Explore topic-wise InterviewSolutions in .

sinA/sinB = √3/2   =>  sinA = (√3/2)sinB   ...............(1)

cosA/cosB = √5/2   =>  cosA = (√5/2)cosB  ...............(2)

Squaring and adding (1) and (2),

1 = (3/4)(sinB)² +  (5/4)(cosB)²

=> 3(sinB)² + 5(cosB)² = 4

=>  3(sinB)² + 3(cosB)²+ 2(cosB)² = 4

=> 3{(sinB)² +  (cosB)²} +  2(cosB)² = 4

=>  (cosB)² = 1/2   

=>  cosB = 1/√2 or -1/√2

=> sinB = 1/√2 or -1/√2

=> sinA = √3/(2√2) or -√3/(2√2)          from (1)

=> cosA = √5/(2√2) or -√5/(2√2)         from (2)

Hence,  tanA + tanB =  √3/√5 + 1

(i) (1/4)[4 – 3(a² – 1)²]

(ii) √(2 − a²)

Given,

\(\frac{sin(A-B)}{sin(A+B)}=\frac{a^2-b^2}{a^2+b^2}\)

Applying componendo and dividendo

\(\frac{sin(A-B)+sin(A+B)}{sin(A-B)-sin(A+B)} \)\(=\frac{a^2-b^2+a^2+b^2}{a^2-b^2-a^2-b^2}\)

⇒ \(-\frac{2\,sin\,A\,cos\,B}{2\,cos\,A\,sin\,B}=\frac{2a^2}{-2b^2}\)

⇒ \(\frac{sin\,A\,cos\,B}{cos\,A\,sin\,B}=\frac{a^2}{b^2}\)

⇒ b² sin A cos B = a² cos A sin B

Putting sin A = ka, sin B = kb

⇒ b²ka cos B = a²kb cos A

⇒ b cos B = a cos A

Applying cosine formula,

\(b(\frac{a^2+c^2-b^2}{2ac})\) = \(a(\frac{b^2+c^2-b^2}{2bc})\)

⇒ b² (a² + c² – b² ) = a² (b² + c² – a²)

⇒ a²b² + b²c² – b⁴ = a²b² + a²c² – a⁴

⇒ b²c² – a²c² – b⁴ + a⁴ = 0

⇒ c² (b² – a²) – (b² – a² ) (b² + a² ) = 0

⇒ (b² – a²) (c² – b² – a² ) = 0

⇒ Either b² – a² = 0 

or, c² – b² – a² = 0

⇒ b = a 

or, c² = b² + a²

∴ Triangle is isosceles if b = a

∴ Triangle is right angled if c² = b² + a²

Correct option is: (d) (√2 ,√2)

Given tan(A +B) = x And tan(A –B) = y 

Consider tan 2A = tan(A +A) 

= tan(A +B +A –B) 

We know that tan(A+B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ tan(A + B + A - B) = \(\frac{tan(A+B)+tan(A-B)}{1-tan(A+B)tan(A-B)}\)

= \(\frac{x+y}{1-xy}\)

Consider tan 2B = tan(B +B) 

= tan(B +A +B –A)

= \(\frac{tan(B+A)+tan(B-A)}{1-tan(B+A)tan(B-A)}\)

We know that tan(-θ) = - tan θ

= \(\frac{tan(A+B)-tan(A-B)}{1+tan(A+B)tan(A-B)}\)

= \(\frac{x-y}{1+xy}\)

Correct option is: (a) 2 :\(\cfrac{\pi}{2}\) : \(\cfrac{\pi}{3}\)+ 1 

α + β = π + γ 

Sin (α + β) =sin (π + γ) 

Sin (α) cos (β) + sin (β) cos (α)=-sin(γ) 

