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we have

cos2θ  = sin 4θ 

⇒ sin(90° - 2θ) = sin 4θ 

comparing both sides, we get

90° - 2° = 4θ

⇒ 2θ + 4θ = 90°

⇒ 6θ= 90°

⇒ θ = \(\frac{90^\circ}6\)

∴ θ = 15°

Hence, the value of θ is 15°

Correct option is: D) 4

Given that Cosec \(\theta\) – Cot \(\theta\) =  \(\frac 14\) ...(1)

\(\therefore\) ( cosec \(\theta\) - cot \(\theta\)) (cosec \(\theta\) + cot \(\theta\)) = \(\frac 14\)(cosec \(\theta\) + cot \(\theta\))

(On multiplying both sides by cosec \(\theta\) + cot \(\theta\))

= \(cosec^2\theta - cot^2\theta = \frac 14 \) (cosec \(\theta\) + cot \(\theta\)) (\(\because\) (a-b) (a+b) = \(a^2 - b^2\))

= \(\frac 14\) (cosec \(\theta\) + cot \(\theta\)) = 1 (\(\because\) \(cosec^2\theta - cot^2\theta = 1\))

= cosec \(\theta\) + cot \(\theta\) = 4

Correct option is: D) 4

sec 5θ = cosec (θ-36°) …(i)

We know that

sec θ = cosec (90° - θ)

So, Eq. (i) become

Cosec (90° - 5θ) = cosec (θ -36°)

On Equating both the sides, we get

90° - 5θ = θ -36°

⇒ -5θ - θ = -36° -90°

⇒ -6θ = -126°

⇒ θ = 21°

(i) LHS = (sec² θ − 1) cot² θ

= tan² θ × cot² θ      (∵ sec ²θ − tan² θ = 1)

= \(\frac{1}{cot^2θ}\times{cot^2θ}\)

= 1

= RHS

(ii) LHS = (sec² θ − 1)(cosec² θ − 1)

= tan² θ × cot² θ (∵ sec² θ − tan² θ = 1and cosec² θ − cot² θ = 1)

= \(tan^2θ \times\frac{1}{tan^2θ}\)

= 1

= RHS

(iii) LHS = (1 − cos² θ) sec² θ

= sin² θ × sec² θ    (∵ sin² θ + cos² θ = 1)

= \(sin^2θ \times\frac{1}{cos^2θ}\)

= \(\frac{sin^2θ}{cos^2θ}\)

= tan²θ

= RHS

Correct option is: B) Cos θ

Sin (90° – \(\theta\)) = cos \(\theta\)

Correct option is: B) Cos θ

Correct answer = (b) 25°

We have

[sin3A =  cos(A -10°)]

=> cos(90° − 3A) = cos(A − 10°) [∵ sin θ = cos(90° − θ)] 

=> 90° − 3A = A − 10° 

=> −4A = −100 

=> A = 100/4 

=> A = 25°

If sin A = cos A 

\(\frac{sin\,A}{cos\,A}=\frac{cos\,A}{cos\,A}\)= 1 

⇒ tan A = 1 = tan 45° 

⇒ A = 45° 

(OR) 

sin 45° = cos 45° 

\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

So, A = 45°

sin3A = cos(A - 26°)

⇒ cos(90° - 3A) = cos(A - 26°)     [∵sin θ = cos(90° - θ)]

⇒ 90° - 3A = A - 26°

⇒ 116° = 4A

⇒ A = \(\frac{116°}4\) = 29°

 sin3θ = cos(θ - 6°)

⇒ cos( 90° - 3θ) = cos(θ - 6°)

⇒ 90° - 3θ =  (θ - 6°)

⇒ 4θ = 96°

⇒ θ = \(\frac{96°}{4} = 24°\)

