Trigonometric Functions

LHS = sec⁴x – sec²x

= (sec²x) ² – sec²x

We know sec² θ = 1 + tan² θ.

= (1 + tan²x) ² – (1 + tan²x)

= 1 + 2tan²x + tan⁴x – 1 - tan²x

= tan⁴x + tan²x = RHS

Hence proved.

LHS = sin⁶x + cos⁶x

= (sin²x)³ + (cos²x)³

We know that a³ + b³ = (a + b) (a² + b² – ab)

= (sin²x + cos²x) [(sin²x)² + (cos²x)² – sin²x cos²x]

We know that sin²x + cos²x = 1 and a² + b² = (a + b)² – 2ab

= 1 × [(sin²x + cos²x)² – 2sin²x cos²x – sin²x cos²x

= 1² - 3sin²x cos²x

= 1 - 3sin²x cos²x = RHS

Hence proved.

LHS = (cosecx – sinx) (secx – cosx) (tanx + cotx)

We know that

We know that sin²x + cos²x = 1.

= 1 = RHS

Hence proved.

LHS = cosecx (secx – 1) – cotx (1 – cosx)

We know that

We know that 1 – cos²x = sin²x.

= RHS

Hence proved.

LHS

We know that

We know that a³ + b³ = (a + b) (a² + b² – ab)

We know that sin²x + cos²x = 1.

= sinx

= RHS

Hence proved.

LHS

We know that

We know that a³ - b³ = (a - b) (a² + b² + ab)

We know that sin²x + cos²x = 1.

We know that

= cosecx × secx + 1

= secx cosecx + 1

= RHS

Hence proved.

LHS

We know that a³ ± b³ = (a ± b) (a² + b²∓ab)

We know that sin²x + cos²x = 1.

= 1 - sinx cosx + 1 + sinx cosx

= 2

= RHS

Hence proved.

LHS = (secx sec y + tanx tan y)² – (secx tan y + tanx sec y)²

= [(secx sec y)² + (tanx tan y)² + 2 (secx sec y) (tanx tan y)] – [(secx tan y)² + (tanx sec y)² + 2 (secx tan y) (tanx sec y)]

= [sec²x sec² y + tan²x tan² y + 2 (secx sec y) (tanx tan y)] – [sec²x tan² y + tan²x sec² y + 2 (sec²x tan² y) (tanx sec y)]

= sec²x sec² y - sec²x tan² y + tan²x tan² y - tan²x sec² y

= sec²x (sec² y - tan² y) + tan²x (tan² y - sec² y)

= sec²x (sec² y - tan² y) - tan²x (sec² y - tan² y)

We know that sec²x – tan²x = 1.

= sec²x × 1 – tan²x × 1

= sec²x – tan²x

= 1

= RHS

Hence proved.

RHS

We know that sin²x + cos²x = 1.

We know that 1 – cos²x = sin²x.

We know that 1 – sin²x = cos²x.

= LHS

Hence proved.

LHS

We know that 1 + tan²x = sec²x and 1 + cot²x = cosec²x

We know that a² + b² = (a + b)² – 2ab

We know that sin²x + cos²x = 1.

= RHS

Hence proved.

LHS

We know that

We know that a³ + b³ = (a + b) (a² + b²-ab)

= 1 – (sin²x + cos²x) + sinx cosx

We know that sin²x + cos²x = 1.

= 1 – 1 + sinx cosx

= sinx cosx

= RHS

Hence proved.

LHS

We know that

We know that sin²x + cos²x = 1.

= RHS

Hence proved.

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)

We know that 1 + tan² θ = sec² θ

= sec² α sec² β

= RHS

Hence proved.

LHS

We know that

We know that a³ - b³ = (a - b) (a² + b²+ab)

= sin²x cos^2x

= RHS

Hence proved.

LHS

We know that 1 – cos²x = sin²x

= cotx

= RHS

Hence proved.

LHS = cosx (tanx + 2) (2 tanx + 1)

= cosx (2 tan²x + 5 tanx + 2)

We know that

= 2 secx + 5 sinx

= RHS

Hence proved.

Given

Rationalizing the denominator,

Hence proved.

Given

We know that sin²x + cos²x = 1→cos²x =1 – sin²x

Given tanx = b/a

Given tanx = a/b

LHS

Dividing by b cosx,

Substituting value of tanx,

= RHS

Hence proved.

Given cosecx – sinx = a³

We know that cosecx = 1/ sinx.

