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We know that the angle of an n sided regular Polygon is equal to \((\frac{2n-4}{n})\) right angles. 

Let θ be the angle of a regular decagon. Then, 

\(θ=(\frac{2\times10-4}{10})=\frac{8}{5}\)  right angles 

 \(\therefore θ=(\frac{8}{5}\times90)^°=144^°\)         [∴ 1 right angle = 90°] 

Also, \(\thereforeθ=(\frac{8}{5}\times100)=160^g\)      [∴ 1 right angle = 100^g ]

And \(θ=(\frac{8}{5}\times\frac{\pi}{2})^c=(\frac{4\pi}{2})^c\) 

[∴ 1 right angle = \((\frac{\pi}{2})^c\) ]

We know that, 60” = 1’ ⇒ 30” = \(\frac{30}{60}\) = 0.5′ 

Now, we have 5° 37’30” = 5° 37.5’ 

We know that, 60’ = 1° 

⇒ 37.5’ = \(\frac{37.5}{60}\) = 0.625 

Now, we have 5° 37.55’ = 5.625° 

We know that 180° = π radian 

⇒ 5.625° = 0.03125π radian.

– 37° 30’ = – 37.5°          [∵ 1° = 60′ ⇒ 0.5° = 30′] 

Now, \(-37.5^°=-(37.5\times\frac{\pi}{180})\) radian 

[∴ \(1^°=\frac{\pi}{180}\) radian ]

= − ( \(\frac{375}{1800}\times\pi\)) radian 

= − \(\frac{5\pi}{24}\) radian

sin (– 1125°) = – sin(1125°) 

= – sin (3 × 360° + 45°) 

= – sin 45° 

= − \(\frac{1}{\sqrt{2}}\)

Answer is (A) 

3A= A+ 2A

=> tan 3A = tan (A + 2A)

=> tan 3 A = (tanA + tan2A)/( 1 – tan A . tan 2A)

=> tan A + tan 2A = tan 3A – tan 3A• tan 2A . tan A

=> tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Consider cot 3x = cot(2x + x)

= (cos2xcotx - 1)/(cot2x + cotx)

⇒ cot 3x(cot 2x+cot x) – cot 2x cot x -1 

⇒ 1 = cot 2x cot x – cot 3x cot 2x – cot 3x cot x

Correct option is: (a) 63/65

Correct answer is A.

Given A = 2 sin² x – cos 2x

[using cos 2x = 1 – 2 sin² x]

so A = 2 sin² x – cos 2x = 2 sin² x –[ 1 – 2 sin² x]

= 2 sin² x -1 + 2 sin² x]

= 4 sin² x – 1

Now A = 2 sin² x – cos 2x = 4 sin² x – 1

As we know sin x lies between -1 and 1

-1 ≤ sin x ≤ 1

0 ≤ sin²x ≤ 1

Multiplying the inequality by 4

0 ≤ 4 sin² x ≤ 4

Subtracting 1 from the inequality

-1 ≤ (4 sin² x – 1) ≤ 3

From the above inequation, we can say that

A = (4 sin² x – 1) belongs to the closed interval [-1, 3]

Correct option is C. \(\cfrac{\pi}6\) 

Given: tan x + sec x = √3

squaring on both sides

(tan x + sec x)² = √3²

tan²x+sec²x + 2 tan x sec x = 3

Also, sec²x - tan²x = 1

tan²x + 1+ tan²x+ 2tan x sec x = 3

2tan²x + 2tan x sec x = 3 - 1

tan²x + tan x sec x = 2/2

tan²x + tan x sec x = 1

tan x sec x = 1 - tan²x

again, squaring on both sides

tan²x sec²x = 1 + tan⁴x - 2 tan² x

(1+ tan²x) tan²x = 1 + tan⁴x - 2 tan²x

Tan⁴x + tan²x = 1 + tan⁴x - 2 tan²x

3 tan²x = 1

tan x = 1/√3

x = π/6.

We know that the hour hand completes one rotation in 12 hours while the minute hand completes one rotational in 60 minutes.

∴ Angle traced by the hour hand in 7 hours 20 minutes i.e., \(\frac{22}{3}\) Hours

\(= (\frac{360}{12}\times\frac{22}{3})\)° = 220°

Also, the angle traced by the minute hand in 60 minutes = 360°

⇒ The angle traced by the minute hand in 20 minutes = \( (\frac{360}{60}\times20)\)= 120°

Hence, the required angle between two hands = 220° - 120°

= 100°

The maximum and minimum value of cos x is 1 or – 1. 

