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Hence, L.H.S. = R.H.S.
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Hence, LHS = RHS
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Hence, LHS = RHS
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Hence, L.H.S. = R.H.S.
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Hence, LHS = RHS
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Hence, LHS = RHS
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Hence, LHS = RHS
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Hence, LHS= RHS
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Hence, L.H.S. = R.H.S.
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Hence, LHS = RHS
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Hence, LHS = RHS
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Hence, LHS = RHS
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Hence, LHS = RHS
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Disclaimer: The question given in the textbook is incorrect. There should be instead . The solution provided here is of the same.
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Now,
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Given that A and B are acute angles such that sin A = cos B.
Hence, the correct answer is option C.
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(d) cos 2β
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Hence, the correct answer is option A.
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(b)
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(c)
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(a) 30o
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(c) 20o
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(b) 30o
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(b) 60o
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(a) 1
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(c)
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(b) 3
sec260o − 1 = (2)2 − 1 = 4 − 1 = 3
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(d)
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(b)
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(b)
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(d)
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(b)
Let us first draw a right ABC right angled at B and .
Given: cosec = , but sin = =
Also, sin = =
So, =
Thus, BC = k and AC = k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = ( k)2 (k)2
⇒ AB2 = 9k2
⇒ AB = 3k
∴ sec = =
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(a)
Let us first draw a right ABC right angled at B and .
Given: tan = , but tan =
So, =
Thus, BC = 8k and AB = 15k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec =
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(b)
Let us first draw a right ABC right angled at B and .
Given: sin θ = , but sin θ =
So, =
Thus, BC = ak and AC = bk
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (bk)2 (ak)2
⇒ AB2 = k2
⇒ AB = ()k
∴ cos θ = = =
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(d) 2
Let us first draw a right ABC right angled at B and .
Given: tan θ =
But tan θ =
So, =
Thus, BC = k and AB = k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = ( k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ =
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(c)
Let us first draw a right ABC right angled at B and .
Given: sec θ =
But cos θ = = =
Thus, AC = 25k and AB = 7k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (25k)2 (7k)2
⇒ BC2= 576k2
⇒ BC = 24k
∴ sin θ =
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(b)
Given: sinθ = , but sinθ =
So, =
Thus, BC = k and AC = 2k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 BC2
AB2 = (2k)2 (k)2
AB2 = 3k2
AB = k
So, tanθ = = =
∴ cotθ = =
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(a)
Since cosθ = but cosθ =
So, =
Thus, AB = 4k and AC = 5k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (5k)2 (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = =
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(c)
Given: 3x = cosec θ and = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and in the given expression, we get:
3 = 3
= 3
=
= [By using the identity: ]
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(a)
Given: 2x = sec A and = tan A
Also, we can deduce that x = and .
So, substituting the values of x and in the given expression, we get:
2 = 2
= 2
=
= [By using the identity: ]
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(c)
Let us first draw a right ABC right angled at B and .
tan θ =
So, AB = 3k and BC = 4k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = (3k)2 + (4k)2
⇒ AC2 = 25k2
⇒ AC= 5k
Thus, sin θ = =
and cos θ =
∴ (sin θ + cos θ) = ( + ) =
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(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25 [∵ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23
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(b)
We have (cos θ +sec θ) =
Squaring both sides, we get:
(cos θ + sec θ)2 = ()2
cos2 θ + sec2 θ + 2 cos θ sec θ =
cos2 θ + sec2 θ + 2 = [∵ sec θ = ]
cos2 θ + sec2 θ = − 2 =
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(d)
=
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(a)
7 tan θ = 4
Now, dividing the numerator and denominator of the given expression by cos θ, we get:
=
= [∵ 7 tan θ = 4]
=
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(d) 9
We have .
Dividing the numerator and denominator of the given expression by sin θ, we get:
=
= = 9 [∵ 3 cot θ = 4]
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(b)
We have tan θ =
Now, dividing the numerator and denominator of the given expression by cos θ, we get:
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(b) 1
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(a) 1
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(b) (sec A − tan A)
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Hence, the correct answer is option B.
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(c)
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(b)
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(b) cos A
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(b) 4
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(d) 2
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(d)
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= (cosec A + cot A).
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(cosec A − cot A)2 =
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(sin 32° cos 58° + cos 32° sin 58°) = 1
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= (sec θ + tan θ)2
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