We know that 1 – sin²x = cos²x

… (1)

Also given secx – cosx = b³

We know that secx = 1/ cosx

We know that 1 – cos²x = sin²x

… (2)

Consider LHS = a²b² (a² + b²)

We know that cos² + sin²x = 1

= 1

= RHS

Hence proved.

Given 4m = cotx (1+ sinx) and 4n = cotx (1 – sinx)

Multiplying both equations, we get

⇒ 16mn = cot²x (1 – sin²x)

We know that 1 – sin²x = cos²x

⇒ 16mn = cot²x cos²x

… (1)

Squaring the given equations and then subtracting,

⇒ 16m² = cot²x (1+ sinx)² and 16n² = cot²x (1 – sinx)²

⇒ 16m² – 16n² = cot²x (4 sinx)

Squaring both sides,

… (2)

From (1) and (2),

⇒ (m² – n²) = mn

Hence proved.

Given sinx + cosx = m

We have to prove that

Proof:

LHS = sin⁶x + cos⁶x

= (sin²x)³ + (cos²x)³

We know that a³ + b³ = (a + b) (a² + b²-ab)

= (sin²x + cos²x)³ – 3sin²x cos²x(sin²x + cos²x)

= 1 – 3 sin²x cos²x

RHS

= 1 – 3 sin²x cos²x

LHS = RHS

Hence proved.

Given a = secx – tanx and b = cosecx + cotx

a and b

LHS = ab + a – b + 1

= 0 = RHS

Hence proved.

LHS

[∵ π/2 <x < π and in second quadrant, cosx is negative]

= RHS

Hence proved.

Given T_n = sinⁿx + cosⁿx

LHS

= sin²x cos²x

RHS

= sin²x cos²x

LHS = RHS

Hence proved.

Given T_n = sinⁿx + cosⁿx

LHS = 2T₆ – 3T₄ + 1

= 2 (sin⁶x + cos⁶x) – 3 (sin⁴x + cos⁴x) + 1

= 2 (sin²x + cos²x) (sin⁴x + cos⁴x – cos²x sin²x) – 3 (sin⁴x + cos⁴x) + 1

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 cos²x)] + 10 (1 – 3 sin²x cos²x) – 1

= 6 (1 – 5 sin²x cos²x + 5 sin⁴x cos⁴x) – 15 (1 – 4 sin²x cos²x + 2 sin⁴x 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.

Given cotx = 12/5 andx is in quadrant III

In third quadrant, tanx and cotx are positive and sinx, cosx and secx & cosecx are negative.

We know that

Given cotx = -1/2 andx is in quadrant II

In second quadrant, sinx and cosecx are positive and tanx, cotx and cosx & secx are negative.

We know that

Given tanx = 3/4 andx is in quadrant III

In third quadrant, tanx and cotx are positive and sinx, cosx, secx and cosecx are negative.

We know that

Given sinx = 3/5 andx is in first quadrant.

In first quadrant, all trigonometric ratios are positive.

We know that

Given sinx = 12/13 andx lies in the second quadrant.

In second quadrant, sinx and cosecx are positive and all other ratios are negative.

We know that

We know that tanx = sinx /cosx and secx = 1/cosx

⇒

⇒

∴

Given sinx = 3/5, tan y = 1/2 and

Thus, x is in second quadrant and y is in third quadrant.

In second quadrant, cosx and tanx are negative.

In third quadrant, sec y is negative.

We know that

⇒

⇒

We know that

Given sinx + cosx = 0 andx lies in fourth quadrant.

⇒sinx = -cosx

∴ tanx = -1

In fourth quadrant, cosx and secx are positive and all other ratios are negative.

We know that

Given cosx= -3/5 and π <x < 3π/2

It is in the third quadrant. Here, tanx and cotx are positive and all other rations are negative.

We know that

⇒

⇒

⇒

⇒

Given sin

⇒

300° lies in fourth quadrant in which sine function is negative.

∴

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 tan(11π/6)

⇒

=330°

330° lies in fourth quadrant in which tangent function is negative.

∴

=tan (90° × 3+60°)

=-cot 60°

=

Given

⇒

⇒ cos (-1125°) = cos (1125°)

= cos (90° × 12 + 45°)

1125° lies in first quadrant in which cosine function is positive.

∴ cos (1125°) = cos (90°× 12 + 45°)

= cos (45°)

= 1/√2

Given tan 7π/4

⇒ 315° = (90° × 3 + 45°)

315° lies in fourth quadrant in which tangent function is negative.