Take square both side 

[sin(α)cos(β)+sin(β)cos(α)]²=sin²(γ) 

sin²(α)cos²(β)+sin²(β)cos²(α)+2 Sin(α)cos(β)sin(β)cos(α)= sin²(γ)

sin²(α)[1-sin²(β)]+sin²(β)[1-sin²(α)]+2 Sin(α)cos(β)sin(β)cos(α) = sin²(γ) 

sin²(α)-Sin²(α)sin²(β)+sin²(β)-sin²(β)sin²(α) -sin²(γ)=- 2Sin(α)cos(β)sin(β)cos(α) 

sin²(α)+sin²(β)-sin²(γ) = 2Sin²(α)sin²(β) - 2Sin(α)cos(β)sin(β)cos(α)

sin²(α)+sin²(β)-sin²(γ) = - 2Sin(α)sin(β)[ cos(β) cos(α)- Sin(α)sin(β)]

sin²(α)+sin²(β) - sin²(γ) = - 2Sin(α)sin(β) cos(α+ β) 

sin²(α)+sin²(β) - sin²(γ) = 2Sin(α)sin(β) sin(γ)

C. 0

Given that, sin θ + cos θ = 1

Squaring both sides, we get

⇒ (sin θ + cos θ)² = 1

⇒ sin²θ + cos²θ + 2sin θ cos θ = 1

⇒ 1 + sin2θ = 1

⇒ sin2θ = 1-1

= 0

Let us consider the LHS

sin 2x/(1 – cos 2x)

As we know that cos 2x = 1 – 2 sin² x

Sin 2x = 2 sin x cos x

Therefore,

sin 2x/(1 – cos 2x) = (2 sin x cos x)/(1 – (1 – 2sin² x))

= (2 sin x cos x)/(1 – 1 + 2sin² x)]

= [2 sin x cos x/2 sin² x]

= cos x/sin x

= cot x

= RHS

Thus proved.

Let us consider the LHS

sin 2x/(1 + cos 2x)

As we know that cos 2x = 1 – 2 sin² x

= 2 cos² x – 1

Sin 2x = 2 sin x cos x

Therefore,

sin 2x/(1 + cos 2x) = [2 sin x cos x/(1 + (2cos²x – 1))]

= [2 sin x cos x/(1 + 2cos² x – 1)]

= [2 sin x cos x/2 cos² x]

= sin x/cos x

= tan x

= RHS

Thus proved.

Let us consider the LHS

√[(1 – cos 2x)/(1 + cos 2x)]

As we know that cos 2x = 1 – 2 sin² x

= 2 cos² x – 1

Therefore,

√[(1 – cos 2x)/(1 + cos 2x)] = √[(1 – (1 – 2sin² x))/(1 + (2cos²x – 1))]

= √[(1 – 1 + 2sin² x)/(1 + 2cos² x – 1)]

= √[2 sin² x/2 cos² x]

= sin x/cos x

= tan x

= RHS

Thus proved.

Let us consider the LHS

2(sin⁶ x + cos⁶ x) – 3(sin⁴ x + cos⁴ x) + 1

As we know, (a + b)² = a² + b² + 2ab

a³ + b³ = (a + b) (a² + b² – ab)

Therefore,

2(sin⁶ x + cos⁶ x) – 3(sin⁴ x + cos⁴ x) + 1 = 2{(sin² x)³ + (cos² x)³} – 3{(sin² x)² + (cos² x)²} + 1

= 2{(sin² x + cos² x) (sin⁴ x + cos⁴ x – sin² x cos² x} – 3{(sin² x)² + (cos² x)² + 2sin² x cos² x – 2sin² x cos² x} + 1

= 2{(1) (sin⁴ x + cos⁴ x + 2 sin² x cos² x – 3 sin² x cos² x} – 3{(sin² x + cos² x)² – 2sin² x cos² x} + 1

As we know, sin² x + cos² x = 1

= 2{(sin² x + cos² x)² – 3 sin² x cos² x} – 3{(1)² – 2sin² x cos² x} + 1

= 2{(1)² – 3 sin² x cos² x} – 3(1 – 2sin² x cos² x) + 1

= 2(1 – 3 sin² x cos² x) – 3 + 6 sin² x cos² x + 1

= 2 – 6 sin² x cos² x – 2 + 6 sin² x cos² x

= 0

= RHS

Thus proved.

