JEE Advanced Level Test: Trigonometric Ratio- 2 - JEE MCQ

# JEE Advanced Level Test: Trigonometric Ratio- 2 - JEE MCQ

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## 25 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced Level Test: Trigonometric Ratio- 2

JEE Advanced Level Test: Trigonometric Ratio- 2 for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced Level Test: Trigonometric Ratio- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Trigonometric Ratio- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Trigonometric Ratio- 2 below.
JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 1

### If tan A and tan B are the roots of the quadratic equation x2 – ax + b = 0, then the value of sin2 (A + B)

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 1

We know that,
If α and β are the roots of the quadratic equation :

We have,

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 2

### If A = tan 6º tan 42º and B = cot 66º cot 78º, then

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 2

A=tan6°tan42° and B=cot66°cot78°
A/B=tan6°tan42°tan66°tan78°
Using tan(60-x)tanxtan(60+x)=tan3x for x=18° we get
tan42°tan18°tan78°=tan54°...(1)
for x=6° we get
tan54°tan6°tan66°=tan18°...(2)
Eliminating tan54°between (1) and (2)
we get
tan6°tan42°tan66°tan78°=1
Hence A/B=1.

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JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 3

### =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 3

We have : cos 290° = cos (270°+20°) = sin 20°
. . . . . . . . sin 250° = sin (270°-20°) = - cos 20°.
∴ ( 1 / cos 290° ) + (1 / √3 sin 250° )
= ( 1 / sin 20° ) + { 1 / [ √3 ( - cos 20° ) ] }
= ( √3 cos 20° - sin 20° ) / ( √3 sin 20° cos 20° )
= 2 [ (√3 /2 ) cos 20° - (1/2) sin 20° ] / [ (√3/2) ( 2 sin 40°]
= 2 ( sin 60° cos 20° - cos 60° sin 20° ) / ((√3/2) sin 40° )
= 2 sin (60° - 20°) / ((√3 /2) sin 40° )
= 4 / √3
Rationalise it, we get 4(3)½/3

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 4

If A + B + C = p & sin  = k sin C/2,then tan A/2 tan B/2 =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 4

Applying componendo and dividendo
[sin(A+C/2) + sin (C/2)]/[sin(A+C/2) - sin (C/2)] = (k+1)/(k-1)
⇒ [2sin(A+C)cos A/2]/[2cos(A+C)sin A/2] ​= (k+1)/(k−1)
⇒ [tan(A+C)/2]/(tan A/2) ​= (k+1)/(k-1)
​⇒ [tan(π−B)/2]/(tan A/2) = (k+1)/(k-1)
​⇒ 1/(tan A/2 tan B/2) = (k+1)/(k-1)
⇒ tan A/2 tan B/2 = (k-1)/(k+1)

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 5

In any triangle ABC, which is not right angled S cos A . cosec B . cosec C is equal to

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 6

If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = sin y, then the value of cos 2x is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 6

⇒3cosx + 2cos3x = cosy

Squaring

⇒ 9+4+12cos(3x−x) = 1

⇒ 12cos(2x) =−12

⇒ cos2x =−1

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 7

If cos a + cos b = a, sin a + sin b = b and a - b = 2q, then  =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 7

we have
cosα + cosβ = a
sinα + sinβ = b
α − β = 2θ
so, (cosα + cosβ)2 = a2
cos2α + cos2β + 2cosαcosβ = a2-----(i)
and (sinα + sinβ)2 = b2
sin2α + sin2β + 2sinαsinβ = b2 -----(ii)
2 + cos(α−β) = a2 + b2
[sin2α + cos2α = 1sin2β + cos2β = 1cosαcosβ + sinαsinβ = cos(α−β)
a2 + b2 = 2 + 2cos(α−β) = 2 + 2cos2θ
(cosθ)/(cos3θ) = (4cos3θ.3cosθ)/cosθ
= 4cos2θ − 3 → 4cos2θ=3------(iii)
a2 + b2 = 2 + 2(2cos2θ−1)
= 2 + 4cos2θ − 2
∴ 4cos2θ = a2 + b2
from (iii)
∴ 3 = a2 + b2
a2 + b2 − 3 = 0

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 8

If A + B + C = p & cos A = cos B . cos C then tan B . tanC has the value equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 8

Given cosA = cosB * cosC  ------- (1)

Apply cos on both sides, we get

We know that Cos(A + B) = cosAcosB - sinA sinB and cos(180 - A) = -cosA.

