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Answer is (B) 

Since cos 90° = 0, we have

cos 1° cos 2° cos 3° …cos 90°… cos 179° = 0

Correct option is C. sec x = 1/2

Sec x = 1/2 is incorrect because for no real value of x sec x attains 1/2.

Correct option is B. 0

Cos 1 × cos 2 × cos 3 × ......× cos 179

= cos 1 × cos 2 × cos 3 × .....× cos 90 × ..... × cos 179

= cos 1 × cos 2 × cos 3 × .....× 0 × ..... × cos 179

= 0 × cos 1 × cos 2 × cos 3 × ....... × cos 179

= 0

L.H.S 

= sin 10° sin 30° sin 50° sin 70° 

= \(\frac{1}{2}\) (sin 10° sin 50° sin 70° )

= \(\frac{1}{2}\) (sin 50° sin 70°) sin 10° 

= \(\frac{1}{2}\) [sin (60° − 10°) sin (60° + 10°)] sin 10° 

= \(\frac{1}{2}\) [(\(sin^260°-sin^210°\))] sin 10° 

= \(\frac{1}{2}\)(( \(\frac{3}{2}\))² − \(sin^210°\)) sin 10° 

= \(\frac{1}{2}\) [ \(\frac{3}{4}\) − \(sin^210°\)] sin 10° 

= \(\frac{1}{8}\) (3 − 4 \(sin^210°\))sin 10° 

= \(\frac{1}{8}\)(3 sin 10° − 4 \(sin^210°\)) 

= \(\frac{1}{11}\) (sin 3 × 10°) 

= \(\frac{1}{8}\) (sin 30°) 

= \(\frac{1}{16}\) = R.H.S

Answer is  1

Answer is (B) 23/10

We have, tanθ + sinθ = m ...... (i)

And tanθ -sinθ = n .......... (ii)

Now, m + n = 2 tanθ

And m – n = 2 sinθ.

(m + n)(m -n) = 4 sinθtanθ

 m² -n² = 4 sinθtanθ

Given, triangle ABC is right angle.

So, let ∠ B = 90°

Then as per the property of angles in a triangle

∠ A + ∠ B + ∠ C = 180°

As ∠ B = 90°

∠ A + 90° + ∠ C = 180°

Then ∠ A + ∠ C = 180° - 90° = 90°

Now, consider sin²A + sin²B + sin²C

As ∠ B = 90°

sin²A + sin²B + sin²C = sin²A + sin²(90°) + sin²C

= sin²A + 1 + sin²C

From before, we know that ∠ A + ∠ C = 90° ; ∠ C = 90° - ∠ A

sin²A + sin²B + sin²C = sin²A + 1 + sin²( 90° - A)

= sin²A + cos²(A) + 1

[by using the identity cos x = sin ( 90° - x)]

sin²A + sin²B + sin²C = (sin 2A + cos 2A) + 1

= 1 + 1

= 2

[by using the identity sin²θ + cos²θ = 1]

Therefore, sin²A + sin²B + sin²C = 2.

LHS = abcosC - ca cosB

= (a² + b² - c²)/2 - (c² + a² - b²)/2

= b² - c²

= RHS

We know,

Range of cos θ is -1.

If cos θ = x + 1/x

if x > 1 then 1/x < 1

Ultimately x + 1/x > 1

Which shows, cos θ > 1 which is not possible.

Angle subtended at the centre by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian. 

Keen eye:

1rad = 180°/π = 57°16'.

If a rotation from the initial side to terminal side is (1/360)^th of a revolution, the angle is said to have a measure of one degree, written as 1°.