Let us consider the LHS

3(sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶ x + cos⁶ x)

As we know, (a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a³ + b³ = (a + b) (a² + b² – ab)

Therefore,

3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x) = 3{(sin x – cos x) ²}² + 6 {(sin x)² + (cos x)² + 2 sin x cos x)} + 4{(sin² x)³ + (cos² x)³}

= 3{(sin x)² + (cos x)² – 2 sin x cos x)}² + 6(sin² x + cos² x + 2 sin x cos x) + 4{(sin² x + cos² x) (sin⁴ x + cos⁴ x – sin² x cos² x)}

= 3(1 – 2 sin x cos x)² + 6 (1 + 2 sin x cos x) + 4{(1) (sin⁴ x + cos⁴ x – sin² x cos² x)}

As we know, sin² x + cos² x = 1

Therefore,

= 3{1² + (2 sin x cos x)² – 4 sin x cos x} + 6(1 + 2 sin x cos x) + 4{(sin² x)² + (cos² x)² + 2 sin² x cos² x – 3 sin² x cos² x)}

= 3{1 + 4 sin² x cos² x – 4 sin x cos x} + 6(1 + 2 sin x cos x) + 4{(sin² x + cos² x) ² – 3 sin² x cos² x)}

= 3 + 12 sin² x cos² x – 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)² – 3 sin² x cos² x)}

= 9 + 12 sin² x cos² x + 4(1 – 3 sin² x cos² x)

= 9 + 12 sin² x cos² x + 4 – 12 sin² x cos² x

= 13

= RHS

Thus proved.

Let us consider the LHS

cos 4x

As we know, cos 2x = 2 cos² x – 1

Therefore,

cos 4x = 2 cos² 2x – 1

= 2(2 cos² 2x – 1)² – 1

= 2[(2 cos² 2x)² + 1² – 2 × 2 cos² x] – 1

= 2(4 cos⁴ 2x + 1 – 4 cos² x) – 1

= 8 cos⁴ 2x + 2 – 8 cos² x – 1

= 8 cos⁴ 2x + 1 – 8 cos² x

= RHS

Thus proved.

Let us consider the LHS

sin 4x

As we know, sin 2x = 2 sin x cos x

cos 2x = cos² x – sin² x

Therefore,

sin 4x = 2 sin 2x cos 2x

= 2(2 sin x cos x) (cos² x – sin² x)

= 4 sin x cos x (cos² x – sin² x)

= 4 sin x cos³ x – 4 sin³ x cos x

= RHS

Thus proved.

Let us consider the LHS

(1 + tan α tan β)² + (tan α – tan β)²

1+ tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

1 + tan² α tan² β + tan² α + tan² β

tan² α (tan² β + 1) + 1 (1 + tan² β)

(1 + tan² β) (1 + tan² α)

As we know, 1 + tan² θ = sec² θ

Therefore,

sec² α sec² β

= RHS

∴ LHS = RHS

Thus proved.

Let us consider the LHS

(cos α + cos β)² + (sin α + sin β)²

Now, upon expansion, we get,

(cos α + cos β)² + (sin α + sin β)² 

= cos² α + cos² β + 2 cos α cos β + sin² α + sin² β + 2 sin α sin β

= 2 + 2 cos α cos β + 2 sin α sin β

= 2 (1 + cos α cos β + sin α sin β)

= 2 (1 + cos (α – β)) [since, cos (A – B) = cos A cos B + sin A sin B]

= 2 (1 + 2 cos² (α – β)/2 – 1) [since, cos2x = 2cos² x – 1]

= 2 (2 cos² (α – β)/2)

= 4 cos² (α – β)/2

= RHS

Thus proved.

