Trignometric Ratios, Functions and Equations Test

Result

Trignometric Ratios, Functions and Equations Test - 9

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Weekly Quiz Competition

Question 1
2 / -0.83

If cot a + tan a = m and 1/cos α – cos a = n, then

A
m (mn² )^1/3 – n(nm² )^1/3 = 1

B
m(m² n)^1/3 – n(nm² )^1/3 = 1

C
n (mn² )^1/3 – m(nm² )^1/3 = 1

D
n(m² n)^1/3 – m(mn² )^1/3 = 1
Question 2
2 / -0.83

If 2 sec² a – sec⁴ a – 2 cosec² a + cosec⁴ a = 15/4, then tan a is equal to

A
1/√2

B
1/2

C
1/2 √2

D
1/4
Question 3
2 / -0.83

If , 0

A

B

C
1

D
Question 4
2 / -0.83

If 3 sin x + 4 cos x = 5 then 4 sin x – 3 cos x is equal to

A
0

B
1

C
5

D
None of these

Solution
Question 5
2 / -0.83

If sin 2q = k, then the value of is equal to

A

B

C
k² + 1

D
2 – k²
Question 6
2 / -0.83

If f(q) = sin² q + sin² + sin² , then f is equal to

A
2/3

B
3/2

C
1/3

D
1/2
Question 7
2 / -0.83

The value of =

A
1+tanx + tan² x –tan³ x

B
1+tan x+tan² x+tan³ x

C
1 –tanx + tan² x +tan³ x

D
(1 + tan x) sec² x

Solution

(sin x + cos x)/cos ³x
= (sin x/cos ³x + cos x/cos ³x)
= tan x sec ²x + sec ²x
= sec ²x (tan x + 1)
= (tan ²x + 1)(tan x + 1)
= tan ³x + tan ²x + tan x + 1
Question 8
2 / -0.83

If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C) then each side is equal to

A
1

B
–1

C
Both A &B

D
none of these

Solution

(secA + tanA)(secB + tanB)(secC + tan C)

=>(secA - tanA)(secB - tanB)(secC - tanC)

{ Mulitply both sides with }
(secA + tanA)(secB + tanB)(secC + tan C)",

we get,

(secA + tanA)2(secB + tanB)2(secC + tan C)2

=>(sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C)

= (1)(1)(1) = 1

=>[(secA + tanA)(secB + tanB)(secC + tanC)]2=1

(secA + tanA)(secB + tanB)(secC + tan C) = ±1

Similarly, we get

(secA –tanA)(secB –tanB)(secC –tan C) = ±1
Question 9
2 / -0.83

The value of is

A
–tan 304 º

B
tan 56 º

C
cot 214 º

D
cot 34 º
Question 10
2 / -0.83

If tan² q = 2 tan² f + 1, then the value of cos 2q + sin² f is

A
1

B
2

C
–1

D
Independent of f