Keen eye: A degree is divided into 60 minutes and a minute is divided into 60 seconds. One sixtieth of a degree is called a minute, written as 1′ and one sixtieth of a minute is called second, written as 1″ 

∴ 1 ° – 60′ 

1’= 60″

Number of revolutions in one minute = 360

Number of revolutions in one second = 360/60 = 6

But, in one revolutions angle traced = 360°

= 2π rad

⇒ Angle traced in 6 revolutions

= 6 x 2π = 12π radians

Angle traced in one second = 12π radians

Given, cosx = cosα cosβ 

  ∴ cos β = \(\frac{cosx}{cos\alpha}\)    ...(i) 

We know that, \(tan^2\frac{θ}{2}\) = \(\frac{1-cosθ}{1+cosθ}\) 

R. H. S = \(tan^2\frac{θ}{2}\) = \(\frac{1-cos\beta}{1+cos\beta}\) 

⇒ \(\frac{1-\frac{cos\alpha}{cosx}}{1+\frac{cosx}{cos\alpha}}\)

⇒\(\frac{cos\alpha-cosx}{cos\alpha +cosx}\) 

= \(\frac{2sin(\frac{a+x}{2})sin(\frac{x-a}{2})}{2cos(\frac{x+a}{2})sin(\frac{x-a}{2}))}\)

= \(tan(\frac{x+a}{2}).tan(\frac{x-a}{2})\) 

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

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

= L. H. S    Hence Proved

Given, α tanθ + b secθ = c    ...(i) 

⇒ (α tanθ − c)² = b² (1 + tan²θ) 

⇒ (a² tan²b² ) − 2ac tanθ + c² = b² + b² tan²θ 

⇒ (a² − b²)  tan²θ − 2ac tanθ + c² − b² = 0 

Since, α and β are roots of the equation (i), 

we have 

tan α + tan β = \(\frac{2ac}{a^2-b^2}\) and tan α.tan β = \(\frac{c^2-b^2}{a^2-b^2}\) 

Therefore, tan (α + β) = \(\frac{tan\alpha tan\beta}{1-tan\alpha tan\beta}\)

=\(\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\)

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

Given, tan A – tan B = x    ...(i) 

and  cot B – cot A = y    ...(ii) 

From (ii), cot B – cot A = y 

⇒ \(\frac{1}{tanB}-\frac{1}{tanA}=y\)

⇒ \(\frac{tanA - tanB}{tanAtanB}=y\) 

⇒ \(\frac{x}{tanAtanB}=y\) 

⇒ tanA tanB = \(\frac{x}{y}\) 

Now, ⇒ cot (A − B) = \(\frac{1}{tan(A-B)}\)= \(\frac{1+tanAtanB}{tanA-tanB}\)

⇒ cot (A − B) = \(\frac{1+\frac{x}{y}}{x}\)= \(\frac{y+x}{xy}\) 

⇒ cot (A − B) = \(\frac{1}{x}+\frac{1}{y}\)Hence Proved

Here,

2 tan² x + sec² x = 2

which gives tan x = ±1/√3

If we take  tan x = 1/√3, then

x = π/6 or 7π/6

Again, if we take  tan x = -1/√3, then

x = 5π/6 or 11π/6

Therefore, the possible solutions of above equations are

x = 5π/6, 11π/6, 5π/6 and 11π/6 

where 0 ≤ x ≤ 2π

According to the question,

tan θ + sin θ = m …(i)

tan θ – sin θ = n …(ii)

Adding equation i and ii,

2 tan θ = m + n …(iii)

Subtracting equation ii from i,

We get,

2sin θ = m – n …(iv)

Multiplying equations (iii) and (iv),

2sin θ (2tan θ) = (m + n)(m – n)

⇒ 4 sin θ tan θ = m² – n²

Hence,

m² – n² = 4 sin θ tan θ

LHS =  = cos 24° + cos 55° + cos 125° + cos 204° + cos 300°

= cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) + cos (90° × 2 + 24°) + cos (90° × 3 + 30°)

We know that when n is odd, cos → sin.

= cos 24° + sin 35° - sin 35° - cos 24° + sin 30°

= 0 + 0 + 1/2 = 1/2

= RHS

Hence proved.

Let us consider the LHS: sin⁶ x + cos⁶ x

(sin² x)³ + (cos² x)³

On using the formula, a³ + b³ = (a + b) (a² + b² – ab)

(sin² x + cos² x) [(sin² x)² + (cos² x)² – sin² x cos² x]

On using the formula, sin² x + cos² x = 1 and a² + b² = (a + b)² - 2ab

1 × [(sin² x + cos² x)² – 2sin² x cos² x – sin² x cos² x

1² – 3sin² x cos² x

1 – 3sin² x cos² x

= RHS

∴ LHS = RHS

Thus proved.