Let us consider the LHS

1 + cos² 2x

As we know, cos2x = cos² x – sin² x

cos² x + sin² x = 1

Therefore,

1 + cos² 2x = (cos² x + sin² x)² + (cos² x – sin² x)²

= (cos⁴ x + sin⁴ x + 2 cos² x sin² x) + (cos⁴ x + sin⁴ x – 2 cos² x sin² x)

= cos⁴ x + sin⁴ x + cos⁴ x + sin⁴ x

= 2 cos⁴ x + 2 sin⁴ x

= 2(cos⁴ x + sin⁴ x)

= RHS

Thus proved.

Let us consider the LHS

sin²(π/8 + x/2) – sin²(π/8 – x/2)

As we know, sin² A – sin² B = sin(A + B) sin(A - B)

Therefore,

sin²(π/8 + x/2) – sin²(π/8 – x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))

= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)

= sin π/4 sin x

= 1/√2 sin x [since, π/4 = 1/√2]

= RHS

Thus proved.

LHS 

= 2 sin² β + 4 cos (α + β) sin α sin β + cos 2(α + β)

= 2 sin² β + 4 (cos α cos β – sin α sin β) sin α sin β

+ (cos 2α cos 2β – sin 2α sin 2β)

= 2 sin² β + 4 sin α cos α sin β cos β – 4 sin² α sin² β + cos 2α cos 2β – sin 2α sin 2β

= 2 sin² β + sin 2α sin 2β – 4 sin² α sin² β + cos 2α cos 2β – sin2α sin2β

= (1 – cos 2β) – (2 sin² α) (2 sin² β) + cos 2α cos 2β 

= (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β

= cos 2α

Answer is (B) sin 4β 

Answer is True

Given as 

sin A = 1/2, cos B = 12/13, where π/2 < A < π and 3π/2 < B < 2π

As we know that, A is in second quadrant, B is in fourth quadrant.

Since, in the second quadrant, sine function is positive, cosine and tan functions are negative.

Since, in the fourth quadrant, sine and tan functions are negative, cosine function is positive.

On using the formulas,

cos A = – √(1 – sin² A) and sin B = -√(1 – cos² B)

Therefore let us find the value of cos A and sin B

cos A = – √(1 – sin² A)

= – √(1 – (1/2)²)

= – √(1 - 1/4)

= – √((4 - 1)/4)

= – √(3/4)

= -√3/2

sin B = -√(1 – cos² B)

= – √(1 - (12/13)²)

= – √(1 - 144/169)

= – √((169 - 144)/169)

= – √(25/169)

= – 5/13

As we know, tan A = sin A/cos A and tan B = sin B/cos B

tan A = (1/2)/( -√3/2) = -1/√3 and

tan B = (-5/13)/(12/13) = -5/12

Therefore, tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

= ((-1/√3) – (-5/12))/(1 + (-1/√3) × (-5/12))

= ((-12 + 5√3)/12√3)/(1 + 5/12√3)

= ((-12 + 5√3)/12√3)/((12√3 + 5)/12√3)

= (5√3 – 12)/(5 + 12√3)

Given as

cos x = -3/5 and π < x < 3π/2

As we know that in the third quadrant, tan x and cot x are positive and all other rations are negative.

On using the formulas,

Sin x = – √(1 - cos² x)

Tan x = sin x/cos x

Cot x = 1/tan x

Sec x = 1/cos x

Cosec x = 1/sin x

Then,

Sin x = – √(1 - cos² x)

= – √(1 - (-3/5)²)

= – √(1 - 9/25)

= – √((25 - 9)/25)

= – √(16/25)

= – 4/5

Tan x = sin x/cos x

= (-4/5)/(-3/5)

= -4/5 × -5/3

= 4/3

Cot x = 1/tan x

= 1/(4/3)

= 3/4

Sec x = 1/cos x

= 1/(-3/5)

= -5/3

Cosec x = 1/sin x

= 1/(-4/5)

= -5/4

∴ (cosec x + cot x)/(sec x - tan x)
= [(-5/4) + (3/4)]/[(-5/3) – (4/3)]

= [(-5 + 3)/4]/[(-5 - 4)/3]

= [-2/4]/[-9/3]

= [-1/2]/[-3]

= 1/6

LHS = (sinx)cosx[tanx + cotx]

= sinxcosx[sinx/cosx + cosx/sinx]

= sin²x + cos²x

= 1 = RHS.

