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No. of Questions 
10 
Time 
10 min 
Medium 
english 
Marks 
10 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionMark review to see the question later 
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Question 1 of 10
1. Question
1 pointsIf cosx + secx = 3; then (tanx)² – (sinx)² =9
Correct
solution:
tanx² – sinx²
= (secx² – 1) – (1 – cosx²)
= secx² + cosx² – 2
= (secx + cosx)² – 2secx cosx – 2
= 3² – 2 – 2= 9 – 4
= 5 (option ‘B’)Incorrect
solution:
tanx² – sinx²
= (secx² – 1) – (1 – cosx²)
= secx² + cosx² – 2
= (secx + cosx)² – 2secx cosx – 2
= 3² – 2 – 2= 9 – 4
= 5 (option ‘B’) 
Question 2 of 10
2. Question
1 pointsIf tanx*tan²x = 1; then sin² 2x + tan² 2x = ?
Correct
solution:
tanx*tan²x = 1
=> tanx = 1/tan²x
=> tanx = cot²x
=> tanx = tan (90 – 2x)
=> x = 90 – 2x
=> 3x = 90
=> x = 30
Put this value in (sin² x)² + (tan² x
= (sin 2*30)² + (tan 2*30)²
= (sin 60)² + (tan 60)²
= (√3/2)² + (√3)²
= 3/4 + 3
= 15/4 (option ‘C’)Incorrect
solution:
tanx*tan²x = 1
=> tanx = 1/tan²x
=> tanx = cot²x
=> tanx = tan (90 – 2x)
=> x = 90 – 2x
=> 3x = 90
=> x = 30
Put this value in (sin² x)² + (tan² x
= (sin 2*30)² + (tan 2*30)²
= (sin 60)² + (tan 60)²
= (√3/2)² + (√3)²
= 3/4 + 3
= 15/4 (option ‘C’) 
Question 3 of 10
3. Question
1 pointsIf 2 sin X + 15 cos² X = 7; then the value of tan X is?
Correct
solution:
2sin X + 15cos² X + 7
=> 2sin X + 15 (1 – sin² X) = 7
=> 2sin X = 7 – 15 + 15 sin² X
=> 15sin² X – 2sin X – 8 = 0
=> 15sin² X – 12sin X + 10sin X – 8 = 0
=> 3sin X (5sin X – 4) + 2 (5sin X – 4) = 0
=> sin X = 2/3 or sin X = 4/5
For sin X = 2/3; hypotenuse = 3 and the sides are 2 and √5. For sin X = 4/5, we can say that one side of the triangle is 4 and the hypotenuse is 5; so the other
side must be 3 as 3, 4 & 5 is a triplet.
Hence cos X = 3/5 or √5/3
Now tan X = sin X/cos X = (4/5)*(5/3) = 4/3 or (2/3)*(3/√5) = 2/√5 = 2√5/5
So tan X = 4/3 or 2√5/5 (option ‘A’)Incorrect
solution:
2sin X + 15cos² X + 7
=> 2sin X + 15 (1 – sin² X) = 7
=> 2sin X = 7 – 15 + 15 sin² X
=> 15sin² X – 2sin X – 8 = 0
=> 15sin² X – 12sin X + 10sin X – 8 = 0
=> 3sin X (5sin X – 4) + 2 (5sin X – 4) = 0
=> sin X = 2/3 or sin X = 4/5
For sin X = 2/3; hypotenuse = 3 and the sides are 2 and √5. For sin X = 4/5, we can say that one side of the triangle is 4 and the hypotenuse is 5; so the other
side must be 3 as 3, 4 & 5 is a triplet.
