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SinA = 1/√10 , SinB= 1/√5 If A and B are both acute angles,then , A+B=?- a)300
- b)750
- c)600
- d)450
Correct answer is option 'D'. Can you explain this answer?
SinA = 1/√10 , SinB= 1/√5 If A and B are both acute angles,then , A+B=?
a)
300
b)
750
c)
600
d)
450
|
Riya Banerjee answered |
We know that:
Sin θ = Opposite / Hypotenuse
Sin θ = Opposite / Hypotenuse
∴ SinA = 1/√10
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5
⇒ Multiply:
Cos(A+B)= CosA x CosB - SinA x SinB
Cos(A+B)= CosA x CosB - SinA x SinB
Substituting the value in above equation we get:
= 3/√10 x 2/√5 - 1/√10 x 1/√5
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2
we know that, sin 45 = 1/ √ 2 therefore
sinθ / cosθ = 45
sinθ / cosθ = 45
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cos 68° cos 8° + sin 68° sin 8° = ?
- a)1/2
- b)1/4
- c)1
- d)0
Correct answer is option 'A'. Can you explain this answer?
cos 68° cos 8° + sin 68° sin 8° = ?
a)
1/2
b)
1/4
c)
1
d)
0
|
Lavanya Menon answered |
We know,
cosA cosB + sinA sinB = cos(A-B)
cosA cosB + sinA sinB = cos(A-B)
cos 68° cos 8° + sin 68° sin 8° = Cos (68-8) = Cos60°
=1/2
=1/2
Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=- a)sinA
- b)sin2A
- c)cosA
- d)cos2A
Correct answer is option 'C'. Can you explain this answer?
Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=
a)
sinA
b)
sin2A
c)
cosA
d)
cos2A
|
Gaurav Kumar answered |
sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A
Can you explain the answer of this question below:sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to- A:sin(A + B)
- B:sin(A – B)
- C:cos(A – B)
- D:cos(A + B)
The answer is c.
sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to
A:
sin(A + B)
B:
sin(A – B)
C:
cos(A – B)
D:
cos(A + B)
|
|
Lavanya Menon answered |
L.H.S. = sin(60+A)cos(30−B)+cos(60+A)sin(30−B)
= sin[(60+A)+(30−B)] (Using, sin(A+B)sinAcosB+cosAsinB)
= sin(90+A−B)
= sin(90+(A−B))
= cos(A−B) (Using, sin(90+θ)=cosθ) = R.H.S.Hence Proved.
= sin[(60+A)+(30−B)] (Using, sin(A+B)sinAcosB+cosAsinB)
= sin(90+A−B)
= sin(90+(A−B))
= cos(A−B) (Using, sin(90+θ)=cosθ) = R.H.S.Hence Proved.
cos(π/4 -x) cos ( π/4 -y)-sin(π/4-x) sin( π/4 -y)=- a)cos(x-y)
- b)sin(x-y)
- c)cos(x+y)
- d)sin(x+y)
Correct answer is option 'D'. Can you explain this answer?
cos(π/4 -x) cos ( π/4 -y)-sin(π/4-x) sin( π/4 -y)=
a)
cos(x-y)
b)
sin(x-y)
c)
cos(x+y)
d)
sin(x+y)
|
|
Riya Banerjee answered |
Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)
= CosA*Cos B - Sin A*Sin B
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)
Chose which of the following expressions equals sinA + cosA.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Chose which of the following expressions equals sinA + cosA.
a)
b)
c)
d)
|
|
Lavanya Menon answered |
sinA + cosA = √2(sinA/√2+cosA/√2)
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula
if cosθ = √3/2How many solutions does this equation have between -π and π ?- a)2
- b)1
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
if cosθ = √3/2
How many solutions does this equation have between -π and π ?
a)
2
b)
1
c)
3
d)
4
|
|
Gaurav Kumar answered |
The general solution of the given question is theta= 2nπ± π/6 but it is mentioned that they are lies between -π to π. So when we put n=0 we get theta =±π/6. And when we put n= 1 we get theta does not lies between -π to ÷π. So we get only two values of theta.
If sinθ = 1/2
How many solutions does this equation have between 0 and π ?- a)2
- b)4
- c)1
- d)3
Correct answer is option 'A'. Can you explain this answer?
If sinθ = 1/2
How many solutions does this equation have between 0 and π ?
How many solutions does this equation have between 0 and π ?
a)
2
b)
4
c)
1
d)
3
|
Pooja Shah answered |
Two results are possible as π/6 & 5π/6 comes by subtracting π/6from π, In general we have Sin∆=nπ+(-1)^n(A) here A is π/6 and by putting different values of n in integer 2 values can be obtained
Which of the following is correct:
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Which of the following is correct:
a)
b)
c)
d)
|
Arnav Chaudhari answered |
All the options are correct except d) as there should be negative sign before 4sin³A
The general solution of sin = 0 is- a)nπ where n is a real number
- b)nπ, where n is an integer
- c)2π
- d)π
Correct answer is option 'B'. Can you explain this answer?