Let, r sinα = √3 – 1 and r cosα = √3 + 1

Therefore, r = √{(√3 – 1)² + (√3 + 1)²} = √8 = 2√2

And, tan α = (√3 – 1) / (√3 + 1)

Therefore, r(sinα cos θ + cosα sin θ) = 2

⇒ r sin (θ+α) = 2

⇒ sin (θ+α) = 1/√2

⇒ sin (θ+α) = sin (π/4)

⇒ θ+α = nπ + (– 1)ⁿ (π/4), n ∈ Z

⇒ θ = nπ + (– 1)ⁿ (π/4) – (π/12), n ∈ Z

We have, (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x

=> 2 sin 2x cos x – 3 sin 2x = 2 cos 2x.cos x – 3 cos 2x

=> sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)

=> sin 2x = cos 2x (As cos x ≠ 3/2)

=> tan 2x = 1 => tan 2x = tan π/4

=> 2x = nπ + π/4 , n∈Z

x = nπ/2 +π/8 , n∈Z

Given A, B, C and D are the angles of a cyclic quadrilateral.

∴ A + C = 180° and B + D = 180°

⇒ A = 180° – C and B = 180° - D

Now, LHS = cos(180°– A) + cos (180°+ B) + cos (180°+ C) – sin (90°+ D)

= -cos A + [-cos B] + [-cos C] + [-cos D]

= -cos A – cos B – cos C – cos D

= -cos (180° - C) – cos (180° - D) – cos C – cos D

= -[-cos C] – [-cos D] – cos C – cos D

= cos C + cos D – cos C – cos D

= 0

= RHS

Hence proved.

We have sin(x + y) = sin xcosy + cosxsiny

Replacing y by x we get 

sin 2x = 2sinx cosx

Again sin2x = \(\frac{2\sin \mathrm \cos \mathrm x}{\sin^2\mathrm x+\cos^2\mathrm x}\)

Dividing numerator and denominator by cos²x,

We get

sin2x = \(\cfrac{\frac{2\sin \mathrm x\cos \mathrm x}{\cos^2\mathrm x}}{1+\frac{\sin^2\mathrm x}{\cos^2\mathrm x}}\)

= \(\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\)

(i) \(\because\) sin(x + y) = sin x cos y + cos x sin y

put y = x, we obtain

sin 2x = sin x cos x + cos x sin x

= 2sin x cos x

(ii) \(\because\) sin 2x = 2sin x cos x

= 2 \(\frac{\sin \mathrm x}{\cos \mathrm x}\) cos²x

= \(\frac{2\tan\mathrm x}{\sec^2\mathrm x}\)  (\(\because\) sec x = \(\frac{1}{\cos\mathrm x}\) & tan x = \(\frac{\sin\mathrm x}{\cos\mathrm x}\))

\(=\frac{2\tan\mathrm x}{1+\tan^2\mathrm x}\)   (\(\because\) sec²x = 1 + tan²x)

L.H.S. = cos 6x = cos 3(2x) [∵ cos 3x = 4 cos³ x – 3 cos x]

= 4 cos³ 2x – 3 cos 2x

= cos 2x (4 cos² 2x – 3)

= cos 2x [4(cos 2x)² – 3] [∵ cos 2x = 2 cos² x – 1]

= cos 2x [4(2 cos² x – l)² – 3]

= cos 2x [4(4 cos⁴ x + 1 – 4 cos² x) – 3]

= cos 2x (16 cos⁴ x + 4 – 16 cos² x – 3)

= (2 cos² x – 1) (16 cos⁴ x + 4 – 16 cos² x – 3)

= 32 cos⁶ x + 8 cos² x – 32 cos⁴ x – 6 cos² x – 16 cos⁴ x – 4 + 16 cos² x + 3

= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

= R.H.S.  Hence Proved.

