Maharashtra Board Class 11 Maths Chapter 2 Trigonometry 1 Solution

Maharashtra Board Class 11 Maths Solution Chapter 2 – Trigonometry 1

Balbharati Maharashtra Board Class 11 Maths Solution Chapter 2: Trigonometry 1. Marathi or English Medium Students of Class 11 get here Trigonometry 1 full Exercise Solution.

Std	Maharashtra Class 11
Subject	Maths Solution
Chapter	2
Chapter Name	Trigonometry 1

Exercise 2.1

(1) Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, −30°, −45°, −60°, −90°, −120°, −225°, −240°, −270°, −315°

Solution:

0°: Sin0° = 0, cos0° = 1, tan0° = 0, cosec0° = undefined, sec 0° = 1, cot0° = undefined

 

30°: Sin 30° = ½, Cos 30° = √3/2, tan 30° = 1/√3, cosec 30° = 2, sec 30° = 2/√3, cot 30° = √3

 

45°: sin 45° = 1/√2, cos45° = 1/√2, tan45° = 1, cosec 45° = √2, sec 45° = √2, cot 45° = 1

 

60°: Sin60° = √3/2, cos60° = 1/2, tan60° = √3, cosec60° = 2/√3, sec60° = 2, cot60° = 1/√3

 

150°: sin150° = sin (180° – 30°) = sin 30° = 1/2

cos150° = cos (180° – 30°) = – cos30° = – √3/2

tan150° = sin150°/cos150° = (1/2)/(- √3/2) = -1/√3

cosec150° = 2

sec150° = -2/√3

cot150° = – √3

 

180° : sin180° = 0, cos180° = cosec180° = undefined, sec180° = -1, cot180° = undefined, tan180° = 0

 

210°: sin210° = sin (180° + 30° = sin30° = -1/2

cos210° = (180° + 30°) = -cos30° = -√3/2

tan210° = sin210°/cos210° = (-1/2)/(-√3/2) = 1/√3

cosec210° = -2

sec210° = -2√3

cot210° = √3

 

300°: sin300° = sin (360° – 60°) = -sin60° = -√3/2

cos300° = cos (360° – 60°) = cos60° = 1/2

tan300° = sin°300°/cos300° = (-√3/2)/(1/2) = -√3

cosec300° = -2/√3

sec300° = 2

cot300° = -1/√3

 

330°: sin330° = sin (360° – 30°) = -sin30° = -1/2

cos330° = cos (360° – 30°) = cos30° = √3/2

tan330° = sin330°/cos330° = (-1/2)/(-√3/2) = -1/√3

cosec330° = -2

sec330° = 2/√3

cot330° = -√3

 

-30° = sin (-30°) = -sin30° = -1/2

cos (-30°) = cos30° = √3/2

tan (-30°) = sin (-30°)/cos (-30°) = (-1/2)/√3/2 = -1/√3

cosec (-30°) = -2

sec (-30°) = 2/√3

cot (-30°) = -√3

 

-45°: sin (-45°) = -sin45° = -1/√2

cos (-45°) = cos45° = 1/√2

tan (-45°) = 1

cosec (-45°) = – √2

sec (-45°) = √2

cot (-45°) = -1

 

-60°: sin (-60°) = -sin60° = – √3/2

cos (-60°) = cos60° = 1/2

tan (-60°) = (-√3/2)/(1/2) = √3

cosec (-60°) = -2/√3

sec (-60°) = 2

cot (-60°) = – 1/√3

 

-90° : sin (-90°) = -sin90° = -1

cos (-90°) = cos90° = 0

tan (-90°) = sin(-90°)/cos(-90°) = -1/0 = undefined

cosec (-90°) = -1

sec (-90°) = undefined

cot (-90°) = 0

 

-120°: sin (-120°) = -sin (180° – 60°) = -sin60° = -√3/2

cos (-120°) = cos (180° – 60°) = -cos60°= -1/2

tan (-120°) = sin (-120°)/cos (-120°) = -√3/2/(1/2) = √3

cosec (-120°) = -2/√3

sec (-120°) = -2

cot (-120°) = 1/√3

 

-225°: sin (-225°) = -sin (180° + 45°) = sin45° = 1/√2

cos (-255°) = cos (180° + 45°) = -cos45° = -1/√2

tan (-225°) = sin (-225°)/cos (-225°) = 1/√2/-1√2  = -1

cosec (-225°) = √2

sec (-225°) = -√2

cot (-225°) = -1

 

