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Can you explain the answer of this question below:
- A:
4
- B:
1/4
- C:
2
- D:
none of these.
The answer is a.
4
1/4
2
none of these.
|
Divey Sethi answered |
sin θ is 3/5.
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
Can you explain the answer of this question below:Value of 
is
- A:
π/2
- B:
1/x
- C:
x
- D:
π
The answer is a.
Value of is
π/2
1/x
x
π
|
Case Studies answered |
Tan-1(1/x) = cot-1(x) and tan-1(x) + cot-1(x) =90degree
Evaluate sin(3 sin–10.4)a)0.56b)0.31c)0.64d)0.9Correct answer is 'D'. Can you explain this answer?
|
Subhankar Choudhary answered |
3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so
Definitely 0.4 comes in this range of x and so
3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]
Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944
Value of 
- a)
- b)
- c)-1
- d)
Correct answer is option 'A'. Can you explain this answer?
Value of
a)
b)
c)
-1
d)
|
Aravind Rane answered |
Cot-1[sin(-90)] =cot-1(-1) =π-cot-1(1) =π-π/4 =3π/4
If sin-1 x + sin-1 y + sin-1z = 
, then - a)
- b)x22 + y42 + z62 - x220- y420 - z620 = 0
- c)x50 + y25 + z5 = 0
- d)
Correct answer is option 'A,B'. Can you explain this answer?
If sin-1 x + sin-1 y + sin-1z = , then
, then a)
b)
x22 + y42 + z62 - x220- y420 - z620 = 0
c)
x50 + y25 + z5 = 0
d)
|
Vikash Yadav answered |
Put x = y = z = 1
This will satisfy the eqns
The maximum value of sin x + cos x is- a)1
- b)2
- c)√2
- d)
Correct answer is option 'C'. Can you explain this answer?
The maximum value of sin x + cos x is
a)
1
b)
2
c)
√2
d)
|
Shreya Gupta answered |
sinx + cosx=sinx + sin(90-x)=2sin{(x+90-x)/2}cos{(x-90+x)/2}using the formula
then x + y + z is equal to 
If xy + yz + zx = 1, then, tan–1x + tan–1y + tan–1z =- a)π
- b)π/2
- c)1
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
If xy + yz + zx = 1, then, tan–1x + tan–1y + tan–1z =
a)
π
b)
π/2
c)
1
d)
None of these
|
Aryan Khanna answered |
If xy + yz + zx = 1,
Then, 1 – xy – yz – zx = 0 ----- eqn1
tan-1x + tan-1y = tan-1((x+y)/(1-xy))
tan-1x + tan-1y + tan-1z = tan-1((x+y)/(1-xy)) + tan-1z
= tan-1( ( ((x+y)/(1-xy)) + z ) / (1 – ((x+y)*z)/(1+xy)))
= tan-1((x+y+z-xyz)/(1-xy-yz-zx))
= tan-1((x+y+z-xyz)/0) [From eqn1]
= π/2
Then, 1 – xy – yz – zx = 0 ----- eqn1
tan-1x + tan-1y = tan-1((x+y)/(1-xy))
tan-1x + tan-1y + tan-1z = tan-1((x+y)/(1-xy)) + tan-1z
= tan-1( ( ((x+y)/(1-xy)) + z ) / (1 – ((x+y)*z)/(1+xy)))
= tan-1((x+y+z-xyz)/(1-xy-yz-zx))
= tan-1((x+y+z-xyz)/0) [From eqn1]
= π/2
The simplest form of 
for x > 0 is …- a)x
- b)-x/2
- c)2x
- d)x/2
Correct answer is option 'D'. Can you explain this answer?
The simplest form of for x > 0 is …
for x > 0 is …a)
x
b)
-x/2
c)
2x
d)
x/2
|
Mira Joshi answered |
tan-1(1-cosx/1+cosx)½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}
= tan-1(tan x/2)
= x/2
= tan-1{(2sin2 x/2) / (2cos2 x/2)}½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}
= tan-1(tan x/2)
= x/2
Evaluate :cos (tan–1 x)- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Evaluate :cos (tan–1 x)
a)
b)
c)
d)
|
Sushil Kumar answered |
tan−1 x = θ , so that x=tanθ . We need to determine cosθ .
sec2θ = 1 + tan2θ = 1 + x2
∴s ecθ = ±√(1+x2)
Then, cos(tan−1x) = cosθ=1/secθ = ±1/√(1+x2)
sec2θ = 1 + tan2θ = 1 + x2
∴s ecθ = ±√(1+x2)
Then, cos(tan−1x) = cosθ=1/secθ = ±1/√(1+x2)
is given by
then the values of x is
- a)4
- b)1/4
- c)2
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
a)
4
b)
1/4
c)
2
d)
none of these.
|
Krishna Iyer answered |
sin θ is 3/5.
