0 .

Given: sin A = 8 11 We know that, sin A = opposite side H y p o t e n u s e = B C A C = 8 11 In △ ABC , by Pythagoras theorem: ⇒ AC 2 = AB 2 + BC 2 ⇒ 11 2 = A B 2 + 8 2 ⇒ A B 2 = 121 - 64 = 57 A B = 57 From the above triangle, cot A = Adjacent side Opposite side = AB BC = 57 8 cosec A = Hypotenuse Opposite side = AC BC = 11 8 Consider L.H.S: cot 2 A - cosec 2 A + 1 = 57 8 2 - 11 8 2 + 1 = 57 64 - 121 64 + 1 = 57 - 121 + 64 64 = 0 64 = 0 Hence, proved.

E M B I B E

Concise Mathematics I.C.S.E - Class IX>Chapter 30 - Chapterwise Revision-(Stage 2)>CHAPTERWISE REVISION>Q 86

HARD

9th ICSE

IMPORTANT

Earn 100

Equal chords of a circle subtend equal angles at the centre, prove it.

Important Questions on Chapterwise Revision-(Stage 2)

HARD

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 30 - Chapterwise Revision-(Stage 2)>CHAPTERWISE REVISION>Q 88

Two circles of radii

10 cm

and

17 cm

intersecting each other at two points and the distance between their centres is

21 cm

, Find the length of the common chord.

MEDIUM

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 30 - Chapterwise Revision-(Stage 2)>CHAPTERWISE REVISION>Q 109

A rectangular container, whose base is a square of side $12 cm$ , contains sufficient water to submerge a rectangular solid $8 cm \times 6 cm \times 3 cm$ . Find the rise in level of the water in the container when the solid is completely immersed in it.