-240°: sin (-240°) = -sin (180° + 60°) = sin60° = √3/2

cos (-240°) = cos (180° + 60°) = -cos60° = -1/2

tan (-240°) = sin (-240°)/cos (-240°) = (√3/2)/(-1/2) = -√3

cosec (-240°) = 2/√3

sec (-240°) = -2

cot (-240°) = -1/√3

 

-270°: sin (-270°) = -sin (180° + 90°) = – sin90° = 1

cos (-270°) = -cos (180° + 90°) = -cos90° = 0

tan (-270°) = undefined

cosec (-270°) = 1

sec (-270°) = undefined

cot (-270°) = 0

 

-315°: sin (-315°) = -sin (360° – 45°) = sin 45° = 1/√3

cos (-315°) = cos (360° – 45°) = cos45° = 1/√2

tan (-315°) =  sin (-315°)/cos (-315°) = (1/√2)/(1√2) = 1

cosec (-315°) = √2

sec (-315°) = √2

cot (-315°) = 1

 

(2) State the signs of

(i) tan380°

(ii) cot230°

(iii) sec468°   

Solution:

(i) tan380°

380° = 360° + 20° which is in first quadrant

All trigonometric ratios are positive in first quadrant

∴ tan380° is positive

 

(ii) cot230°

230° = 180° + 50° which is in third quadrant

tan & cot are positive in third quadrant

∴ cot 230° is positive

 

(iii) sec468°

468° = 5×90° + 15° which is in second quadrant

cos & sec are negative in second quadrant

∴ sec468° in negative

 

(3) State the signs of cos 4^c and cos4°. Which of these two functions is greater?

Solution:

4^c = 4×180°/π = 4×180×7/22 = (229 1/11)°

4^c lies in third quadrant

∴ cos 4^c is negative

4° lies in first quadrant

∴ cos 4° is positive

∴ cos 4° is a greater function

(4) State the quadrant in which θ lies if

(i) sinθ < 0 and tanθ >0

(ii) cosθ < 0 and tanθ >0

Solution: 

(i) sinθ < 0 and tanθ > 0

Third quadrant

 

(ii) cosθ < 0 and tan θ > 0

Third quadrant

 

(5) Evaluate each of the following:

(i) sin30° + cos 45° + tan180°

(ii) cosec45° + cot 45° + tan0°

(iii) sin30° × cos 45° × tan360°

Solution:

(i) sin30° + cos45° + tan180°

= 1/2 + 1/√2 = 0 = 2 + √2/2√2

 

(ii) cosec45° + cot45° + tan0°

= √2 + 1 + 0 = 1 + √2

 

(iii) sin30° × cos45° × tan360°

= 1/2 × 1/√2 × 0 = 0

 

(6) Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, −4).

Solution:

At point (3, -4), x = 3, y = -4

∴ r = √x² + y² = √3² + (-4)² = √9+16 = 5

Sinθ = -4/5, cosθ = 3/5, tanθ = -4/3, cosec θ = -5/4, secθ = 5/3, cotθ = -3/4

 

(7) If cosθ =12/13, 0 < θ < π/2 , find the value of  

sin²θ – cos²θ/2 sinθ cosθ, 1/tan²θ

Solution:

cosθ = 12/13

=) r = √x² + y²

=) 13² = 12² + y²

=) y² = 169 – 140

=) y = 5

sinθ = 5/13, tanθ = 5/12, catθ = 12/5

sin²θ – cos²θ/2sinθ cosθ = 1/2 [sinθ/cosθ – cosθ/sinθ]

= 1/2 [tanθ – cotθ]

= 1/2 [5/12 – 12/5]

= 1/2 [25-144/60]

= -1/2 [119/60] = – 119/120

1/tan²θ = 1/(25/144) = 144/25

 

(8) Using tables evaluate the following:

(i) 4cot45° − sec2 60° + sin 30°  

(ii) cos² 0 + cos² π/6 + cos² π/3 + cos² π/2

Solution:

(i) 4cot 45° – sec² 60° + sin 30°

= 4×1 – 4 + 1/2

= 1/2

 

(ii) cos² 0 + cos² π/6 + cos² π/3 + cos² π/2

= 1 + 3/4 + 1/4 + 0

= 2

 

(9) Find the other trigonometric functions if

(i) If cosθ = 3/5 − and 180° < θ < 2700.

