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

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

We have,

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.

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

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)

⇒3cosx + 2cos3x = cosy

Squaring


 

⇒ 9+4+12cos(3x−x) = 1

⇒ 12cos(2x) =−12

⇒ cos2x =−1

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)
adding (i) and (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

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.

ANSWER :- b

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

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

Correct Answer : a

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°

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

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

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

(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

(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