The general solution of sin = 0 is
a)
nπ where n is a real number
b)
nπ, where n is an integer
c)
2π
d)
π
|
Neha Joshi answered |
sinθ = 0
⇒ sin-1 θ = 0,π,2π,...
⇒ θ = 0 + nπ → n∈Z
The general solution for sin x = 0 will be, x = nπ, where n∈I.
⇒ sin-1 θ = 0,π,2π,...
⇒ θ = 0 + nπ → n∈Z
The general solution for sin x = 0 will be, x = nπ, where n∈I.
If cos a + 2cos b + cos c = 2 then a, b, c are in- a)2b = a+c
- b)b2 = a x c
- c)a = b = c
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If cos a + 2cos b + cos c = 2 then a, b, c are in
a)
2b = a+c
b)
b2 = a x c
c)
a = b = c
d)
none of these
|
Hansa Sharma answered |
cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ cos A + cos C = 2(1 – cos B)
2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
2 sin(A/2) sin(C/2) = sin(B/2)
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
2(s – b) = b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
The value of tan 3A – tan 2A – tan A is- a) tan 3A . tan 2A . tan A
- b)- tan 3A . tan 2A . tan A
- c)tan A . tan 2A – tan 2A . tan 3A – tan 3A . tan A
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The value of tan 3A – tan 2A – tan A is
a)
tan 3A . tan 2A . tan A
b)
- tan 3A . tan 2A . tan A
c)
tan A . tan 2A – tan 2A . tan 3A – tan 3A . tan A
d)
None of these
|
Shoubhya Sinha answered |
The correct answer is c option . first we calculate the formula of tan 3A and put the formula of tan 3Athen we have put the formulas in the equation that is given the question and then we solve it and then we get the correct option c.
In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2 is equal to:
- a)2/5
- b)100/222
- c)34/37
- d)5/2
Correct answer is option 'A'. Can you explain this answer?
In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2 is equal to:
a)
2/5
b)
100/222
c)
34/37
d)
5/2
|
Baishali Sarkar answered |
Given:
tan(A/2) = 5/6
tan(B/2) = 20/37
To find:
tan(C/2)
Solution:
Since we know the values of tan(A/2) and tan(B/2), we can use the following formula to find tan(C/2):
tan(C/2) = [ tan(A/2) + tan(B/2) ] / [ 1 - tan(A/2) * tan(B/2) ]
Substituting the given values, we get:
tan(C/2) = [ 5/6 + 20/37 ] / [ 1 - (5/6) * (20/37) ]
Simplifying the above expression, we get:
tan(C/2) = [ (185 + 120) / (6 * 37) ] / [ (37 - 20) / (6 * 37) ]
tan(C/2) = (305/222) / (17/222)
tan(C/2) = 305/17
This is not equal to option 'A', which is 2/5. Therefore, the correct answer is not given in the options.
However, we can check whether tan(C/2) = 2/5 by solving the triangle.
Using the formula for the sum of angles in a triangle, we have:
A + B + C = 180
Substituting the values of A and B, we get:
(180 - C) + 2 * atan[(5/6)/(1 + (5/6)^2)] + 2 * atan[(20/37)/(1 + (20/37)^2)] = 180
Simplifying the above expression, we get:
C - atan(25/36) - atan(40/9) = 0
Using the identity tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x) * tan(y)), we can simplify the above expression to:
C - atan[(25/36) * (9/40)] = atan(25/36) + atan(40/9)
C - atan(15/64) = atan(25/36) + atan(40/9)
Using the identity tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y)), we can further simplify the above expression to:
C = atan[(25/36 + 40/9) / (1 - (25/36) * (40/9))] + atan(15/64)
C = atan(305/222) + atan(15/64)
Now, using the identity tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y)), we can find tan(C):
tan(C) = (305/222 + 15/64) / (1 - (305/222) * (15/64))
tan(C) = 2/5
Therefore, the correct answer is option 'A', which is 2/5.
tan(A/2) = 5/6
tan(B/2) = 20/37
To find:
tan(C/2)
Solution:
Since we know the values of tan(A/2) and tan(B/2), we can use the following formula to find tan(C/2):
tan(C/2) = [ tan(A/2) + tan(B/2) ] / [ 1 - tan(A/2) * tan(B/2) ]
Substituting the given values, we get:
tan(C/2) = [ 5/6 + 20/37 ] / [ 1 - (5/6) * (20/37) ]
Simplifying the above expression, we get:
tan(C/2) = [ (185 + 120) / (6 * 37) ] / [ (37 - 20) / (6 * 37) ]
tan(C/2) = (305/222) / (17/222)
tan(C/2) = 305/17
This is not equal to option 'A', which is 2/5. Therefore, the correct answer is not given in the options.