Let us consider the LHS:

(sec x sec y + tan x tan y)² – (sec x tan y + tan x sec y)²

By expanding the above equation we get,

[(sec x sec y)² + (tan x tan y)² + 2(sec x sec y) (tan x tan y)] – [(sec x tan y)² + (tan x sec y)² + 2(sec x tan y) (tan x sec y)] [sec² x sec² y + tan² x tan² y + 2(sec x sec y) (tan x tan y)] – [sec² x tan² y + tan² x sec² y + 2(sec² x tan² y) (tan x sec y)]

sec² x sec² y – sec² x tan² y + tan² x tan² y – tan² x sec² y

sec² x(sec² y – tan² y) + tan² x(tan² y – sec² y)

sec² x(sec² y – tan² y) – tan² x(sec² y – tan² y)

As we know, sec² x – tan² x = 1.

sec² x × 1 – tan² x × 1

sec² x – tan² x

1 = RHS

∴ LHS = RHS

Thus proved.

Let us consider the LHS: sec⁴ x – sec² x

(sec² x)² – sec² x

On using the formula, sec² θ = 1 + tan² θ.

(1 + tan² x)² – (1 + tan² x)

1 + 2tan² x + tan⁴ x – 1 – tan² x

tan⁴ x + tan² x

= RHS

∴ LHS = RHS

Thus proved.

LHS = sin \(\cfrac{10\pi}3\) cos \(\cfrac{13\pi}6\) + cos \(\cfrac{8\pi}3\) sin \(\cfrac{5\pi}6\) 

= sin 600° cos 390° + cos 480° sin 150°

= sin (90° × 6 + 60°) cos (90° × 4 + 30°) + cos (90° × 5 + 30°) sin (90° × 1 + 60°)

We know that when n is odd, sin → cos and cos → sin.

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

= -sin 60° cos 30° - sin 30° cos 60°

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

We know that sin A cos B + cos A sin B = sin (A + B)

= -sin (60° + 30°)

= -sin 90°

= -1

= RHS

Hence proved.

Let us consider the LHS

sin² 2π/5 – sin² π/3 = sin² (π/2 – π/10) – sin² π/3

As we know, sin (90°– A) = cos A

Therefore, sin² (π/2 – π/10) = cos² π/10

Sin π/3 = √3/2

Now the above equation becomes,

= Cos² π/10 – (√3/2)²

As we know, cos π/10 = √(10+2√5)/4

Then, the above equation becomes,

= [√(10 + 2√5)/4]² – 3/4

= [10 + 2√5]/16 – 3/4

= [10 + 2√5 – 12]/16

= [2√5 – 2]/16

= [√5 – 1]/8

= RHS

Thus proved.

Given,

A + B = \(\frac{\pi}{4}\)

Taking tangent both sides, we get

tan(A + B) = tan\(\frac{\pi}{4}\)

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

or, tan A + tan B = 1 – tan A tan B

or, tan A + tan B + tan A tan B = 1

Now, adding 1 both sides, we get

tan A + tan B + tan A tan B + 1 = 2

(tan A + 1) + tan B (1 + tan A) = 2

or, (tan A + 1) (1 + tan B) = 2

or, (1 + tan A) (1 + tan B) = 2

We have:

2 cos²x + 3 sinx = 0

2 cos²x= – 3 sinx

Squaring both sides, we get 

4 cos⁴x = 9 sin²x

4 cos⁴x – 9 (1 – cos²x) = 0

Let cos²x = y

4y² + 9y – 9 = 0

4y² + 12y – 3y – 9 = 0

4y (y + 3) – 3 (y + 3) = 0

(y + 3) (4y – 3) = 0

y = – 3 or, y = \(\frac{3}{4}\)

cos²x = – 3 or cos²x = \(\frac{3}{4}\)