∴ tan (315°) = tan (90° × 3 + 45°)

=-cot 45°

=-1

Given

⇒ 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

Given

⇒ 570° = (90° × 6 + 30°)

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

∴ cos (570°) = cos (90° × 6 + 30°)

= -cos (30°)

Given

⇒ -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

⇒ 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°

Given

⇒ -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

⇒ 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°

Given

⇒ 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°

Given

⇒ 1755° = 90° × 19 + 45°

1755° lies in the fourth quadrant in which the cosine function is positive.

∴ cos 1755° = cos (90° × 19 + 45°)

= sin 45°

Given

⇒ 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 = 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.

LHS

= 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°]

= 3/4 - 1/4

= 2/4

= 1/2

= RHS

Hence proved.

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.

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

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.

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

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

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

= cos 570^o sin 510^o - sin (330^o) cos (390^o)

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

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

= RHS

Hence proved.

LHS

= 3 sin 30° sec 60° - 4 sin 150° cot 45°

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

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

Hence proved.

LHS

= 1

= RHS

Hence proved.

LHS

We know that when n is odd, cosec→sec and also sec (-x) = secx.

= 1 + 1

= 2

= RHS

Hence proved.

LHS

We know that cosec (-x) = -cosecx.

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

= 1

= RHS

Hence proved.

LHS

= {1 + cotx – (-cosecx)} {1 + cotx + (-cosecx)}

= {1 + cotx + cosecx} {1 + cotx – cosecx}

= {(1 + cotx) + (cosecx)} {(1 + cotx) – (cosecx)}

We know that (a + b) (a – b) = a² – b²

= (1 + cotx)² – (cosecx)²

= 1 + cot²x + 2 cotx – cosec²x

We know that 1 + cot²x = cosec²x

= cosec²x + 2 cotx – cosec²x

= 2 cotx

= RHS

Hence proved.

LHS

We know that sin (-x) = -sinx.

We know that when n is odd, tan→cot and cosec→sec.

= 1

= RHS

Hence proved.

LHS

We know that when n is odd, sin→cos.

We know that sin² + cos²x = 1.

= 1 + 1

= 2

= RHS

Hence proved.

LHS

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

We know that when n is odd, sec→cosec and tan→cot.

= [-cosecx] [cosecx] - [-cotx] [cotx]

= -cosec²x + cot²x

= - [cosec²x – cot²x]

We know that cosec²x – cot²x = 1

= -1

= RHS

Hence proved.

We know that in ΔABC, A + B + C = π

(i) Here A + B = π – C

LHS = cos (A + B) + cos C

= cos (π – C) + cos C

We know that cos (π – C) = -cos C

= -cos C + cos C

= 0

= RHS

Hence proved.

(ii) ⇒ A + B = π – C

LHS

We know that

= RHS

Hence proved.

(iii)

⇒ A + B = π – C

LHS

We know that

= RHS

Hence proved

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^o – A) + cos (180^o + B) + cos (180^o + C) – sin (90^o + 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.

⇒ cosec (90° + θ) +x cos θ cot (90° + θ) = cos θ

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

⇒ sec θ +x cos θ [-tan θ] = cos θ

⇒ sec θ –x cos θ tan θ = cos θ

⇒ sec θ –x cos θ (sin θ/ cos θ) = cos θ

⇒ sec θ –x sin θ = cos θ

⇒ sec θ – cos θ =x sin θ

We know that 1 – cos² θ = sin² θ

⇒ tan θ =x

∴x = tan θ

Given

⇒x cot (90° + θ) + tan (90° + θ) sin θ + cosec (90° + θ) = 0

⇒x [-tan θ] + [-cot θ] sin θ + sec θ = 0

⇒ -x sin θ – cos² θ + 1 = 0

We know that 1 – cos² θ = sin² θ

⇒ -x sin θ + sin² θ = 0

⇒x sin θ = sin² θ

⇒x = sin² θ/ sin θ

∴x = sin θ

LHS

= tan 720° - cos 270° - sin 150° cos 120°

= tan (90° × 8 + 0°) – cos (90° × 3 + 0°) – sin (90° × 1 + 60°) cos (90° × 1 + 30°)

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

= tan 0° - sin 0° - cos 60° [-sin 30°]

= tan 0° - sin ° + cos 60° sin 30°

= 0 – 0 + 1/2 (1/2)

= 1/4

= RHS

Hence proved.