Hence, cos (cos x) has minimum value at x = 0° 

i. e., cos (cos 0°) = cos (1) = cos 1

Answer is False

Answer is (C)

Given,

 sin θ + cos θ = 1

Sq. both side

(sin θ + cos θ)² = 1²

sin² θ + cos² θ + 2sin θ .cos θ = 1

1 + 2sin θ .cos θ = 1

2sin θ .cos θ = 1 -1 = 0

sin 2θ = 0   (sin 2θ = 2sin θ .cos θ )

sin 4A = sin (2A + 2A)

We know that,

sin(A + B) = sin A cos B + cos A sin B

Therefore, sin 4A = sin 2A cos 2A + cos 2A sin 2A

⇒ sin 4A = 2 sin 2A cos 2A

From T-ratios of multiple angle,

We get,

sin 2A = 2 sin A cos A and cos 2A = cos²A – sin²A

⇒ sin 4A = 2(2 sin A cos A)(cos²A – sin²A)

⇒ sin 4A = 4 sin A cos³A – 4 cos A sin³A

Hence, sin 4A = 4 sin A cos³A – 4 cos A sin³A

L.H.S. = sin 4A

= 2 sin 2A- cos 2A 

= 2(2 sin A cosA)(cos² A – sin² A)

= 4 sin A • cos³ A – 4 cos A sin³ A = R.H.S.

 sin(π/2 - x)

= cos[π/2 - (π/2 - x)]

∵ cos(π/2 - x) = sinx

= cosx 

LHS = cos 570° sin 510°+ sin (-330°) cos (-390°)

We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

= cos 570° sin 510°+ [sin (-330°)] cos (-390°)

= cos 570° sin 510° - sin (-330°) cos (-390°)

= cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)

We know that cos is negative at 90° + θ i.e. in Q₂ and when n is odd, sin → cos and cos → sin.

= -cos 30° cos 60° - [-cos 60°] cos 30°

= -cos 30° cos 60° + cos 60° cos 30°

= 0

= RHS

Hence proved.

LHS = tan (-225°) cot (-405°) – tan (-765°) cot (675°) 

We know that  tan (-x) = -tan (x) and cot (-x) = -cot (x).

= [-tan (225°)] [-cot (405°)] – [-tan (765°)] cot (675°)

= tan (225°) cot (405°) + tan (765°) cot (675°)

= tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°) 

= tan 45° cot 45° + tan 45° [-tan 45°]

= 1 × 1 + 1 × (-1)

= 1 – 1

= 0

= RHS

Hence proved.

The vales must be following:

Cos 510° = cos (90° × 5 + 60° = - sin 60°) = \(\frac{-\sqrt 3}{2}\)

Cos 330° = cos (90° × 3 + 60° = sin 60°) = sin 60° = \(\frac{-\sqrt 3}{2}\)

Sin 390° = sin (90° × 4 + 30°) = sin 30° = \(\frac{1}{2}\)

Cos 120° = cos (90° × 1 + 30°) = – sin 30° = \(\frac{-1}{2}\)

Now,

L.H.S = cos 510° cos 330° + sin 390° cos 120°

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

= \((\frac{-3}{4})+(\frac{-1}{4})\)

= \((\frac{-3}{4}+\frac{-1}{4})\)

= − 1 = R. H. S

LHS = ksinAsin(B - C) + ksinBsin(C – A) + ksinCsin(A – B) 

= k[sin(B + C)sin(B – C)+sin(C + A)sin(C – A) + sin(A + B) sin(A – B)] 

= K[sin2B – sin2C + sin2  C -sin2A + sin2A – sin2B] 

= k x 0 =0

= RHS.

Answer is (B) 

We know that, in first quadrant if θ is increasing, then sin θ is also increasing.

∴sin 1° < sin 1 [∵ 1 radian = 57◦30′]

Answer is (D) π/4

Answer is (D) 

3 cos x + 4sin x + 8 = 5 (3/5 cos x + 4/5sin x) + 8

= 5(sin α cos x + cos α sin x) + 8

= 5 sin(α + x) + 8, where tan α = 3/4

Answer is (D) No value of θ is possible

Given, cos (p sinx) = sin (p cosx) 

⇒ sin (\(\frac{\pi}{2}\)− p sinx ) = sin (p cosx) 

⇒ \(\frac{\pi}{2}\) − p sinx = πcosx 

⇒ p (sinx + cosx) = \(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p (sinx \(\frac{1}{\sqrt{2}}\) + cosx \(\frac{1}{\sqrt{2}}\) ) =\(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p [sinx cos\(\frac{\pi}{4 }\)+ cosx sin \(\frac{\pi}{4 }\) ] = \(\frac{\pi}{2}\)