(ii) If secA = − 25/7 and A lies in the second quadrant.

(iii) If cot x = 3/4, x lies in the third quadrant.

(iv) tan x = −5/12, x lies in the fourth quadrant.

Solution:

(i) cosθ = -3/5

∴ y² = 25-9 = 16 => y = 4

sinθ = -4/5, tanθ = 4/5, cosecθ = -5/4 secθ = -5/3, cotθ = 3/4

 

(ii) sec A = -25/7

∴ y² = 625 – 49 = 576 => y = 24

sin A = 24/25, cosec A = 25/24, tan A = -24/7, cot A = -7/24

CosA = -7/25

 

(iii) cot x = 3/4

∴ x = √9+16 = 5

∴ sin x = -4/5, cos x = -3/5, tan x = 4/3, cosec x = -5/4, sec x = -5/3

 

(iv) tan x = -5/12

∴ r = √25 + 144 = 13

∴ sin x = – 5/13, cos x = 12/13 cosec x = -13/5, sec x = 13/12, cot x = -12/5

 

Ex – 2.2

(1) If 2 sinA = 1 = √2 cosB and π/2 < A < π, 3π/2 <B <2π, then find the value of

tanA + tanB/cosA – cosB

Solution:

2 sin A = 1 = √2 cosB

∵ sinA = 1/2, cosB = 1/√2

sinA = 1/2, A = 30°

cosA = √3/2, tanA = 1/√3

∵ cosB = 1/√2, B = 45°

∴ SinB = 1/√2, tan B = 1

∴ tanA + tanB/cosA – cosB

(2) If sinA/3 = sinB/4 = 1/5 and A, B are angles in the second quadrant then prove that 4cosA + 3cosB = -5.

Solution:

SinA/3 = sinB/4 = 1/5

∴ sinA = 3/5, sinB = 4/5

For A, x = √25 – 9 = 4

For B, x = √25 – 16 = 3

∴ cosA = -4/5, cosB = -3/5 (∵ A & B are in 2^nd quadrant)

∴ 4cosA + 3cosB = 4 × -4/5 + 3 × -3/5 = -16/5 -9/5 = -25/5 = -5

 

(3) If tanθ = 1/2, evaluate 2sinθ + 3cosθ/4cosθ + 3sinθ

Solution:

(4) Eliminate θ from the following:

(i) x = 3secθ , y = 4tanθ

(ii) x = 6cosecθ , y = 8cotθ

(iii) x = 4cosθ − 5sinθ, y = 4sinθ + 5cosθ

(iv) x = 5 + 6cosecθ, y = 3 + 8cotθ

(v) 2x = 3 − 4tanθ, 3y = 5 + 3secθ

Solution:

(i) x = 3 secθ, y = 4tanθ

∴ secθ = x/3, tanθ = y/4

∴ 1 + tan²θ = sec²θ

= 1 + y²/1 = x²/9

= x²/9 – y²/16 = 1

 

(ii) x =  6 cosecθ, y = 8 cotθ

∴ cosecθ = x/6, cotθ = y/8

∴ cosec²θ = cot² θ + 1

=) x²/36 = y²/64 + 1

=) x²/36 – y²/64 = 1

 

(iii) x = 4cosθ – 5sinθ, y = 4sinθ + 5cosθ

x² + y² = (4cosθ – 5sinθ)² + (4sinθ + 5cosθ)²

= 16cos²θ + 25 sin²θ – 40cosθ sinθ + 16sin²θ + 25cos²θ + 40sinθ cosθ

= 16 (sin²θ + cos²θ) + 25 (sin²θ + cos²θ)

= 16 + 25

= 41

 

(iv) x = 5 + 6 cosecθ, y = 3 + 8 cotθ

∴ cosecθ = x-5/6 cotθ = y-3/8

cosec²θ = cot²θ + 1

= (x-5/6)² – (y-3/8)² = 1

 

(v) 2x = 3 – 4 tanθ, 3y = 5 + 3 secθ

tanθ = 3 – 2x/4, secθ = 3y-5/3

∴ 1 + tan² θ = sec²θ

= (3y – 5/3)² – (3- 2x/4)² = 1

 

(5) If 2sin²θ + 3sinθ = 0, find the permissible values of cosθ. 