LHS

= sin 780° sin 480° + cos 120° sin 150°

= sin (90° × 8 + 60°) sin (90° × 5 + 30°) + cos (90° × 1 + 30°) sin (90° × 1 + 60°)

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

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

= sin 60° cos 30° - sin 30° cos 60°

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

= sin (60° - 30°)

= sin 30°

= 1/2

= RHS

Hence proved.

LHS

= sin 780° sin 120° + cos 240° sin 390°

= sin (90° × 8 + 60°) sin (90° × 1 + 30°) + cos (90° × 2 + 60°) sin (90° × 4 + 30°)

We know that when n is odd, sin→cos.

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

= sin 60° cos 30° - sin 30° cos 60°

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

= sin (60° - 30°)

= sin 30°

= 1/2

= RHS

Hence proved.

LHS

= 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.

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.

Let cos x = t

Range of t =(-1,1)

∴ Maximum and Minimum value of cos x is 1 and -1 respectively.

Now,

cos(-x) = cos x

∴ Range of cos(cos x) = [cos(1),cos(0)]

⇒ cos(cos x) = [cos1,0]

sin(x) has maximum value at x = π/2 and its minimum at

x = -π/2 which are 1 and -1 respectively.

As 1 < π/2;

so, the argument of the outer sin always lies within the interval

[-π/2, π/2]

So the maximum and minimum of the given function are

sin 1 and - sin 1.

Value of cos(x) varies from -1 to 1 for all R and sin(x) is increasing in [-π/2,π/2]

∴ sin(cos x) has max value of sin1.

Given sin x = cos²x

To find the value of cos² x (1 + cos² x).

⇒ cos² x (1 + cos² x).

⇒ cos² x + cos⁴ x.

As cos²x = 1- sin²x the above equation becomes

⇒ 1- sin²x + sin²x

⇒ 1.

(Question might be different)

sin x+ cosec x = 2

⇒

⇒ sin²x +1 = 2sin x

⇒ sin²x - 2sin x+1 = 0

⇒ (sin x-1)²=0

⇒ sin x = 1

As sin x = 1

sinⁿx = 1

∴ sinⁿ x + cosecⁿx

⇒

⇒ 2.

Given: sin x + sin²x = 1

To find the value of cos¹² x + 3 cos¹⁰ x + 3 cos⁸ x + cos⁶ x.

⇒ sin x = 1 – sin²x

⇒ sin x = cos²x

⇒ cos¹²x = sin⁶x, cos¹⁰x = sin⁵x, cos⁸x = sin⁴x, cos⁶x = sin³x.

Substituting above values in given equation we get

⇒ sin⁶x + 3sin⁵x + 3sin⁸x + sin³x [(a+b)³ = a³+3a²b+3ab²+b³]

⇒ (sin x + sin² x)³ = (1)³

⇒ 1.

Given: sin x + sin²x = 1

To find the value of cos⁸ x + 2 cos⁶ x + cos⁴ x.

⇒ sin x = 1 – sin²x

⇒ sin x = cos²x

⇒ cos⁸x = sin⁴x, cos⁶x = sin³x, cos⁴x = sin²x .

Substituting above values in given equation we get

⇒ sin⁴x+2 sin³x+ sin²x [(a+b)² = a²+2ab+b²]

⇒ (sin x + sin² x)² = (1)²

⇒ 1

Given that sin θ₁ + sin θ₂ + sin θ₃ = 3

We know that in general the maximum value of sin θ = 1 when θ = π/2

As sin θ₁ + sin θ₂ + sin θ₃ = 3

⇒ θ₁= θ₂ = θ₃ = π/2.

The above case is the only possible condition for the given condition to satisfy.

∴ cos θ₁ + cos θ₂ + cos θ₃

⇒ cos π/2 + cos π/2 + cos π/2

⇒ 0+0+0

⇒ 0.

We know sin(180+θ ) = -sin θ

Also, sin(360-θ ) = -sin θ

Given all angles are complementary in nature.

sin350 = sin(360-10) = -sin10°

so finally each of them cancel each other and finally we get

the sum equal to 0.

Let the angle subtended be θ.

For calculating we have the formula

⇒ θ = 45°

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.

The given expression can be rearranged as:

(Cos 1 + cos 179) + (cos2 + cos 178) + (cos3+ cos177) +…. + (cos89 + cos 91) + (cos90) + cos180

We know that: cos(180 - x)= - cos x.