∴  cos²x = \(\frac{3}{4}\) = cos²\((\frac{\pi}{6})\) 

x = nπ ±  \(\frac{\pi}{6}\)

y = 7 cos x + 24 sin x

y² = (7 cos x + 24 sinx)²

= 49 cos²x + 576 sin²x + 2 × 7× 24 cos x sin x

= 49 – 49 sin² x + 576 – 576 cos² x + 2 × 7 × 24 cos x sin x

= 625 – (7 sin x – 24 cos x)²

∴ Maximum value = 25

For maximum value

Cos x = \(\frac{-7}{25}\) and sin x =\(\frac{-24}{25}\)

∴ Minimum value = 7(\(\frac{-7}{25}\)) + 24 (\(\frac{-24}{25}\))

\(=\frac{-49-576}{25}\) = -25

∴ Minimum value = – 25

Answer is True

Answer is (A) (√5 + 1)/8

L.H.S=sin(12°)sin(48°)sin(54°)

= [sin(12°)sin(48°)]sin(54°)

= sin(54°) * [cos(48° - 12°) - cos(48° + 12°)]/2

= sin(54°) * [cos(36°) - cos(60°)]/2

= sin(54°) * [cos(36°) - 1/2]/2

= cos(36°) * [cos(36°)/2 - 1/4]

= cos²(36°)/2 - cos(36°)/4.

Since cos(36°) = (1 + √5)/4:

cos²(36°)/2 - cos(36°)/4

= [(1 + √5)/4]²/2 - [(1 + √5)/4]/4

= (1 + √5)²/32 - (1 + √5)/16

= (6 + 2√5)/32 - (1 + √5)/16

= (6 + 2√5)/32 - (2 + 2√5)/32

= 4/32

= 1/8 R.H.S

(D) is the correct choice, since

sinx cosx = 1/2, sin 2x ≤1/2, since |sin2x | ≤ 1.

Answer is (B) equal to 90 degree

(i) Let us consider the LHS:

(cos 11^° + sin 11^°)/(cos 11^° – sin 11^°)

Let us divide the numerator and denominator by cos 11^° we get,

(cos 11^° + sin 11^°)/(cos 11^° – sin 11^°) = (1 + tan 11^°)/(1 – tan 11^°)

= (1 + tan 11^°)/(1 - 1 × tan 11^°)

= (tan 45^° + tan 11^°)/(1 – tan 45^° × tan 11^°)

As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 45^° + tan 11^°)/(1 – tan 45^° × tan 11^°) = tan (45^° + 11^°)

= tan 56^°

= RHS

∴ LHS = RHS

Thus proved.

(ii) (cos 9^° + sin 9^°)/(cos 9^° – sin 9^°) = tan 54^°

Let us consider the LHS:

(cos 9^° + sin 9^°)/(cos 9^° – sin 9^°)

Let us divide the numerator and denominator by cos 9^° we get,

(cos 9^° + sin 9^°)/(cos 9^° – sin 9^°) = (1 + tan 9^°)/(1 – tan 9^°)

= (1 + tan 9^°)/(1 – 1 × tan 9^°)

= (tan 45^° + tan 9^°)/(1 – tan 45^° × tan 9^°)

As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 45^° + tan 9^°)/(1 – tan 45^° × tan 9^°) = tan (45^° + 9^°)

= tan 54^°

= RHS

∴ LHS = RHS

Thus proved.

(iii) Let us consider the LHS:

(cos 8^° – sin 8^°)/(cos 8^° + sin 8^°)

Let us divide the numerator and denominator by cos 8^° we get,

(cos 8^° – sin 8^°)/(cos 8^° + sin 8^°) = (1 – tan 8^°)/(1 + tan 8^°)

= (1 – tan 8^°)/(1 + 1 × tan 8^°)

= (tan 45^° – tan 8^°)/(1 + tan 45^° × tan 8^°)

As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 45^° – tan 8^°)/(1 + tan 45^° × tan 8^°) = tan (45^° – 8^°)

= tan 37^°

= RHS

∴ LHS = RHS

Thus proved.