Solution:

2 sin² θ + 3 sinθ = 0

= sinθ (2sinθ + 3) = 0

∴ sinθ = 0 or sinθ = -3/2

∵ -1 ≤ sinθ ≤ 1

∴ sinθ = 0

∴ cosθ = √1 – sin²θ

= cosθ = ±1

 

(6) If 2cos² θ −11cosθ + 5 = 0 then find possible values of cosθ.   

Solution:

2cos² θ – 11cosθ + 5 = 0

∴ cosθ = 5 or cosθ = ½

∵ -1 ≤ cosθ ≤ 1

∴ cosθ = 1/2

(7) Find the acute angle θ such that 2cos² θ = 3sinθ

Solution:

2cos²θ = 3sin θ

= 2 (1 – sin²θ) = 3sin θ

= 2 sin²θ + 3sinθ – 2 = 0

∴ sin θ = – 3 ± √9+16/4

= -3±5/4

∴ sin θ = ½ or sin θ = -2

∵ -1 ≤ sinθ ≤ 1

∴ sin θ = 1/2

∴ θ = 30°

 

(8) Find the acute angle θ such that 5tan2 θ + 3 = 9secθ

Solution:

5tan² θ + 3 = 9 sec θ

= 5 (sec² θ – 1) + 3 = 9 sec θ

= 5 sec² θ – 9 sec θ – 2 = 0

∴ sec θ = 9 ± √81+40/10

= 9±11/10

∴ sec θ = 2 or sec  θ = -1/5

∵ sec θ ≥ 1 or sec θ ≤ -1

∴ sec θ = 2 => cos θ = 1/2

∴ θ = 60°

 

(9) Find sinθ such that 3cosθ + 4sinθ = 4

Solution:

3cosθ + 4sinθ = 4

=> 3cosθ = 4 (1 – sinθ)

=> 9cos²θ = 16 (1 – sinθ)²

=> 9 (1 – sin²θ) = 16 (1 – sinθ)²

=> 9 – 9 sin²θ = 16 + 16 sin²θ – 32sinθ

= 25sin²θ – 32sinθ + 7 = 0

∴ sinθ = (32 ± √1024 – 700) /50 = 32±18/50

∴ sinθ = 1 or sinθ = 14/50 = 7/25

 

(10) If cosecθ + cotθ = 5, then evaluate secθ.

Solution:

Cosecθ + cotθ = 5

=> 1/sinθ + cosθ/sinθ = 5

=> 1 + cosθ = 5 sinθ

=> (1 + cosθ)² = (5sinθ)²

= 1 + cos²θ + 2secθ – 25 sin²θ

= 1 + cos²θ+ 2cosθ = 25 – 25 cos²θ

= 26cos²θ + 2cosθ – 24 = 0

∴ cos θ = – 2 ± √4 + 2496/52

= -2±√2500/52

= -2 ± 50/52

∴ cosθ = -1 or cos θ = 12/13

∴ secθ = 13/12 (∵ if cosθ = -1, cosecθ & cotθ are undefined)

 

(11) If cotθ = 3/4 and π < θ < 3π/4 then find the value of 4cosecθ + 5cosθ.

Solution:

Cotθ = 3/4

∵ cosec²θ = 1 + cot²θ = 1 + 9/16 = 25/16

=> cosecθ = ± 5/4 => cosecθ = -5/4 (∵ θ is 3^rd quadrant)

=> sinθ = -4/5

∴ cosθ = √1 – sin²θ = √1 – 16/25  = ± 3/5 => cosθ = -3/5

= 4 × -5/4 + 5 × -3/5 = -5 -3 = -8

 

(12) Find the Cartesian co-ordinates of points whose polar coordinates are:

(i) (3, 90°)

(ii) (1, 180°)

Solution:

Given, r = 3, θ = 90°

∴ r = √x² + y² and tanθ = y/x

∴ x² + y² = 9 and tan 90° = y/x

∴ y/x = 1/0

=> x = 0

∴ y = √9 – x²

= 3

∴ The Cartesian coordinates are: (0, 3)

 

(ii) Given, r = 1, θ = 180°

∴ r = √x² + y²

=> x² + y² = 1

=> y = 0

∴ x² = √1 – 10 = ±1 = -1 (∵ x is negative in 2^nd & 3^rd quadrant)

∴ Cartesian coordinates are: (-1, 0)

 

(13)   Find the polar co-ordinates of points whose Cartesian co-ordinates are:

(i) (5, 5)

(ii) (1, √3)

(iii) (-1, -1)

(iv) (-√3, 1)

Solution:

(i) (5, 5)