⇒ \(\sqrt{2 }\) p [sin (x + \(\frac{\pi}{4 }\) )] = \(\frac{\pi}{2}\)

⇒ sin (x + \(\frac{\pi}{4 }\)) = \(\frac{\pi}{2\sqrt{2}p}\) = \(\frac{\pi}{(2)\frac{1}{2^p}}\) 

Since, sine lies between – 1 and 1 

Therefore, − 1 ≤ \(\frac{\pi}{(2)\frac{1}{2^p}}\) ≤ 1 

Hence, the minimum value of p is \(\frac{\pi}{(2)\frac{1}{2}}\)

[∴ Positive value of p is to taken]

Given, 5 cosx – 12 sinx + 7 

= 13 (\(\frac{5}{13}cosx - \frac{12}{13}sinx\)) + 7 

Using formula, sin (A + B) = sin A cos B + cos A sin B 

Suppose, sin A = \(\frac{5}{13}\) and cos A = \(\frac{12}{13}\)

= 13 (sin A cosx – cos A sinx) + 7 

= 13 sin (A – x) + 7 

We know the range of sin x is [– 1, 1] 

so, the range of 13 sin (A – x) + 7 is [– 1 × 13 + 7, 1 × 13 + 7] 

= [– 6, 20]

Suppose f(x) = sin 100° – sin 10°

On dividing and multiplying by √(1² + 1²) i.e. by √2,

f(x) = √2(1/√2 sin 100° – 1/√2 sin 10°)

f(x) = √2(cos π/4 sin (90 + 10)° – sin π/4 sin 10°) (since, 1/√2 = cos π/4 and 1/√2 = sin π/4)

f(x) = √2(cos π/4 cos 10° – sin π/4 sin 10°)

As we know that cos A cos B – sin A sin B = cos (A + B)

f(x) = √2 cos (π/4 + 10°)
∴ f(x) = √2 cos 55°

Suppose f(x) = (2√3 + 3)sin x + 2√3 cos x

Here, A = 2√3, B = 2√3 + 3 and C = 0

– √[(2√3)² + (2√3 + 3)²] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)² + (2√3 + 3)²]
– √[12 + 12 + 9 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[12 + 12 + 9 + 12√3]

– √[33 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[33 + 12√3]

– √[15 + 12 + 6 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[15 + 12 + 6 + 12√3]

As we know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5

Therefore, if we replace, (12√3 + 6 with 12√5) the above inequality still holds.

Therefore by rearranging the above expression √(15+12+12√5)we get, 2√3 + √15

– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15

Thus proved.

B. sin 1°< sin 1

We know that, 1 radian = 180°/π = 57°30’ approx.

57° lies between 0 and 90 degrees and since in first quadrant sin θ increases when θ increases.

⇒ sin 1°< sin 1

We have: \(\sqrt{3}cosθ+sinθ=\sqrt{2}\) 

Dividing both sides by 2, we get 

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

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

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

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

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

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

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

Given,

a cos θ + b sin θ = m …(1)

a sin θ – b cos θ = n …(2)

Squaring and adding equation 1 and 2, we get –

(a cos θ + b sin θ)² + (a sin θ – b cos θ)² = m² + n²

⇒ a²cos²θ + b²sin²θ + 2ab sin θ cos θ + a²sin²θ + b²cos²θ - 2ab sin θ cos θ = m² + n²

⇒ a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n²

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = m² + n²

Using: sin²θ + cos²θ = 1, we get –

⇒ a² + b² = m² + n²

Answer is (D) √3/2 + 1

2 sin 5π/12  cos π/12

= 2 sin 75° cos 15°

= 2 sin (90° – 15°) cos 15°

= 2 cos 15° cos 15°

= 2 cos² 15°

= cos² × 15° + 1

= cos 30° + 1

= √3/2 + 1 

Answer is (A) 2π/3

LHS = cos(π/4)cosx - sin(π/4)sinx + cos(π/4)cosx + sin(π/4)sinx

= 2cos(π/4)cosx = 2(1/√2)cosx = √2cosx

= RHS

We have 

\(\sqrt{3}\) cosec 20° − sec 20° 

= \(\frac{\sqrt{3}}{sin20°}-\frac{1}{cos20°}\) 

= \(\frac{\sqrt{3}\space cos20°-sin20°}{sin20°cos20°}\)

= 4 \(\bigg(\frac{\frac{\sqrt{3}}{2}cos20°-\frac{1}{2}sin20°}{2sin20°cos20°}\bigg)\)

= 4 \(\bigg(\frac{sin60°cos20°-cos60°sin20°}{sin40°}\bigg)\)

= 4 \(\bigg(\frac{sin(60°-20)}{sin40°}\bigg)\) 

= 4

The literal meaning of the word trigonometry is the science of measuring the sides and the angles of triangles.