L.H.S. = cos 4x = cos 2(2x)

(cos θ = 1 – 2 sin² θ)

= 1 – 2 sin² 2x

= 1 – 2 (2 sin x cos x)² [∵ sin 2x = 2 sin x cos x]

= 1 – 2 x 4 sin² x cos² x

= 1 – 8 sin² x cos² x = R.H.S.

sin(x – y) = sin[x + (-y)]

= sin x cos(-y) + cos x  sin(-y)

= sin x cos y-cos x sin y .

cos3x = cos(2x + x) 

= cos2x cos x – sin2xsinx 

= (2cos²x - 1)cosx – 2sinxcosxsinx  

= 2cos³x - cosx - 2cosx(1 -  cos²x) 

= 4cos³  x - 3cosx 

LHS = (1 – sin²2x) – (1 – sin²6x)

= sin²6x – sin²2x = sin(6x + 2x)sin(6x – 2x)

= sin8x sin4x

= RHS 

Let f(x) = √3 sin x – cos x 

Dividing and multiplying by √(3 + 1) = 2,

⇒ f(x) = 2\((\frac{\sqrt3}2sinx - \frac{1}2cosx)\)

Sine of expression:

⇒ f(x) = 2\((cos\frac{π }6sinx-sin\frac{π }6cosx)\)

We know that sinA cosB – cosA sinB = sin(A –B)

∴ f(x) = 2sin\((x-\frac{π }6)\)

Cosine of the expression:

⇒ f(x) = 2\((sin\frac{π }6sinx-cos\frac{π }6cosx)\)

We know that cosA cosB – sinA sinB = cos(A +B)

 ∴ f(x) = -2cos\((\frac{π }3+x)\)

We have: 2 tan² x+ sec² x= 2 

2 tan² x + 1 + tan² x = 2 

Which gives tan x = ± \(\frac{1}{\sqrt{3}}\) 

If we take tan x = \(\frac{1}{\sqrt{3}}\) ,then x = \(\frac{\pi}{6}\) or \(\frac{7\pi}{6}\) 

Again, if we take tanx = −\(\frac{1}{\sqrt{3}}\) , then x = \(\frac{5\pi}{6}\) or  \(\frac{11\pi}{6}\) 

Therefore, the possible solutions of above equations are 

x = \(\frac{\pi}{6}\) , \(\frac{5\pi}{6}\) , \(\frac{7\pi}{6}\)  and \(\frac{11\pi}{6}\) Where 0 ≤ x ≤ 2π

We have: \(\sqrt{3 }\)cos x − sin x = 1 

Dividing both sides by 2, we get 

cos x. \(\frac{\sqrt{3}}{2}\) − sin x . \(\frac{1}{2}\) = \(\frac{1}{2}\) 

⇒ cos x . cos \(\frac{\pi}{6}\) − sin x . sin \(\frac{\pi}{6}\) = \(\frac{1}{2}\) 

⇒ cos (x + \(\frac{\pi}{6}\) ) = cos ( \(\frac{\pi}{3}\) ) 

⇒ x + \(\frac{\pi}{6}\) = \(2n\pi\) ± \(\frac{\pi}{3}\)

⇒ x = \(2nx\) ± \(\frac{\pi}{3}\) − \(\frac{\pi}{6}\) 

⇒ x = \(2n\pi\) + \(\frac{\pi}{3}\) − \(\frac{\pi}{6}\) or x = \(2nx\) − \(\frac{\pi}{3}\) −   \(\frac{\pi}{6}\)  