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
Can you explain the answer of this question below:Domain of f(x) = cos–1 x + cot–1 x + cosec–1 x is
- A:
[–1, 1]
- B:
R
- C:
- D:
{–1, 1}
The answer is D.
Domain of f(x) = cos–1 x + cot–1 x + cosec–1 x is
[–1, 1]
R
{–1, 1}
|
Adarsh Kumar Maurya answered |
Writing domain of all three function separately and then take intersection of all three ,then u will get ur answers
If x< 0 then value of tan-1(x) + tan-1 
is equal to - a)π/2
- b)–π/2
- c)0
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If x< 0 then value of tan-1(x) + tan-1 is equal to
is equal to a)
π/2
b)
–π/2
c)
0
d)
None of these
|
Lavanya Menon answered |
Let y = tan-1 x
x = tan y
⇒ 1/x = 1/tan y
⇒ 1/x = cot y
⇒ 1/x = tan(π/2 - y)
⇒ π/2 - y = tan-1 (1/x)
As we know that y = tan-1 x
π/2 = tan-1(x) + tan-1(1/x)
x = tan y
⇒ 1/x = 1/tan y
⇒ 1/x = cot y
⇒ 1/x = tan(π/2 - y)
⇒ π/2 - y = tan-1 (1/x)
As we know that y = tan-1 x
π/2 = tan-1(x) + tan-1(1/x)
The value ofcos150−sin150 is- a)
- b)
- c)0
- d)
Correct answer is option 'D'. Can you explain this answer?
The value ofcos150−sin150 is
a)
b)
c)
0
d)
|
Poojan Angiras answered |
Heyy!!! write cos 15 and sin 15 as cos(60-45) and sin(60-45) respectively.And then apply formula of cos(a+b) and sin(a+b).Proceed as the question u will get correct answer :-):-)^_^
If ab + bc + ca = 0, then find 1/a2-bc + 1/b2 – ca + 1/c2- ab- a)π
- b)0
- c)-1
- d)2π
Correct answer is option 'B'. Can you explain this answer?
If ab + bc + ca = 0, then find 1/a2-bc + 1/b2 – ca + 1/c2- ab
a)
π
b)
0
c)
-1
d)
2π
|
Preeti Iyer answered |
Given ab+bc+ca=0 and asked to find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc)Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab +
then tanθ is equal to -
The number of solutions of the equation sin-1 x - cos-1 x = sin-1(1/2) is
a) 3
b) 1
c) 2
d) infinite.
Correct answer is option 'B'. Can you explain this answer?
|
Raghavendra Ghoshal answered |
Solution:
Given equation is sin^-1(x) - cos^-1(x) = sin^-1(1/2)
We know that sin(x) + cos(x) = √2 cos(x - π/4)
So, sin^-1(x) - cos^-1(x) = π/2 - sin^-1(√2x)
Therefore, the given equation becomes π/2 - sin^-1(√2x) = sin^-1(1/2)
sin(sin^-1(x)) = xsin(π/2 - sin^-1(√2x)) = √[1 - 2x^2]
√[1 - 2x^2] = 1/2
2x^2 = 3/4
x = ±√3/2
Therefore, the given equation has only 1 solution, which is x = √3/2 or x = -√3/2.
Hence, the correct answer is option B.
Given equation is sin^-1(x) - cos^-1(x) = sin^-1(1/2)
We know that sin(x) + cos(x) = √2 cos(x - π/4)
So, sin^-1(x) - cos^-1(x) = π/2 - sin^-1(√2x)
Therefore, the given equation becomes π/2 - sin^-1(√2x) = sin^-1(1/2)
sin(sin^-1(x)) = xsin(π/2 - sin^-1(√2x)) = √[1 - 2x^2]
√[1 - 2x^2] = 1/2
2x^2 = 3/4
x = ±√3/2
Therefore, the given equation has only 1 solution, which is x = √3/2 or x = -√3/2.
Hence, the correct answer is option B.