Given sin 17π

⇒ sin 17π = sin 3060°

⇒ 3060° = 90° × 34 + 0°

3060° is in negative direction of x - axis i.e. on boundary line of II and III quadrants.

∴ sin (3060°)= sin(90° × 34 + 0°)

= -sin 0°

= 0

Given cosec (-\(\cfrac{20\pi}3\))

⇒ cosec (-\(\cfrac{20\pi}3\)) = cosec(-1200°)

⇒ cosec (-1200°) = cosec (1200°)

= cosec (90° × 13 + 30)

1200° lies in second quadrant in which cosec function is positive.

∴ cosec (-1200°) = -cosec (90° × 13 + 30°)

= -sec 30°

= \(-\cfrac{2}{\sqrt3}\)

Given tan(\(-\cfrac{13\pi}4\))

⇒ tan \(-\cfrac{13\pi}4\) = tan -585°

⇒ -tan 585° = -tan (90° × 6 + 45°)

585° lies in the third quadrant in which the tangent function is positive.

∴ tan (-585)° = -tan (90° × 6 + 45°)

= - (tan 45°)

= -1

Given sin(\(\cfrac{41\pi}4\))

⇒ sin(\(\cfrac{41\pi}4\)) = sin 1845°

⇒ sin 1845° = 90° × 20 + 45°

1845° lies in the first quadrant in which the sine function is positive.

∴ sin 1845° = sin (90° × 20 + 45°)

= sin 45°

= \(\cfrac1{\sqrt2}\)

tan(19π/3) = tan(6π + π/3) = tan(π/3) = √3

Given cos \(\cfrac{19\pi}4\)

⇒  cos \(\cfrac{19\pi}4\) = cos 855°

⇒ 855° = 90° × 9 + 45°

855° lies in the second quadrant in which the cosine function is negative.

∴ cos 855° = cos (90° × 9 + 45°)

= -sin 45°

= \(\cfrac{-1}{\sqrt2}\)

Given cos \(\cfrac{19\pi}6\)

⇒  cos \(\cfrac{19\pi}6\) = cos 570°

⇒ 570° = (90° × 6 + 30°)

570° lies in third quadrant in which cosine function is negative.

∴ cos (570°) = cos (90° × 6 + 30°)

= -cos (30°)

= \(-\cfrac{\sqrt3}2\)

To find: Value of sin 405°

We have,

sin 405° = sin [90° × 4 + 45°]

= sin 45°

[Clearly, 405° is in Ist Quadrant and the multiple of 90° is even]

= \(\frac{1}{\sqrt{2}}\) [∵ sin45° = \(\frac{1}{\sqrt{2}}\)]

sin765° = sin(720° + 45°) = sin45° = 1/√2

we know that values of sin x repeats after an interval of 2π

So, sin (765°) = sin (720° + 45°)

= sin (2 x 360° + 45°)

= sin 45°

(∵ After repetition it will comes in 1st quardrant)

= 1/√2

Hence, sin 765° = 1/√2

Answer is (B) tan θ

(i) As we know that sin (A – B) = sin A cos B – cos A sin B

sin 78^° cos 18^° – cos 78^° sin 18^° = sin(78 – 18)°

= sin 60°

= √3/2

(ii) As we know that cos A cos B – sin A sin B = cos (A + B)

cos 47^° cos 13^° – sin 47^° sin 13^° = cos (47 + 13)°

= cos 60°

= 1/2

(iii) As we know that sin (A + B) = sin A cos B + cos A sin B

sin 36^° cos 9^° + cos 36^° sin 9^° = sin (36 + 9)°

= sin 45°

= 1/√2

(iv) As we know that cos A cos B + sin A sin B = cos (A – B)

cos 80^° cos 20^° + sin 80^° sin 20^° = cos (80 – 20)°

= cos 60°

= 1/2

As we know that, 7π/12 = 105°, π/12 = 15°; 5π/12 = 75°

Let us consider the LHS: cos 105° + cos 15°

cos (90° + 15°) + sin (90° – 75°)

-sin 15° + sin 75°

sin 75° – sin 15°

= RHS

∴ LHS = RHS

Thus proved.