⇒ x = \(2n\pi\) + \(\frac{\pi}{6}\) or x = \(2n\pi\) − \(\frac{\pi}{2}\)

We have: 3 tan x + cot x = 5 cosec x 

⇒ \(3(\frac{sinx}{cosx})+(\frac{cosx}{sinx})=\frac{5}{sinx}\)

⇒ \(3\space\frac{sinx}{cosx}=\frac{5-cosx}{sinx}\) 

⇒ \(3sin^2x=5cosx-cos^2x\) 

⇒\(3 (1-cos^2x)=5cosx-cos^2x\)

⇒ \(3-3cos^2x+cos^2x-5cosx=0\)

⇒ \(2cos^2x+5cosx-3=0\) 

⇒ \((2cosx-1)(cosx+3)=0\) 

∴ \(cosx+3=0\)  or  \(cosx+3=0\) 

⇒ \(cosx=\frac{1}{2} \)  or  \(cosx=\frac{-3}{2}\) 

As range of cos x is [– 1, 1] 

⇒ ∴ \(cosx =\frac{1}{2}\) 

⇒ \(cosx=cos\frac{\pi}{3}\) 

⇒\(x=2n\pi ±\frac{\pi}{3},n∈z\)

We have: \(\sqrt{2}secθ-tanθ=\sqrt{3}\)

⇒ \(\frac{\sqrt{2}}{cosθ}-\frac{sinθ}{cosθ}=\sqrt{3}\) 

⇒ \(\sqrt{2}-sinθ =\sqrt{3}cosθ\) 

⇒ \(\sqrt{3}cosθ+sinθ=\sqrt{2}\) 

Divide the equation by 2 

\(\frac{\sqrt{3}}{2}cosθ+\frac{1}{2}sinθ=\frac{\sqrt{2}}{2}\)

⇒ \(cos(\frac{\pi}{6})cosθ+sin(\frac{\pi}{6})sinθ=\frac{1}{\sqrt{2}}\)

⇒ \(cos(θ-\frac{\pi}{6})=cos(\frac{\pi}{4})\)

⇒ \(θ-\frac{\pi}{6}=2n\pi ± \frac{\pi}{4}\)

⇒ \(θ=2n\pi ±\frac{\pi}{4}+\frac{\pi}{6}\) 

⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6}\)  or  \(θ=2n\pi -\frac{\pi}{4}+\frac{\pi}{6}\)

⇒ \(θ=2n\pi +\frac{5\pi}{12}\)  or  \(θ=2n\pi-\frac{\pi}{12}\)

According to the question,

We have,

tan θ = -1

And cos θ =1/√2 .

⇒ θ = – π/4

So, we know that,

θ lies in IV quadrant.

θ = 2π – π/4 = 7π/4

So, general solution is θ = 7π/4 + 2 n π, n∈ Z

We have tan θ = -1 and cos θ =1/√2 .

So, θ lies in IV quadrant.

θ = 7π/4

So, general solution is θ = 7π/4 + 2 n π, n ∈ Z

Here l = 37.4 cm, θ = 60° = π/3 rad

But r = l/θ = (37.4 x 3)/π = (34.4 x 3 x 7)/22

= 35.7cm

Given T_n = sinⁿx + cosⁿx

LHS = 6T₁₀ – 15 T₈ + 10T₆ – 1

= 6 (sin¹⁰x + cos¹⁰x) – 15 (sin⁸x + cos⁸x) + 10 (sin⁶x + cos⁶x) – 1

= 6 (sin⁶x + cos⁶x) (sin⁴x + cos⁴x) – cos⁴x sin⁴x (sin²x + cos²x) - 15 (sin⁶x + cos⁶x) (sin²x + cos²x) – cos²x sin²x (sin⁴x + cos⁴x) + 10 (sin²x + cos²x) (sin⁴x + cos⁴x – cos²x sin²x) – 1

We know that sin²x + cos²x = 1.