= > cosB cosC - sinB sinC = -cosA

= > cosB cosC - sinB sinC = -cosB cosC(From (1))

= > cosB cosC - sinB sinC + cosB cosC = 0

= > 2cosB cosC = sinB sinC

= > 2 = sinB sinC/cosB cosC.
Now,
We know that TanA = sinA/cosA.
Therefore TanB TanC = sinA sinB/cosB cosC
2.

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 9

The value of tan + 2 tan π/8 + 4 is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 9

Solution :- By using the identity : tan A = cot A - 2cot 2A

= { cot π/16 - 2 cot 2(π/16) } + 2{ cot π/8 - 2cot 2(π/8)} + 4

= cot π/16 - 2cot π/8 + 2cot π/8 - 4cot π/4 + 4

= cot π/16 - 4(1) +4

= cot π/16

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 10

The value of cos π/19 +cos+cos+...+ cos 17π/19 is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 10

cos (π/19) + cos (3π/19) + cos (5π/19) +… + cos (17π/19)

Multuiply and divide with 2sin(π/19)

=1/2sin(π/19) [2sin(π/19) cos (π/19) +2sin(π/19) cos (3π/19) + 2sin(π/19) cos (5π/19) +… +2sin(π/19) cos (17π/19)]

We know that

sin(2A) = 2sinAcosA
2sinAcosB = sin(A+B) + sin(A-B)

=1/2sin(π/19)[sin(2π/19) +sin(4π/19) +sin(-2π/19) +sin(6π/19) +sin(-4π/19)+ …+sin(18π/19) +sin(-16π/19)]

=1/2sin(π/19)[sin(18π/19)]

=1/2sin(π/19)[sin(π-π/19)]

=1/2sin(π/19)[sin(π/19)]

=1/2

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 11

Consider a right angle triangle PQR, if PQ is 27 and QR is 17 then the value of angle P is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 11

Explanation : Since ∆PQR is right angled at Q, PR is it's hypotenuse.

QR is the opposite side of angle P and PQ is the adjacent side for the angle P.

So QR/PQ = opp/adj. = tan P

=> 17/27 = tan P

=> 0.62 = tan P

=> angle P = 32.0°

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 12

If f(q) = sin4 q + cos2 q, then range of f(q) is

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 13

If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 13

Cosx = (1-tan2x/2)/(1+tan2x/2)

Sin x = (2tanx/2)/(1+tan2x/2)

2cosx + sinx=1

2(((1-tan2x/2)/(1+tan2x/2)) + ((2tanx/2)/(1+tan2x/2))=1

Solving this we get quadratic in tanx/2

3tan2x/2 – 2tanx/2 -1=0

(3tanx/2+1)(tanx/2 -1)=0

Tanx/2 =1 or -1/3

7cosx + 6sinx

7((1-tan2x/2)/(1+tan2x/2)) + 6((2tanx/2)/(1+tan2x/2))

Putting value of Tanx/2, we get

When

Tanx/2 =1 Ans=6

When

Tanx/2=-1/3 Ans=2

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 14

If cosec A + cot A = 11/2, then tan A is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 14

cosecA +cotA =11/2
or 1/sinA + cosA/sinA = 11/2
or 1+ cosA/sinA = 11/2
2 cos^2A/ 2 / 2sinA/ cosA/2 = 11/2
cotA/2 =11/2
so tanA/2 = 2/11
so tanA = 2 tanA/2/( 1- tan^2A/2
={ 2* 2/11}/( 1- 4/121)
= (4/11)/ (117/121)
=4/11* 121/117
=44/117

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 15

If 0° < x < 90° & cos x = 3/√10, then the value of log10 sin x + log10 cos x + log10 tan x is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 15

0 < x < 900
cos x = 3/√10

sin x =  1/√10

log sin x + log cos x + log tan x  [all logs with base 10]
= log sinx cos x tanx
= log sin2 x = 2 log sin x
= 2 log (10)-1/2

= 2X-1/2 log10
= -1 log1010

= –1

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 16

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

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 17

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

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 18

If , 0 < A, B < π/2, then tan A + tan B is equal to

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 19

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

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 19

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 20

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

JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 21

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

*Multiple options can be correct
JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 22

The value of  =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 22

(sin x + cos x)/cos³ x
= (sin x/cos³ x + cos x/cos³ x)
= tan x sec² x + sec² x
= sec² x (tan x + 1)
= (tan² x + 1)(tan x + 1)
= tan³ x + tan² x + tan x + 1

*Multiple options can be correct
JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 23

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

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 23

(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

*Multiple options can be correct
JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 24

The value of  is

*Multiple options can be correct
JEE Advanced Level Test: Trigonometric Ratio- 2 - Question 25

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

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