Range of f(x) = sin–1 x + tan–1 x + sec–1 x is- a)
- b)
- c)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Range of f(x) = sin–1 x + tan–1 x + sec–1 x is
a)
b)
c)
d)
None of these
|
Tanuja Kapoor answered |
The function f is defined for x = ±1
Now, f(1) = sin-1(1) + sec-1(1) + tan-1(1)
= π/2 + 0 + π/4 = 3π/4
Also, f(-1) = sin-1(-1) + sec-1(-1) + tan-1(-1)
= - π/2 + π - 0 - π/4
= 3π/4
Range = { π/4, 3π/4}
Now, f(1) = sin-1(1) + sec-1(1) + tan-1(1)
= π/2 + 0 + π/4 = 3π/4
Also, f(-1) = sin-1(-1) + sec-1(-1) + tan-1(-1)
= - π/2 + π - 0 - π/4
= 3π/4
Range = { π/4, 3π/4}
The complete solution set of the inequality [cot–1 x]2 – 6[cot–1 x] + 9 ≤0 is (where [ * ] denotes the greatest integer function)- a)(– ∞, cot 3]
- b) [cot 3, cot 2]
- c) [cot 3, ∞)
- d)None of these
Correct answer is 'A'. Can you explain this answer?
The complete solution set of the inequality [cot–1 x]2 – 6[cot–1 x] + 9 ≤0 is (where [ * ] denotes the greatest integer function)
a)
(– ∞, cot 3]
b)
[cot 3, cot 2]
c)
[cot 3, ∞)
d)
None of these
|
Janani Roy answered |
Since $\cot x>0$ for $0< /><\pi $,="" the="" given="" inequality="" is="" equivalent="" to=""><\frac{\pi}{4}$, or=""><\frac{\pi}{2}$. thus,="" the="" solution="" set="" is="">< /><\frac{\pi}{2}.$$>
- a) tan-1 (9/5√10)
- b)cos-1(9/5√10)
- c)tan-1(9/14)
- d)sin-1(9/5√10)
Correct answer is option 'D'. Can you explain this answer?
a)
tan-1 (9/5√10)
b)
cos-1(9/5√10)
c)
tan-1(9/14)
d)
sin-1(9/5√10)
|
|
Aisha Gupta answered |
Given α = cos-1(⅗)
cos α = 3/5
sin α = ⅘
tan α = 4/3
Also β = tan-1(⅓)
So tan β = 1/3
tan (α-β) = (tan α – tan β)/(1 + tan α tan β)
= (4/3 – ⅓)/(1 + 4/9)
= 1/(13/9)
= 9/13
So (α-β) = tan-1 (9/13)
= sin-1(9/5√10)
Hence option d is the answer.
If sin A + cos A = 1, then sin 2A is equal to- a)2
- b)1/2
- c)0
- d)1
Correct answer is option 'C'. Can you explain this answer?
If sin A + cos A = 1, then sin 2A is equal to
a)
2
b)
1/2
c)
0
d)
1
|
|
Suresh Iyer answered |
(Sin A + Cos A)2 = sin2A + cos2A + 2 sinAcosA
1 = 1 + Sin 2A
so, Sin 2A = 0
Hence A = 0
The value of sin–1 (sin 12) + cos–1 (cos 12) is equal to- a)Zero
- b)24 – 2π
- c) 4π – 24
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The value of sin–1 (sin 12) + cos–1 (cos 12) is equal to
a)
Zero
b)
24 – 2π
c)
4π – 24
d)
None of these
|
Nandini Iyer answered |
sin-1(sin12)+cos-1(cos12)
Here, we have to convert sin12 into a value from -π/2 to π/2
and cos12 into a value from 0 to π.
We can write,
sin-1(sin12)+cos-1(cos12)=sin-1(sin(12-4π))+cos-1(cos(4π-12))
=12-4π+4π-12=0
∴sin-1(sin12)+cos-1(cos12)=0
Here, we have to convert sin12 into a value from -π/2 to π/2
and cos12 into a value from 0 to π.
We can write,
sin-1(sin12)+cos-1(cos12)=sin-1(sin(12-4π))+cos-1(cos(4π-12))
=12-4π+4π-12=0
∴sin-1(sin12)+cos-1(cos12)=0
Evaluate: sin (2 sin–10.6)- a)0.6
- b)0.66
- c)0.36
- d)0.96
Correct answer is option 'D'. Can you explain this answer?
Evaluate: sin (2 sin–10.6)
a)
0.6
b)
0.66
c)
0.36
d)
0.96
|
|
Nikita Singh answered |
Let, sin-1(0.6) = A…………(1)
( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )
( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )When x = π/2, then tan x, is- a)∞ or −∞
- b)−∞−∞
- c)∞
- d)not defined.
Correct answer is option 'D'. Can you explain this answer?
When x = π/2, then tan x, is
a)
∞ or −∞
b)
−∞−∞
c)
∞
d)
not defined.
|
Poulomi Datta answered |
tan π/2 = n.d.i.e. not defined.
cos2θ is not equal to- a)1−2sin2θ
- b)2cos2θ−1
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
cos2θ is not equal to
a)
1−2sin2θ
b)
2cos2θ−1
c)
d)
|
Priyanka Sharma answered |
cos 2θ is equal to cos2θ - sin2θ = 2cos2θ - 1
- a)±√3.