However, we can check whether tan(C/2) = 2/5 by solving the triangle.
Using the formula for the sum of angles in a triangle, we have:
A + B + C = 180
Substituting the values of A and B, we get:
(180 - C) + 2 * atan[(5/6)/(1 + (5/6)^2)] + 2 * atan[(20/37)/(1 + (20/37)^2)] = 180
Simplifying the above expression, we get:
C - atan(25/36) - atan(40/9) = 0
Using the identity tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x) * tan(y)), we can simplify the above expression to:
C - atan[(25/36) * (9/40)] = atan(25/36) + atan(40/9)
C - atan(15/64) = atan(25/36) + atan(40/9)
Using the identity tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y)), we can further simplify the above expression to:
C = atan[(25/36 + 40/9) / (1 - (25/36) * (40/9))] + atan(15/64)
C = atan(305/222) + atan(15/64)
Now, using the identity tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y)), we can find tan(C):
tan(C) = (305/222 + 15/64) / (1 - (305/222) * (15/64))
tan(C) = 2/5
Therefore, the correct answer is option 'A', which is 2/5.
The principal solution for tan x = √3 is,- a)π , 2π
- b)π/3 , 4π/3
- c)2π/3 , π/6
- d)π/6 , 2π/6
Correct answer is option 'B'. Can you explain this answer?
The principal solution for tan x = √3 is,
a)
π , 2π
b)
π/3 , 4π/3
c)
2π/3 , π/6
d)
π/6 , 2π/6
|
Syed Naqeeb answered |
Because it lies only in third quadrant
If acos x + bsin x = c, then the value of (asin x – bcos x)² is:
- a)a² + b² + c²
- b)a² - b² - c²
- c)a² - b² + c²
- d)a² + b² - c²
Correct answer is option 'D'. Can you explain this answer?
If acos x + bsin x = c, then the value of (asin x – bcos x)² is:
a)
a² + b² + c²
b)
a² - b² - c²
c)
a² - b² + c²
d)
a² + b² - c²
|
|
Preeti Iyer answered |
(acos x + bsin x)² + (asin x – bcos x)² = a² + b²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²
If tan x = tan α , then the general solution of the equation is- a)nπ – α
- b)2nπ- α
- c)nπ + α
- d)2nπ + α
Correct answer is option 'C'. Can you explain this answer?
If tan x = tan α , then the general solution of the equation is
a)
nπ – α
b)
2nπ- α
c)
nπ + α
d)
2nπ + α
|
|
Anand Sen answered |
If tan x = tan y, then x and y are either equal or they differ by an integer multiple of pi.
In other words, if two angles have the same tangent value, then they are either equal or they are "coterminal" angles, which means they differ by a multiple of 180 degrees or pi radians.
In other words, if two angles have the same tangent value, then they are either equal or they are "coterminal" angles, which means they differ by a multiple of 180 degrees or pi radians.
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals:
- a)c/a
- b)a/c
- c)1
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals:
a)
c/a
b)
a/c
c)
1
d)
none of these
|
|
Neha Sharma answered |
cosec A (sin B cos C + cos B sin C)= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1
If 3 × tan(x – 15) = tan(x + 15), then the value of x is:
- a)30º
- b)45º
- c)60º
- d)90º
Correct answer is option 'B'. Can you explain this answer?
If 3 × tan(x – 15) = tan(x + 15), then the value of x is:
a)
30º
b)
45º
c)
60º
d)
90º
|
|
Nandini Iyer answered |
3 × tan (x – 15) = tan (x + 15)
⇒ tan(x + 15) / tan(x – 15) = 3/1
⇒ {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = (3 + 1) / (3 – 1)
⇒ {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = 2
⇒ sin(x + 15 + x – 15) / sin(x + 15 – x + 15) = 2
⇒ sin 2x / sin 30 = 2
⇒ sin 2x / (1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45
⇒ tan(x + 15) / tan(x – 15) = 3/1
⇒ {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = (3 + 1) / (3 – 1)
⇒ {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = 2
⇒ sin(x + 15 + x – 15) / sin(x + 15 – x + 15) = 2
⇒ sin 2x / sin 30 = 2
⇒ sin 2x / (1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45
tan(π + x)=- a)-tan x
- b)tan π + tan x
- c)0
- d)tan x
Correct answer is option 'D'. Can you explain this answer?
tan(π + x)=
a)
-tan x
b)
tan π + tan x
c)
0
d)
tan x
|
|
Geetika Shah answered |
(180+theta ) lies in third quadrant...where tantheta is positive....tan(180+x)=tanx
The value of Cos75°is equal to- a)√3/2
- b)
- c)1/2
- d)
Correct answer is option 'B'. Can you explain this answer?
The value of Cos75°is equal to
a)
√3/2
b)
c)
1/2
d)
|
|
Anagh Gupta answered |
Use cos(A+B) And assume. A=30• B= 45•
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