Let us consider the LHS

sin² 42° – cos² 78° = sin² (90° – 48°) – cos² (90° – 12°)

= cos² 48° – sin² 12° [since, sin (90 – A) = cos A and cos (90 – A) = sin A]

As we know, cos (A + B) cos (A – B) = cos²A – sin²B

Now the above equation becomes,

= cos² (48° + 12°) cos (48° – 12°)

= cos 60° cos 36° [since, cos 36° = (√5 + 1)/4]

= 1/2 × (√5 + 1)/4

= (√5 + 1)/8

= RHS

Thus proved.

Let us consider the LHS

cos 78° cos 42° cos 36°

Now let us multiply and divide by 2 we get,

cos 78° cos 42° cos 36° = 1/2 (2 cos 78° cos 42° cos 36°)

As we know, 2 cos A cos B = cos (A + B) + cos (A – B)

Now the above equation becomes,

= 1/2 (cos (78° + 42°) + cos (78° – 42°)) × cos 36°

= 1/2 (cos 120° + cos 36°) × cos 36°

= 1/2 (cos (180° – 60°) + cos 36°) × cos 36°

= 1/2 (-cos (60°) + cos 36°) × cos 36° [since, cos(180° – A) = – A]

= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36° = (√5 + 1)/4]

= 1/2 (√5 + 1 – 2)/4 ((√5 + 1)/4)

= 1/2 (√5 – 1)/4) ((√5 + 1)/4)

= 1/2 ((√5)² – 1²)/16

= 1/2 (5 - 1)/16

= 1/2 (4/16)

= 1/8

= RHS

Thus proved.

Let us consider the LHS

sin² 24° – sin² 6°

As we know, sin (A + B) sin (A – B) = sin²A – sin²B

Now the above equation becomes,

sin² 24° – sin² 6° = sin (24° + 6°) – sin (24° – 6°)

= sin 30° – sin 18°

= sin 30° – (√5 – 1)/4 [since, sin 18° = (√5 – 1)/4]

= 1/2 × (√5 – 1)/4

= (√5 – 1)/8

= RHS

Thus proved.

We know that values of sinx repeats after an interval of 2π. sin(31π/3) = sin(10π + π/3) = sin(π/3) = √3/2

Given sin \(\cfrac{151\pi}6\)

⇒ sin \(\cfrac{151\pi}6\) = sin 4530°

⇒ sin 4530° = 90° × 50 + 30°

4530° lies in the third quadrant in which the sine function is negative.

∴ sin 4530° = sin (90° × 50 + 30°)

= - sin 30°

= -1/2

LHS = \(sin\cfrac{8\pi}3cos\cfrac{23\pi}6+cos\cfrac{13\pi}3sin\cfrac{35\pi}6\)

= sin 480° cos 690° + cos 780° sin 1050°

= sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)

We know that when n is odd, sin → cos and cos → sin.

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

= \(\cfrac{\sqrt3}2\times\cfrac{\sqrt3}2-\cfrac12\times\cfrac12\)

= 3/4 - 1/4

= 2/4

= 1/2

= RHS

Hence proved.

(i) Given as

sin (-11π/6)

sin (-11π/6) = sin (-330^°)

= – sin (90 × 3 + 60)^°

Here, 330^° lies in the IV quadrant in which the sine function is negative.

sin (-330^°) = – sin (90 × 3 + 60)^°

= – (-cos 60^°)

= – (-1/2)

Thus, 1/2

(ii) Given as

cosec (-20π/3)

cosec (-20π/3) = cosec (-1200)^°

= – cosec (1200)^°

= – cosec (90 × 13 + 30)^°

Here, 1200^° lies in the II quadrant in which cosec function is positive.