(C) √(b²-a²)/b

According to the question,

sin θ =a/b

We know, sin² θ +cos² θ =1

sin² A = 1-cos² A

sin A = √(1-cos² A

So, cos θ = √(1-a²/b² ) = √((b²-a²)/b² ) = √(b²-a² )/b

Hence, cos θ = √(b² – a² )/b

tan² A – sin² A 

= \(\frac{sin^2\,A}{cos^2\,A}\) – sin² A 

sin² A ( \(\frac{1}{cos^2\,A}\) – 1 ) 

= sin² A (sec² A – 1) 

= sin² A. tan² A

Given A + B + C = 180°; A + B = 180° – C; cot(A + B) = cot (180° – C)

\(\frac{cot(A).cot(B) - 1}{cotA + cotB} = - cotC\)

cotA . cotB – 1 = -cotA · cotC – cot B . cotC. 

∴ cotA . cotB + cotB cotC + cotC . cotA = 1.

Correct option is (A) sin θ = cos (90 – θ)

Correct option is (D) 1

cos (60° + θ) – sin (30° - θ)

= sin {90°-(60°+θ)} – sin (30° - θ) [∵ cos θ = sin (90° - θ)]

= sin ( 90° - 60°-θ) – sin (30° - θ)

= sin (30° - θ) - sin (30° - θ)

= 0

2 tan 45° + cos 45° – sin

= 2(1) + 1/√2 - 1/√2 = 2

Correct option is (C) 2

cosec (65° + θ) – sec (25° - θ)

= sec {90°-(65°+θ)} – sec (25° - θ)

[∵ cosec θ = sec (90° - θ)]

= sec ( 90° - 65°-θ) – sec (25° - θ)

= sec (25° - θ) - sec (25° - θ)

= 0

Correct option is (D) 1  \(\frac{cos \,28° }{sin\,62° }\) 

=  \(\frac{sin(90​° - 28° )}{sin62° }\) 

= \(\frac{sin62° }{sin62° }\) 

= 1

L.H.S (sinA – cosA)² 

= sin²A + cos²A – 2sinA cosA 

= 1 – sin2A = R.H.S.

(b) sin x < sin y

Since sin θ in cosec from θ = 0° to θ = 90°, i.e., where 

0° < θ < 90° then 0 < sin θ < 1. ∴ sin x < sin y.

(b) q

q cosec θ = p

∴ \(\sqrt{p^2-q^2}\) tan θ

= \(\sqrt{q^2\,cosec^2\theta-q^2}.tan\,\theta\) = \(\sqrt{q^2\,(cosec^2\theta-1)}.tan\,\theta\)

= \(\sqrt{q^2\,cot^2\,\theta}.tan\,\theta\) = q cot θ . tan θ = q.

cos² 75° + cos² 15° = 1

LHS = cos² 75° + cos² 15°

= cos² 75° + cos² (90 – 75)°

= cos² 75° + sin² 75°

= 1= RHS

LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10°

= 0

= RHS

Correct option is: A) \(\frac{17}{13}\)

We have cot A = \(\frac 5{12}\)

\(\therefore\) \(1 + cot^2A = 1 + (\frac 5 {12})^2 = 1 + \frac {25}{144} \)

= \(\frac {144+25}{144} = \frac {169}{144} = (\frac {13}{12})^2\)

= \(cosec^2A = (\frac {13}{12})^2\) (\(\because\) \(1 + cot^2A = cosec^2A\))

= cosec A = \(\frac {13}{12}\)

\(\therefore\) Sin A = \(\frac {1}{cosec \,A} = \frac 1{\frac {13}{12}} = \frac {12}{13}\)

\(\therefore\) cot A = \(\frac {cos\, A}{sin \, A} = \frac 5{12}\)

= cos A = \(\frac 5{12}\) sin A = \(\frac 5{12}\) \(\times\) \(\frac {12}{13}\) = \(\frac 5{13}\)

\(\therefore\) Sin A + cos A = \(\frac {12}{13}\) + \(\frac 5{13}\) = \(\frac {17}{13}\)