Hence cos X = 3/5 or √5/3
Now tan X = sin X/cos X = (4/5)*(5/3) = 4/3 or (2/3)*(3/√5) = 2/√5 = 2√5/5
So tan X = 4/3 or 2√5/5 (option ‘A’) 
Question 4 of 10
4. Question
1 pointsIf cosec X – sin X = a³ and sec X – cos X = b³; then what is the value of a²b²(a² + b²)
Correct
solution:
a³ = cosec X – sin X and b³ = sec X – cos X (given)
=> a³ = 1/sin X – sin X and b³ = 1/cos X – cos X
=> a³ = (1 – sin² X)/sin X and b³ = (1 – cos² X)/cos X
=> a³ = (cos² X/sin X) and b³ = (sin² X/cos X)
=> a = ∛(cos² X/sin X) and b = ∛(sin² X/cos X)
Now the given expression
a²b²(a² + b²)
= a b + a b
Now by putting the values of a and b in it, the expression
= [(cos² X/sin X) ][(sin² X/cos X) ] + [(cos² X/sin X) ][(sin² X/cos X) ]
= cos² X + sin² X
= 1 (option ‘B’)Incorrect
solution:
a³ = cosec X – sin X and b³ = sec X – cos X (given)
=> a³ = 1/sin X – sin X and b³ = 1/cos X – cos X
=> a³ = (1 – sin² X)/sin X and b³ = (1 – cos² X)/cos X
=> a³ = (cos² X/sin X) and b³ = (sin² X/cos X)
=> a = ∛(cos² X/sin X) and b = ∛(sin² X/cos X)
Now the given expression
a²b²(a² + b²)
= a b + a b
Now by putting the values of a and b in it, the expression
= [(cos² X/sin X) ][(sin² X/cos X) ] + [(cos² X/sin X) ][(sin² X/cos X) ]
= cos² X + sin² X
= 1 (option ‘B’) 
Question 5 of 10
5. Question
1 pointsIf (a² – b²) sin X + 2ab cos X = a² + b²; then tan X = ?
Correct
solution:
(a² – b²) sin X + 2ab cos X = a² + b²
=> a² sin X – b² sin X – 2ab cos X = a² + b²
=> a² sin X – b² sin X – 2ab cos X – a² – b² = 0
=> a² (1 – sin X) – b² (1 + sin X) – 2ab cos X = 0
=> a² (1 – sin X) + b² (1 + sin X) + 2ab cos X = 0 (By change of signs)
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab cos X = 0
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√cos² X
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√(1 – sin² X)
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√[(1 + sin X) (1 – sin X)]
=> [(a√(1 – sin X) + b√(1 + sin X)]² = 0
=> a√(1 – sin X) = b√(1 + sin X)
=> a²(1 – sin X) = b²(1 + sin X)
=> (a² + b²) sin X = a² – b²=> sin X = (a² – b²)/(a² + b²)
=> cos X = 2ab/(a² + b²) [cos X = √(1 sin² X)]
=> tan X = (a² – b²)/(2ab) (option ‘C’)Incorrect
solution:
(a² – b²) sin X + 2ab cos X = a² + b²
=> a² sin X – b² sin X – 2ab cos X = a² + b²
=> a² sin X – b² sin X – 2ab cos X – a² – b² = 0
=> a² (1 – sin X) – b² (1 + sin X) – 2ab cos X = 0
=> a² (1 – sin X) + b² (1 + sin X) + 2ab cos X = 0 (By change of signs)
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab cos X = 0
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√cos² X
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√(1 – sin² X)
=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√[(1 + sin X) (1 – sin X)]
=> [(a√(1 – sin X) + b√(1 + sin X)]² = 0
=> a√(1 – sin X) = b√(1 + sin X)
=> a²(1 – sin X) = b²(1 + sin X)
=> (a² + b²) sin X = a² – b²=> sin X = (a² – b²)/(a² + b²)
=> cos X = 2ab/(a² + b²) [cos X = √(1 sin² X)]
=> tan X = (a² – b²)/(2ab) (option ‘C’) 
Question 6 of 10
6. Question
1 pointsIf sin A + cos A = 17/13, 0 < x < 90; then sin A – cos A is ?
Correct
solution:
Squaring sin A + cos A = 17/13 on both sides
sin² A + cos² A + 2sin A cos A = 289/169
=> 2sin A cos A = 289/169 – 1 (by putting sin² A + cos² A = 1)
=> 2sin A cos A = 120/169
=> sin A cos A = 60/169
Now sin A – cos A
=√(sin A – cos A)²
= √(sin² A + cos² A – 2 sin A cos A)²
= √[1 – 2*(60/169)]² (by putting sin² A + cos² A = 1 & sin A cos A = 80/169)
= √(49/169)
= 7/13 (option ‘D’)Incorrect
solution:
Squaring sin A + cos A = 17/13 on both sides
sin² A + cos² A + 2sin A cos A = 289/169
=> 2sin A cos A = 289/169 – 1 (by putting sin² A + cos² A = 1)
=> 2sin A cos A = 120/169
=> sin A cos A = 60/169
Now sin A – cos A
=√(sin A – cos A)²
= √(sin² A + cos² A – 2 sin A cos A)²
= √[1 – 2*(60/169)]² (by putting sin² A + cos² A = 1 & sin A cos A = 80/169)
= √(49/169)
= 7/13 (option ‘D’) 
Question 7 of 10
7. Question
1 pointsIf tan² A = 2tan² B + 1; then Önd cos² A + sin² B?