∴ r = √5² + 5² = 5√2 and tanθ = 5/4 = 1 => θ = 45°

∴ Polar coordinates are: (5√2, 45°)

 

(ii) (1, √3)

∴ r = √1² + (√3)² = 2

tanθ = √3/1 = θ = 60°

∴ Polar coordinates are: (2, 60°)

 

(iii) (-1, -1)

r = √(-1)² + (-1)² = √2

tanθ = -1/-1 = 1

∴ θ = 45° or 225° (∵ (-1, -1) is in 3^rd quadrant)

∴ Polar coordinates are: (√2, 225°)

 

(iv) (-√3, 1)

r = √3+1 = 2

tan θ = -1/√3

∴ θ = 150°

∴ Polar coordinates are: (2, 150°)

 

(14) Find the value of

(i) sin 19π^c/3

(ii) cos 1140°

(iii) cot 25π^c/3

Solution:

(i) sin 19π^c/3

= sin (6π + π/3)

= sin π/3 = √3/2

 

(ii) cos 1140°

= cos (12×90° + 60°

= cos60° = 1/2

 

(iii) cot 25π^c/3

= cot (8π + π/3)

= cot π/3 = 1/√3

 

(15) Prove the following identities:

(i) (1 + tan² A) + (1 + 1/tan²A) = 1/sin² A-sin⁴ A

(ii) (cos² A – 1) (cot² + A + 1) = – 1  

(iii) (sinθ + secθ)² + (cosθ + cosecθ)² = (1 + cosecθ secθ)²

(iv)  (1 + cotθ – cosecθ secθ) = 2

(v) tan³ θ/1 + tan² θ + cot³ θ/1 + cot² θ

= secθ cosecθ – 2sinθ cosθ

(vi) 1/secθ+tanθ – 1/cosθ = 1/cosθ – 1/secθ – tanθ

(vii) sinθ/1 + cosθ + 1 + cosθ/sinθ = 2cosecθ

(viii) tanθ/secθ – 1 = secθ + 1/tanθ

(iv) cotθ/cosecθ – 1 = cosecθ + 1/cotθ 

(x) (secA + cosA) (secA – cosA) = tan² A + sin²A 

(xi) 1 + 3cosec²θ, cot²θ + cot⁶θ = cosec⁶θ

(xii) 1-secθ + tanθ/1 + secθ – tanθ = secθ + tanθ – 1/secθ + tanθ + 1

Solution:

(ii) (cos²A – 1) (cot²A + 1)

= (Cos²A – 1) (cos²A/sin²A + 1)

= (cos²A – 1) (cos²A + sin²A/sin²A)

= cos²A – 1/sin²A

= -sin²A/sin²A

= -1 = RHS

 

(iii) (sinθ + secθ)² + (cosθ + cosecθ)²

= sin²θ + sec²θ + 2sinθ secθ + cos²θ + cosec²θ + 2cosθ cosecθ

= (sin²θ + cos²θ) + (1/sin²θ + 1/cos²θ) + 2 (sinθ/cosθ + cosθ/sinθ)

= 1 + 1/sin²θ cos²θ + 2/sinθ cosθ

= 1 + cosec²θ sec²θ + 2coseθ secθ

= (1 + cosecθ secθ)²

= RHS

 

(x) (SecA + cosA) (secA – cosA)

= sec²A – cos²A

= 1 + tan²A – 1 + sin²A

= tan²A + sin²A = RHS

(xi) 1 + 3 cosec²θ cot²θ + cot⁶θ

= 1 + 3 cosec²θ (cosec²θ – 1) + (cosec²θ – 1)³

= 1 + 3 cosec⁴θ – 3cosec²θ + (cosec⁶θ – 1 + 3cosec⁴θ + 3cosec²θ)

= cosec⁶θ = RHS

MISCELLANEOUS – 2

(I) select the correct option from the given alternatives.

(1) The value of the expression cos1°. cos2°. cos3°. …….. cos179° =

(A) -1

(B) 0

(C) 1/√2

(D) 1

Solution:

(1) (i) Cos1°. cos2°. ……. cos179°

∵ Cos 90° = 0

∴ Cos1°. cos2°. ………. cos179° = 0

A: (B) 0

 

(2) tanA/1+secA + 1+secA/tanA is equal to

(A) 2cosec A

(B) 2sec A

(C) 2sin A

(D) 2cos A

Solution:

TanA/1+secA + 1+secA/tanA = sinA/cosA+1 + cosA+1/sinA

= 2cosA + 1 + 1/sinA (cosA + 1) = 2(cosA + 1)/sinA (cosA + 1) = 2cosecA

A: (A) 2 cosecA.