So all the bracket totals except last 2 terms will be zero.

So given expression is: 0 + (cos90) + ( cos180)

= 0 + 0 + (-1)

=-1.

Given: cot(α + β) = 0

∴

⇒ cotα.cotβ = 1

Now,

Sin(α +2β ) = sinα.cos2β + cosα.sin2β

=sin α (2 cos² β-1)+cos α.2 sin β.cos β

Given: tanA + cotA = 4

⇒

Squaring both sides we get

⇒

⇒

⇒

Squaring both sides we get

⇒

⇒

⇒

We know that cos²x and sec²x ≥ 0

∴ By applying AM and GM we get,

⇒

⇒ cos² x + sec² x≥2

∴ Least value of the given function is 2.

We know the range of sin x is

-1≤ sin x ≤ 1

∴ 0≤ sin¹⁴x ≤ 1

We know the range of cos x is

-1≤ cos x ≤ 1

∴ 0≤ cos²⁰x ≤ 1

0< sin¹⁴x + cos²⁰x≤2

which means that the value of x lies in the interval [0,2]

But there's a problem, when sine is 0 cosine is 1, they might even be 0 and -1 at particular points (not in this case, since they are even powers), so the minimum we would get should be more than 0. Hence the value of x lies in (0,1]

⇒ 3 sin x +5cos x = 5

⇒ 3sin x = 5-5cos x

⇒ 3sin x = 5(1-cos x)

Squaring both sides we get

⇒ 9sin²x = 25(1-cos x)²

⇒ 9sin²x = 25(1+cos²x-2cos x)

⇒ 9sin²x+9cos²x = 25 + 25cos²x - 50cos x + 9cos²x

⇒ 9(sin²x+ cos²x) = 25 +34cos²x-50cos x

⇒ 34cos²x-50cos x+16=0

⇒ 17cos²x-25cos x+8=0

⇒ 17cos²x-17cos x-8cos x+8=0

⇒ 17cos x(cos x-1)-8(cos x-1)=0

When cos x = 1

3sin x + 5cos x = 5

3sin x = 0

Sin x = 0

Substituting the value cos x = 1 and sin x = 0

5(0)-3(1) = 0-3

⇒ -3.

⇒

⇒ sin²x+ cos²x = 1

⇒ sin²x = 1- cos²x

⇒ 5sin x – 3cos x

∴ -3 and 3.

⇒ sec x – tan x

⇒ sec x – tan x

⇒ -2x

∴ the value of sec x – tan x = -2x, 1/2x.

⇒ sec x – tan x

⇒ sec x – tan x

⇒ 2x

∴ the value of sec x – tan x = 2x, 1/2x.

Given:

Now

(Rationalizing we get)

⇒ sec x – tan x

In the given range tan x = -tan x and sec x is –sec x

∴ -sec x-(-tan x)

⇒ tan x – sec x

Given:

⇒ π<x<2π

Now

(Rationalizing we get)

⇒ cosec x + cot x

In the given range cot x = -cot x and cosec x is –cosec x

∴ -cosec x-(+cot x)

⇒ -cosec x – cot x

If 0 < x < π/2 then we take . So, that square root

is open with positive sign.

Adding 1 on both sides

Given:

X = r sin θ cos ϕ

Y = r sin θ sin ϕ

Z = r cos θ

⇒ x²+y²+z²

⇒ (r sin θ cos ϕ )² + (r sin θ sin ϕ) ² + (r cos θ )²

⇒ r²sin²θcos²ϕ + r²sin²θsin²ϕ + r²cos²θ

⇒ r²sin²θ(cos²ϕ+ sin²ϕ) + r²cos²θ

⇒ r²sin²θ + r²cos²θ

⇒ r²(sin²θ + cos²θ)

⇒ r².

∴ It is independent of θ and ϕ .

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 tan²x

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

Tan⁴x + tan²x = 1 + tan⁴x - 2 tan²x

3 tan²x = 1

tan x = 1/√3

x = π/6.