= 6 [(1 – 3 sin²x cos²x) (1 – 2 sin²x cos²x) – sin⁴x cos⁴x] - 15 [(1 – 3 sin²x cos²x) – sin²x cos²x (1 – 2 sin²x cos 2x)] + 10 (1 – 3 sin²x cos²x) – 1

= 6 (1 – 5 sin²x cos²x + 5 sin⁴x cos⁴x) – 15 (1 – 4 sin²x cos²x + 2 sin ⁴x cos⁴x) + 10 (1 – 3 sin²x cos²x) – 1

= 6 – 30 sin²x cos²x + 30 sin⁴x cos⁴x – 15 + 60 sin²x cos²x - 30 sin⁴x cos⁴x + 10 – 30 sin²x cos²x – 1

= 6 – 15 + 10 – 1

= 0

= RHS

Hence proved.

We have, 

(\(1+cos \frac{\pi}{8}\) ) (\(1+cos\frac{3\pi}{8}\) ) (\(1+cos\frac{5\pi}{8}\) ) (\(1+cos\frac{7\pi}{8}\) ) 

= (\(1+cos\frac{\pi}{8}\)) (\(1+sin(\frac{\pi}{2}-\frac{3\pi}{8})\)) (\(1+sin(\frac{\pi}{2}-\frac{5\pi}{8})\)) (\(1+cos(\pi -\frac{\pi}{8})\)) 

= (\(1+cos\frac{\pi}{8}\)) (\(1+sin\frac{\pi}{8}\)) . (\(1-sin\frac{\pi}{8}\)) (\(1-cos\frac{\pi}{8}\) ) 

= (\(1-cos^2\frac{\pi}{8}\) ) (\(1-sin^2\frac{\pi}{8}\) ) 

= \(sin^2\frac{\pi}{8}cos^2\frac{\pi}{8}\) 

= \(\frac{1}{4}(2sin\frac{\pi}{8}cos\frac{\pi}{8})^2\) 

=\(\frac{1}{4}sin^2(\frac{2\pi}{8})\)

= \(\frac{1}{4}sin^2(\frac{2\pi}{8})\)

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

= \(\frac{1}{4}\times \frac{1}{2}\)= \(\frac{1}{8}\)

L.H.S = cos 10° + cos 110° + cos 130° 

= (cos 10° + cos 110°) cos 130° 

= 2 cos (\(\frac{110+10°}{2}\)) cos ( \(\frac{110-10°}{2}\) ) + cos 130° 

[∵ cosC + cosD = 2 cos ( \(\frac{C+D}{2}\) ) cos ( \(\frac{C-D}{2}\))] 

= 2 cos 60° cos 50° + cos 130° 

= 2 × \(\frac{1}{2}\) cos 50° + cos (180° − 50) 

= cos 50° - cos 50° 

= 0 = R. H.S       Hence Proved

False

Explanation:

Let, sin 10° > cos 10°

⇒ sin 10° > cos (90° - 80°)

⇒ sin 10° > sin80°

Which is not possible since value of sin increases with increase in θ.

Hence, statement is false.

False

Explanation:

We know that, maximum value of sin A is 1.

Given that, sin A + sin 2A + sin 3A = 3

It is possible when sin A = sin 2A = sin 3A=1

But sin2A and sin3A is not equal to 1.

Hence, statement is false.

Answer is False

We have 13x = 9x + 4x 

⇒ tan 13x = tan(9x + 4x) 

We know that tan(A + B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ tan 13x = \(\frac{tan9x+tan4x}{1-tan9x\,tan4x}\)

⇒ tan 13x(1 – tan 9x tan 4x) = tan 9x + tan 4x 

⇒ tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x 

∴ tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x 

Hence proved.

B. 0

Explanation:

According to the question,

Since cos90° =0

We get,

⇒ cos 1° cos 2° cos 3°… cos90°… cos 179° = 0

Thus, option (B) 0 is the correct answer.