Correct
solution:
Since both the terms in the “Önd” entry are in squares ………………….so the answer cannot be 1,
again it cannot be 0 also……..because than both the terms should be zero as none can be negative for both the terms to be 0.
again the answer cannot be 2 also because then both the terms in FIND entry should be 1 as it will make tan 90 which is undeÖned
so the correct answer is 1 one only (option ‘B’)Incorrect
solution:
Since both the terms in the “Önd” entry are in squares ………………….so the answer cannot be 1,
again it cannot be 0 also……..because than both the terms should be zero as none can be negative for both the terms to be 0.
again the answer cannot be 2 also because then both the terms in FIND entry should be 1 as it will make tan 90 which is undeÖned
so the correct answer is 1 one only (option ‘B’) 
Question 8 of 10
8. Question
1 pointsIf cot A + cosec A = 3 and A is an acute angle; then the value of cos A is ?
Correct
cosec A + cot A = 3 (given) — (i)
cosec² A – cot² A = 1
=> (cosec A + cot A)(cosec A – cot A) = 1
=> cosec A – cot A) =1 ⁄3 (by putting cosec A + cot A = 3) — (ii)
By adding (i) and (ii)
2cosec A = 10 ⁄ 3=> cosec A = 10 ⁄6
By Subtracting (ii) from (i)2 cot A =8 ⁄3
=> cot A = 8 ⁄6cos A =cot A ⁄ cosec A=
(8/6) ⁄(10/6)= 4⁄ 5
(option ‘D’)
Incorrect
cosec A + cot A = 3 (given) — (i)
cosec² A – cot² A = 1
=> (cosec A + cot A)(cosec A – cot A) = 1
=> cosec A – cot A) =1 ⁄3 (by putting cosec A + cot A = 3) — (ii)
By adding (i) and (ii)
2cosec A = 10 ⁄ 3=> cosec A = 10 ⁄6
By Subtracting (ii) from (i)2 cot A =8 ⁄3
=> cot A = 8 ⁄6cos A =cot A ⁄ cosec A=
(8/6) ⁄(10/6)= 4⁄ 5
(option ‘D’)

Question 9 of 10
9. Question
1 pointsIf sin 7x = cos 11x, then the value of tan 9x + cot 9x is?
Correct
solution:
sin 7x = cos (90 – 7x)
So, cos (90 – 7x) = cos 11x
=> 90 – 7x = 11x
=> x = 5
Now putting this in tan 9x + cot 9x
tan 45 + cot 45
= 1 + 1 = 2 (option ‘C’)Incorrect
solution:
sin 7x = cos (90 – 7x)
So, cos (90 – 7x) = cos 11x
=> 90 – 7x = 11x
=> x = 5
Now putting this in tan 9x + cot 9x
tan 45 + cot 45
= 1 + 1 = 2 (option ‘C’) 
Question 10 of 10
10. Question
1 pointsIf sin (x+y) = cos 3(x+y); then the value of tan 2(x+y) is ?
Correct
solution:
Given sin (x+y) = cos 3(x+y)
=> Sin (x+y) = sin [90 – 3(x+y)] [Using cos@ = sin (90 – @)]
=> x+y = 90 – 3x – 3y
=> 4(x+y) = 90
=> x+y = 90/4
Now substitute the value of x+y in tan 2(x+y)
tan 2(90/4) = tan 45 = 1 (option ‘A’)Incorrect
solution:
Given sin (x+y) = cos 3(x+y)
=> Sin (x+y) = sin [90 – 3(x+y)] [Using cos@ = sin (90 – @)]
=> x+y = 90 – 3x – 3y
=> 4(x+y) = 90
=> x+y = 90/4
Now substitute the value of x+y in tan 2(x+y)
tan 2(90/4) = tan 45 = 1 (option ‘A’)