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

= – sec 30^°

= -2/√3

(iii) Given as

tan (-13π/4)

tan (-13π/4) = tan (-585)^°

= – tan (90 × 6 + 45)^°

Here, 585^° lies in the III quadrant in which the tangent function is positive.

tan (-585)^° = – tan (90 × 6 + 45)^°

= – tan 45^°

= -1

(iv) Given as

cos 19π/4

cos 19π/4 = cos 855^°

= cos (90 × 9 + 45)^°

Here, 855^° lies in the II quadrant in which the cosine function is negative.

cos 855^° = cos (90 × 9 + 45)^°

= – sin 45^°

= – 1/√2

(v) Given as

sin 41π/4

sin 41π/4 = sin 1845^°

= sin (90 × 20 + 45)^°

Here, 1845^° lies in the I quadrant in which the sine function is positive.

sin 1845^° = sin (90 × 20 + 45)^°

= sin 45^°

= 1/√2

(vi) Given as

cos 39π/4

cos 39π/4 = cos 1755^°

= cos (90 × 19 + 45)^°

Here, 1755^° lies in the IV quadrant in which the cosine function is positive.

cos 1755^° = cos (90 × 19 + 45)^°

= sin 45^°

= 1/√2

(vii) Given as

sin 151π/6

sin 151π/6 = sin 4530^°

= sin (90 × 50 + 30)^°

Here, 4530^° lies in the III quadrant in which the sine function is negative.

sin 4530^° = sin (90 × 50 + 30)^°

= – sin 30^°

Thus, -1/2

LHS = tan 225° cot 405° + tan 765° cot 675°

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

We know that when n is odd, cot → tan. =

tan 45° cot 45° + tan 45° [-tan 45°]

= tan 45° cot 45° - tan 45° tan 45°

= 1 × 1 – 1 × 1

= 1 – 1

= 0

= RHS

Hence proved.

(i) Let us consider the LHS:

tan 225° cot 405° + tan 765° cot 675°

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

As we know that when n is odd, cot → tan.

tan 45° cot 45° + tan 45° [-tan 45°]

tan 45° cot 45° – tan 45° tan 45°

1 × 1 – 1 × 1

1 – 1

0 = RHS

∴ LHS = RHS

Thus proved.

(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

Let us consider the LHS:

sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6

sin 480° cos 690° + cos 780° sin 1050°

sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)

As we know that when n is odd, sin → cos and cos → sin.

cos 30° sin 60° + cos 60° [-cos 60°]

√3/2 × √3/2 – 1/2 × 1/2

3/4 – 1/4

2/4

1/2

= RHS

∴ LHS = RHS

Thus proved.

(iii) Let us consider the LHS:

cos 24^° + cos 55^° + cos 125^° + cos 204^° + cos 300^°

cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) + cos (90° × 2 + 24°) + cos (90° × 3 + 30°)

As we know that when n is odd, cos → sin.

cos 24° + sin 35° – sin 35° – cos 24° + sin 30°

0 + 0 + 1/2

1/2

= RHS

∴ LHS = RHS

Thus proved.

(iv) Let us consider the LHS:

tan (-125^°) cot (-405^°) – tan (-765^°) cot (675^°)

As 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

∴ LHS = RHS

Thus proved.

(v) Let us consider the LHS:

cos 570^° sin 510^° + sin (-330^°) cos (-390^°)

As 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°)

As 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

∴ LHS = RHS

Thus proved.

(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6 = (3 – 4√3)/2

Let us consider the LHS:

tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6

tan (11 × 180^°)/3 – 2 sin (4 × 180^°)/6 – 3/4 cosec² 180^°/4 + 4 cos² (17 × 180^°)/6

tan 660^° – 2 sin 120^° – 3/4 (cosec 45^°)² + 4 (cos 510^°)²

tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4 [cosec 45°]² + 4 [cos (90° × 5 + 60°)]²

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

[-cot 30°] – 2 cos 30° – 3/4 [cosec 45°]² + [-sin 60°]²

– cot 30° – 2 cos 30° – 3/4 [cosec 45°]² + [sin 60°]²

-√3 – 2√3/2 – 3/4 (√2)² + 4 (√3/2)²

-√3 – √3 – 6/4 + 12/4

(3 – 4√3)/2

= RHS

∴ LHS = RHS

Thus proved.

(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

Let us consider the LHS:

3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4

3 sin 180^°/6 sec 180^°/3 – 4 sin 5(180^°)/6 cot 180^°/4

3 sin 30° sec 60° – 4 sin 150° cot 45°

3 sin 30° sec 60° – 4 sin (90° × 1 + 60°) cot 45°

As we know that when n is odd, sin → cos.