- b)0
- c)
- d)1
Correct answer is option 'D'. Can you explain this answer?
a)
±√3.
b)
0
c)
d)
1
|
Devendra Singh answered |
Apply apply formula of Sin inverse X + Cos inverse X and the question will be solved
- a)7/25
- b)-24/25
- c)24/25
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
a)
7/25
b)
-24/25
c)
24/25
d)
none of these.
|
Om Desai answered |
Put 
therefore the given expressionis sin2θ = 2sinθcosθ
therefore the given expressionis sin2θ = 2sinθcosθIf ƒ(x) = √(2tan(x)),then f-1(√(2)) =- a)0
- b)1
- c)√(2)
- d)2
Correct answer is option 'B'. Can you explain this answer?
If ƒ(x) = √(2tan(x)),then f-1(√(2)) =
a)
0
b)
1
c)
√(2)
d)
2
|
|
Siddharth Chakraborty answered |
If what? Please provide more information for me to assist you better.
If ƒ(x) = tan(x), then f-1(1/√(3)) =- a)π/6
- b)π/4
- c)π/3
- d)π
Correct answer is option 'A'. Can you explain this answer?
If ƒ(x) = tan(x), then f-1(1/√(3)) =
a)
π/6
b)
π/4
c)
π/3
d)
π
|
Riya Banerjee answered |
f(x) = tanx
Let y = tanx
⇒x=tan−1 y
⇒f − 1(x)= tan−1 implies (−1)(1/√(3)
=tan−1(1/√(3)
⇒ =tan−1tan π/6
=π/6 [∵ tanπ/6 = (1/√(3)]
Let y = tanx
⇒x=tan−1 y
⇒f − 1(x)= tan−1 implies (−1)(1/√(3)
=tan−1(1/√(3)
⇒ =tan−1tan π/6
=π/6 [∵ tanπ/6 = (1/√(3)]
- a)
- b)1
- c)0
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
a)
b)
1
c)
0
d)
none of these.
|
|
Nikita Singh answered |
sin-1√3/5 = A
Sin A = √3/5 , cos A = √22/5
Therefore Cos-1√3/5 = B
Cos B = √3/5 , sin B = √22/5
sin(A+B) = sinA cosB + cosA sinB
= √3/5 * √3/5 + √22/5 * √22/5
= 3/25 * 22/25
= 25/25
= 1
Sin A = √3/5 , cos A = √22/5
Therefore Cos-1√3/5 = B
Cos B = √3/5 , sin B = √22/5
sin(A+B) = sinA cosB + cosA sinB
= √3/5 * √3/5 + √22/5 * √22/5
= 3/25 * 22/25
= 25/25
= 1
If and x + y + z = xyz, then a value of tan−1x+tan−1y+tan−1z is- a)3π/2
- b)π/2
- c)π
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
If and x + y + z = xyz, then a value of tan−1x+tan−1y+tan−1z is
a)
3π/2
b)
π/2
c)
π
d)
none of these.
|
|
Ruchi Kumar answered |
This question is incomplete and cannot be answered. Please provide more information.
The value of sin-1[cos{cos-1 (cos x) + sin-1(sin x)}], where x 
- a)π/2
- b)π/4
- c)–π/4
- d)–π/2
Correct answer is option 'D'. Can you explain this answer?
The value of sin-1[cos{cos-1 (cos x) + sin-1(sin x)}], where x
a)
π/2
b)
π/4
c)
–π/4
d)
–π/2
|
|
Riya Banerjee answered |
x implies(pi/2, pi)
= sin-1(cos(cos-1(cosx) + sin-1(sinx)))
= sin-1(cos(x + π - x)
= sin-1(cos π)
= sin-1(-1)
= -π/2
= sin-1(cos(cos-1(cosx) + sin-1(sinx)))
= sin-1(cos(x + π - x)
= sin-1(cos π)
= sin-1(-1)
= -π/2
If the numerical value of tan (cos–1 (4/5) + tan–1 (2/3)) is a/b then- a)a + b = 23
- b)a – b = 11
- c)3b = a + 1
- d)2a = 3b
Correct answer is option 'A,B,C'. Can you explain this answer?
If the numerical value of tan (cos–1 (4/5) + tan–1 (2/3)) is a/b then
a)
a + b = 23
b)
a – b = 11
c)
3b = a + 1
d)
2a = 3b
|
Geetika Shah answered |
Therefore a + b = (17 + 6) = 23
a - b = (17-6) = 11
3b = a + 1
=> 3(6) = 18,
a + 1 => 17 + 1 = 18
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