Correct option is: A) \(\frac{17}{13}\)

Correct option is: A) \(\frac{5}{13}\)

We have cos A = \(\frac {12}{13}\)

\(\therefore\) Sin A = \(\sqrt {1-cos^2A} = \sqrt {1- (\frac {12}{13})^2} \)

\(= \sqrt {1- \frac {144}{169}} = \sqrt {\frac {25}{169}} = \frac 5{13}\)

Correct option is: A) \(\frac{5}{13}\)

sec 70° sin 20° - cos 20° cosec 70°

= cosec (90°-70°) cos (90° - 20°)- cos 20° cosec 70°

[∵ sec θ = cosec (90° - θ) and Sin θ = cos (90° - θ)]

= cosec 70° cos 20° - cos 20° cosec 70°

= 0

True

Justification:

According to the question,

cos A+cos² A = 1

i.e., cos A = 1- cos² A

Since,

sin² θ+cos² θ = 1

sin² θ = 1- cos² θ)

We get,

cos A = sin² A …(1)

Squaring L.H.S and R.H.S,

cos² A = sin⁴ A …(2)

To find sin²A+sin⁴ A=1

Adding equations (1) and (2),

We get

sin²A + sin⁴ A= cos A + cos² A

Therefore, sin²A + sin⁴ A = 1

Taking LHS = sin54^o + cos67^o

We know that

cos θ = sin (90° - θ)

Here, θ = 67°

⇒ sin 54°+ sin (90° - 67°)

⇒ sin 54°+ sin 23°

We also know that

Sin θ = cos (90° - θ)

Here, θ = 54°

⇒ cos (90° - 54°) + sin 23°

⇒ cos 36°+ sin 23° = RHS

Hence Proved

We know that

cos² θ + sin² θ = 1

⇒ cos² 55° + sin² 55° = 1

⇒ sin² 55° = 1 – cos² 55°

⇒ sin 55° = √(1 – cos² 55°)

And Given that cos 55° = x²

⇒ sin 55° = √{1– (x²)²}

⇒ sin 55° = √(1 – x⁴)

 α + β = \(\frac{π}{2}\) ⇒  α = \(\frac{π}{2}\) - β ⇒ tan α = tan \(\bigg(\frac{π}{2}- β\bigg)\)

⇒ tan α = cot β ⇒ tan α tan β = 1.         …(i) 

Now, β + γ = α ⇒ γ = α - β 

⇒ tan γ = tan (α - β) ⇒ tan γ = \(\frac{tan\,α - tan\,β}{1+tan\,α\,tan\,β}\)

⇒  tan γ = \(\frac{tan\,α-tan\,β}{1+1}\) = \(\frac{tan\,α-tan\,β}{2}\)            (∵ tan α tan β = 1)

⇒ 2 tan γ = tan α - tan β ⇒ tan α = tan β + 2 tan γ .

 In a right angled triangle APQ,

tan A = \(\frac{PQ}{AQ}\) and tan P = \(\frac{AQ}{PQ}\)

∵  tan A =  tan P

\(\frac{PQ}{AQ} = \frac{AQ}{PQ}\)

∵ PQ = AQ

∠P = ∠A = 45°

3cosθ - 4sinθ = 2cosθ + sinθ

3cosθ - 2cosθ = 4sinθ + sinθ

cosθ = 5sinθ

tanθ = \(\frac{1}{5}\)

In a ΔABC, right angled at A,

tan C = √3

i.e ∠C = 60° and ∠B = 90 - 60 = 30° 

Sin C = Sin 60° = √3/2 

Cos C = Cos 60° = 1/2 

Sin B = Sin 30° = 1/2 

Cos B = Cos 30° = √3/2 

According to the question, 

sin B cos C + cos B sin C 

= (1/2) (1/2) + (√3/2) (√3/2) 

= 1/4 + 3/4 

= 1

Correct option is: C) 1

\(Sin^2 \theta + cos^2 \theta\) = 1 for any value of \(\theta\).