3 sin 30° sec 60° – 4 cos 60° cot 45°

3 (1/2) (2) – 4 (1/2) (1)

3 – 2

1

= RHS

∴ LHS = RHS
Thus proved.

cosec (- 1410°)

= – cosec (1410°)

= – cosec[4(360°) – 30°]

= – cosec (- 30°)

= cosec 30° = 2

Given sin \((\cfrac{-11\pi}6)\)

⇒  sin \(\cfrac{-11\pi}6\) = sin -330°

⇒ -sin 330° = -sin (90° × 3 + 60°)

330° lies in the fourth quadrant in which the sine function is negative.

∴ sin (-330)° = -sin (90° × 3 + 60°)

= - (-cos 60°)

= - (-1/2)

= 1/2

Given sin \(\cfrac{17\pi}6\)

 ⇒ sin \(\cfrac{17\pi}6\) = sin 510°

⇒ 510° = (90° × 5 + 60°)

510° lies in second quadrant in which sine function is positive.

∴ sin (510°) = sin (90° × 5 + 60°)

= cos (60°) = 1/2

sin(- 11π/3) = - sin(11π/3) = - sin(4π - π/3)

= sin(π/3) = √3/2

To find: Value of cot\(\frac{13\pi}{4}\)

We have,

cot\(\frac{13\pi}{4}\)

Putting π = 180°

= cot\(\frac{(13\times180^\circ)}{4}\)

= cot (13 × 45°)

= cot (585°)

= cot [90° x 6 + 45°]

= cot 45°

[Clearly, 585° is in IIIrd Quadrant and the multiple of 90° is even]

= 1 [\(\because\) cot 45° = 1]

sin⁶x + cos⁶x = (sin²x)³ + (cos²x)³

=(sin²x + cos²x)(sin⁴x+cos⁴x - sin²xcos²x)

= 1 (sin⁴x + cos⁴x – sin²xcos²x)

Substituting above value in given equation

⇒ 2(sin⁴x + cos⁴x – sin²xcos²x) - 3(sin⁴ x + cos⁴ x) + 1

⇒ 2sin⁴x + 2cos⁴x – 2sin²xcos²x – 3sin⁴x - 3cos⁴x + 1.

⇒ -sin⁴x - cos⁴x - 2sin²xcos²x + 1

⇒ -[(sin²x)² + (cos²x)² - 2sin²xcos²x] + 1

⇒ -[( sin²x + cos²x)²] + 1

⇒ -1 + 1

⇒ 0

We have 

tan (\(\frac{\pi}{8}\)) = tan ( \(\frac{1}{2},\frac{\pi}{4}\) ) 

Using the half angle formulae, 

\(tan(\frac{\alpha}{8})=\frac{sin\alpha}{1+cos\alpha}\)

\(tan(\frac{\pi}{8})=\frac{sin(\frac{\pi}{4})}{1+cos(\frac{\pi}{4})}\) \(=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{1}{1+\sqrt{2}}\)

Given, \(secβ=\frac{-5}{3},\frac{\pi}{2}<β<\pi\)

∴ \(tan^2β=1-sec^2β\) \(=1-\frac{25}{9}=\frac{-16}{9}\) 

⇒ \(tanβ=\frac{-4}{3}\)                 (∵ \(\frac{p}{2}<β<p\)) 

And \(cot\alpha=\frac{1}{2},\)              \(\frac{\pi}{2}<\alpha<\frac{3\pi}{2}\)

Radius of the circular wire = 7.5 cm 

∴ Length of the circular wire = 2π × 7.5 = \(\frac{1}{2}\) 15 π cm 

[∴ Circumference = 2πr] 

Radius of the hoop = 120 cm 

Let θ be the angle subtended by the wire at the centre of the hoop, Then, 

\(θ=\frac{arc}{radius}\Rightarrow θ = (\frac{15\pi}{120})°\)

\(=(\frac{\pi}{8})°=(\frac{\pi}{8}\times\frac{180}{\pi})°=22°30'\)