In IV quadrant cos x is positive

We know

tan²x + 1 =sec²x

Given:

We know csc²α = cot²α +1

⇒ cotα +1

In the given range cot is negative

∴ -1-cotα

sin⁶A + cos⁶A = (sin²A)³ + (cos²A)³

=(sin²A+ cos²A)(sin⁴A+cos⁴A-sin²Acos²A)

= 1 (sin⁴A + cos⁴A – sin²Acos²A)

∴ sin⁴A + cos⁴A – sin²Acos²A+ 3 sin² A cos² A

⇒ sin⁴A + cos⁴A + 2sin²Acos²A

⇒ (sin²A + cos²A)²

= 1²

=1

Given:

Let cosec x =a, cot x = b

∴ According to the question

⇒

But, cosec²x –cot²x =1

⇒ a² – b² = 1

⇒ (a-b) (a + b) = 1

⇒ a + b=2

a-b =1/2 …(1)

a + b =2 …(2)

Adding (1) and (2)

2a = 1/2+2

⇒ a = 5/4

Cos²x = 1 – sin²x

Let cosec x =a, cot x = b

∴ According to the question

⇒

But, cosec²x –cot²x =1

⇒ a² – b² = 1

⇒ (a-b)(a + b) = 1

Adding (1) and (2)

First of all we need to check the condition on x

If x = 0 then sec²x attains to infinity, so that condition must be true i.e x should not be zero

Again if x+y =0 then the RHS part becomes infinity so that condition must be true i.e. x+y should not be zero.

∴ Option B is the correct answer.

Given x is an acute angle and value of tan x =1/√7.

⇒ We know tan²x + 1 = sec²x

⇒ Also, cot²x+1 = cosec²x

⇒ cot² x=7

⇒ cot²x+1 = 7+1

=8

⇒ cosec²x = 8

∴

⇒

⇒

⇒

=sin²5+sin²10+sin²15+.......sin²45+...........+sin²75+sin²80+sin²85+sin²90

We know that sin(90-x) =cos x

So sin²(90-x)= cos²x

=sin²5+sin²10+sin²15+.......sin²45+...........+cos²15+cos²10+cos²5+sin²90

And sin²x+cos²x= 1

So, in given series on rearranging terms we get 8 cases where sin²x+cos²x= 1

So, given changes to

8+sin²45+sin²90

= 9.5

We know that sin(90-x) =cos x

So sin²(90-x)= cos²x

And sin²x+cos²x= 1

Rearranging we get,

=1+1

=2

Given: tan A + cot A = 4

⇒

Squaring both sides we get

Squaring both sides we get

=194

According to the given question:

⇒ x =8.

Given:

3tanA +4 =0

⇒

tan²x + 1 = sec²x

Because in second quadrant cos is negative.

⇒

sin²x+cos²x= 1

∴ The value of 2 cot A − 5 cos A + sin A =

Let cosec x =a, cot x = b

∴ According to the question

But, cosec²x –cot²x =1

⇒ a² – b² = 1

⇒ (a-b)(a + b) = 1

a + b =11/2 …(1)

a-b =2/11 …(2)

Adding (1) and (2)

We know tan²x + 1 = sec²x…(1)

Let tan θ be a and sec θ be b

According to question

a + b =e^x

Manipulating and Substituting in 1 we get

⇒ a² – b² = -1

⇒ (a-b)( a + b) = -1

⇒ (a-b).e^x =-1

⇒ a-b = -e^-x

a + b =e^x

a-b = -e-^x

subtracting above equations we get

2b =e^x +e^-x

We know tan²x + 1 = sec²x…(1)

Let tan x be a and sec x be b

According to question

a + b =k

Manipulating and Substituting in 1 we get

⇒ a² – b² = -1

⇒ (a-b)( a + b) = -1

⇒ (a-b).k =-1

⇒ a-b = -k^-1

a + b =k

a-b = -k-¹

subtracting above equations we get

2b =k +k^-1

tan²x + 1 = sec²x

sin²x+cos²x= 1

∴ cos²x= 1 - sin²x

Substituting in f(x) we get

1 - sin²x+ tan²x + 1

Minimum value of is 0.

∴ f(x)≥ 2.

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

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

tan 1° tan 2° tan 3° ... tan 89° = tan (90° - 89° ) tan(90° - 88° ) tan(90° - 87°) ... tan(90° - 46° ) tan 45° tan 46° …. tan 89°

= cot 89° cot 88° cot 87° ….. cot 46° tan 45° tan 46° ….. tan 89°

(∵ tan(90° - θ ) = cot θ )

= tan 45°

= 1

∴ 1rad = 57.32°

In Range 0 to π/2 sin x is an increasing function

∴ sin1 will always be greater than sin1°

Because sin1 = sin57.32°