So, hence if sin \(\theta\) = \(\frac {20}{29}\) then \(Sin^2 \theta + cos^2 \theta\) = 1

Alternative :

We have sin \(\theta\) =\(\frac {20}{29}\)

\(\therefore\) \(cos^2\, \theta = 1- sin^2 \theta = 1 - (\frac {20}{29})^2 = 1- \frac {400}{841}\)

= \(\frac {841-400}{841} = \frac {441}{841}\)

Now, \(Sin^2 \theta + cos^2 \theta\) = \((\frac {20}{29})^2+ \frac {441}{841} = \frac {400}{841} + \frac {441}{841} = \frac {841}{841} = 1\)

Correct option is: C) 1

Correct option is: C) 4

\(\because\) tan \(45^\circ\) = 1

\(\therefore\) \((1+ tan^2 45^\circ)^2 = (1+1^2)^2 = (1+1)^2 = 2^2 = 4\) 

Correct option is: C) 4

Correct option is: A) 1

\(cosec^2 A - cot^2A = 1 \) for any value of A except those values where sin A = 0

Here, sin A = \(\frac 9{15} \neq 0\)

\(\therefore\) \(cosec^2 A - cot^2A = 1 \)

Correct option is: A) 1

Answer : (d) π/6 

√3 cos θ – sin θ = 1

Dividing throughout by \(\sqrt{(\sqrt{3})^2+(-1)^2}\) = 2 We have 

\(\frac{\sqrt{3}}{2}\) cos θ - \(\frac{1}{2}\) sin θ = \(\frac{1}{2}\) 

⇒ sin π/3 cos θ – cos π/3 sin θ = \(\frac{1}{2}\)  

⇒ sin (π/3 - θ) = sin π/6 

⇒ π/3 - θ = π/6 

⇒ θ = π/6.

Answer : (c) 5

sin x cos 3x = sin 3x cos 5x 

⇒ 2 sin x cos 3x – 2 sin 3x cos 5x = 0 

⇒ [sin (3x + x) – sin (3x – x)] – [sin (3x + 5x) – sin (5x – 3x)] = 0 

⇒ sin 4x – sin 2x – sin 8x + sin 2x = 0 

⇒ sin 4x – sin 8x = 0

⇒ 2 cos \(\big( \frac{4x+8x}{2}\big)\) sin \(\big( \frac{8x-4x}{2}\big)\) = 0 

⇒ 2 cos 6x sin 2x = 0 ⇒ cos 6x = 0 or sin 2x = 0 

⇒ 6x = (2n + 1) π/2 or 2x = nπ

⇒ x = (2n + 1) π/2 or x = \(\frac{n\pi}{2}\) 

⇒ x = 0, \(\frac{\pi}{2}\), \(\frac{\pi}{12}\), \(\frac{3\pi}{12}\),\(\frac{5\pi}{12}\) in [0. \(\frac{\pi}{2}\)] = 5

Answer :  (a) 0

sin⁴ θ – 2 sin² θ – 1 = 0 

⇒ (sin² θ)² – 2 sin² θ – 1 = 0 

⇒ sin² θ = \( {2 \,\pm \sqrt{4-4\times 1 \times (-1)} \over 2\times 1}\) 

\( = {2\, \pm\, \sqrt{8} \over 2}\) = \( {2\, \pm \,2\sqrt{2} \over 2}\) 

= 1- \( \sqrt{2} \)  or 1 + \( \sqrt{2} \) 

sin² θ = 1 – \( \sqrt{2} \) is inadmissible as it is not real. 

∵ –1 ≤ sin θ ≤ 1 

⇒ 0 ≤ sin² θ ≤ 1 

⇒ sin² θ = 1 + \( \sqrt{2} \) is not possible. 

