Trignometric Ratios, Functions and Equations Test - 8

We know that,
If  αand 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
cos α+ cos β= a
sin α+ sin β= b
α−β= 2 θ
so, (cos α+ cos β)² = a²
cos² α+ cos² β+ 2cos αcos β= a² -----(i)
and (sin α+ sin β)² = b²
sin² α+ sin² β+ 2sin αsin β= b² -----(ii)
adding (i) and (ii)
2 + cos(α−β) = a² + b²
[sin² α+ cos² α= 1sin² β+ cos² β= 1cos αcos β+ sin αsin β= cos(α−β)
a² + b² = 2 + 2cos(α−β) = 2 + 2cos² θ
(cos θ)/(cos3 θ) = (4cos3 θ.3cos θ)/cos θ
= 4cos² θ−3 →4cos2 θ=3------(iii)
a² + b² = 2 + 2(2cos² θ−1)
= 2 + 4cos² θ−2
∴4cos² θ= 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 (3 π/19) + cos (5 π/19) +…+ cos (17 π/19)

Multuiply and divide with 2sin(π/19)

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

We know that

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

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

=1/2sin(π/19)[sin(18 π/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-tan² x/2)/(1+tan² x/2)

Sin x = (2tanx/2)/(1+tan² x/2)

2cosx + sinx=1

2(((1-tan² x/2)/(1+tan² x/2)) + ((2tanx/2)/(1+tan² x/2))=1

Solving this we get quadratic in tanx/2

3tan² x/2 –2tanx/2 -1=0

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

Tanx/2 =1 or -1/3

7cosx + 6sinx

7((1-tan² x/2)/(1+tan² x/2)) + 6((2tanx/2)/(1+tan² x/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 0
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 log₁₀ 10

= –1