Hence the given equation has no real root.

Answer : (b) \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) 

sin (x + π/4) = sin 2x 

⇒ x + π/4 = nπ + (–1)n 2x, n ∈ I 

⇒ x – (–1)ⁿ 2x = nπ – π/4, n ∈ I 

⇒ x {1 – (–1)ⁿ .2} = nπ – π/4 

⇒ x = \(\frac{n\pi- \pi/4}{1-(-1)^n \,2}\) , n ∈ I.

Answer : (d)  2nπ + \(\frac{7π}{4}\)  

tan θ = –1 

⇒ tan θ = – tan π/4 = tan (π – π/4) = tan (2π - π/4)

⇒ tan θ = tan \(\frac{3π}{4}\) or tan \(\frac{7π}{4}\) ⇒ θ = \(\frac{3π}{4}\) , \(\frac{7π}{4}\) ...(i)

cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ cos θ = cos \(\frac{π}{4}\)

= cos (2π – π/4)

⇒ cos θ = cos \(\frac{π}{4}\) = cos \(\frac{7π}{4}\) ⇒ θ = \(\frac{π}{4}\) ,  \(\frac{7π}{4}\)...(ii)

∴ From (i) and (ii) θ = \(\frac{7π}{4}\)  is the only value of θ in [0, 2π] which satisfies both the equations. 

∴ The general value of θ satisfying both the equations is

θ =  2nπ + \(\frac{7π}{4}\)  

False

Given: ‘a’ is a positive number and a≠1

⇒ AM > GM

(Arithmetic Mean (AM) of a list of non- negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

If a and b be such numbers, then

AM = (a + b)/2 and GM = √ab

By assuming that a + 1/2 = 2 sin θ statement is be true.

Similarly, AM and GM of a and 1/a are (a+1/a)/2 and √(a.1/a) respectively.

By property, (a+1/a)/2 > √(a.1/a)

a + 1/a > 2

⇒ 2 sin θ > 2 (By our assumption)

⇒ sin θ > 1

But -1 ≤ sin θ ≤ 1

∴ Our assumption is wrong and that 2 sin θ cannot be equal to a + 1/a

(cosec² 45° sec² 30°) (sin² 30° + 4 cot² 45°- sec² 60°)

= \((2\times\frac{2}3)\)\((\frac{1}4+4\times1-4)\)

= \(\frac{4}3\)\((\frac{1}4+4\times1-4)\) = \(\frac{1}3\)

4(sin⁴ 30° + cos² 60°) - 3(cos² 45° - sin² 90°) - sin² 60°

\(4(\frac{1}{16} + \frac{1}{4}) - 3(\frac{1}{2} - 1) - \frac{3}{4}\)

= \(\frac{20}{16} + \frac{3}{2} - \frac{3}{4}\)

= 2

cot² 30° - 2 cos² 60° - \(\frac{3}4\)sec² 45° - 4 sec² 30°

= \((\sqrt3)^2\) - 2\((\frac{1}2)^2\) - \(\frac{3}4\) x \((\sqrt2)^2\) - 4\((\frac{2}{\sqrt3})^2\)

= 3 - \(\frac{1}2\) - \(\frac{3}2\) - \(\frac{16}3\) = - \(\frac{13}3\)

\(\frac{4}{cot^2 30°} + \frac{1}{sin^2 60°} - cos^2 45°\) 

= \(\frac{4}{(\sqrt{3})^2}\)+\(\frac{1}{(\frac{\sqrt{3}}{2})^2}\)-\((\frac{1}{\sqrt{2}})^2\)

= \(\frac{4}{3} + \frac{4}{3} - \frac{1}{2} = \frac{13}{6}\)

sin² A + cos² A = 1

sin² A = 1 - cos² A

= 1- (a/b)²

=1 - a²/b²

= (b² - a²)/b²

False. 

The value of (sinθ + cosθ) for θ = 0° is 1.