 

(3) If α is a root of 25cos² θ + 5cosθ – 12 = 0, π/2 <a <π then sin 2 α is equal to.

(A) – 24/25

(B) – 13/18

(C) 13/18

(D) 24/25

Solution:

25cos² θ + 5cosθ – 12 = 0

cosθ = -5 ± √25+100/50 = -5±35/50

∴ cosθ = 3/5 or cosθ = -4/5

∵ π/2 <∝ <π

cos∝ = -4/5

∴ sin ∝ = √1 – cos²∝ = 3/5

∴ sin 2 ∝ = sin ∝ cos ∝

= 2 × 3/5 × (-4/5)

= -24/25

A: (A) -24/25

 

(4) If θ = 60°, then 1+tan² θ/2tan θ = is equal to

(A) √3/2

(B) 2/√3

(C) 1/√3

(D) √3

Solution:

1 + tan² θ/2tanθ = 1+tan²60°/2tan60° = 1+3/2√3 = 2/√3

A: (B) 2/√3

(5) If secθ = m and tanθ = n, then 1/m {(m+ n) + 1/(m + n)} is equal to

(A) 2

(B) mn

(C) 2m

(D) 2n

(6) If cosecθ + cotθ = 5/2, then the value of tanθ is

(A) 14/25

(B) 20/21

(C) 21/20

(D) 15/16 

Solution:

cosecθ + cotθ = 5/2 —– (i)

cosec²θ – cot²θ = 1

= (cosecθ + cotθ) (cosecθ – cotθ) = 1

= 5/2 (cosecθ – cotθ) = 1

= cosecθ – cotθ = 2/5 —— (2)

(1) – (2) => 2cotθ = 5/2 – 2/5 = 21/10

=> cotθ = 21/10

tanθ = 20/21

A: (B) 20/21

 

(7) 1 – sin²θ/1+cosθ + 1+cosθ/sinθ – sinθ/1-cosθ equals

(A) 0

(B) 1

(C) sinθ

(D) cosθ

Solution:

‘

(8) If cosecθ − cotθ = q, then the value of cot θ is

(A) 2q/1+q²

(B) 2q/1-q²

(C) 1-q²/2q

(D) 1+q²/2q 

Solution:

cosecθ – cotθ = q —— (1)

cosec²θ – cot²θ = 1

= (cosecθ – cotθ) (cosecθ + cotθ) = 1

= q (cosecθ + cotθ) = 1

= cosecθ + cotθ = 1/q —— (2)

(2) —– (1) = 2cotθ =1/q – q

= cotθ = 1-q²/2q

A: (c) 1-q²/2q

 

(9) The cotangent of the angles π/3, π/4 and π/6 are in

(A) A.P.

(B) G.P.

(C) H.P.

(D) Not in progression

Solution:

cot π/3 = 1/√3, cot π/4 = 1, cot π/6 = √3

Let, a = 1/√3, b = 1, c = √3

∴ b/a = √3 and c/b = √3

∴ a, b and c are in G.P.

 

(10) The value of tan1°.tan2°tan3°….. tan89° is equal to

(A) -1

(B) 1

(C) π/2

(D) 2

Solution:

tan1° tan2° ……. tan89°

= [tan1° tan2° …… tan44°] tan45° [tan (90° – 44°) ……. tan (90° – 1°)]

= [tan1° tan2° …… tan44°] 1 [cot44° …. cot1°)]

= 1

A: (B) 1

 

(II) Answer the following

(1) Find the trigonometric functions of: 90°, 120°, 225°, 240°, 270°, 315°, −120°, −150°, −180°, −210°, −300°, −330°

Solution:

90°: sin 90° = 1

cos 90° = 0

tan 90° = undefined

cosec 90° = 1

sec 90° = undefined

cot 90° = 0

 

120°: sin 120° = sin (180° – 60°) = sin 60° = √3/2

cos 120° = cos (180° – 60°) = – cos 60° = – 1/2

tan 120° = – √3

cosec 120° = 2/√3

sec 120° = -2

cot 120° = 1/√3

 

225°: sin 225° = sin (180° + 45°) = -sin 45° = -1/√2

cos 225° = cos (180° + 45°) = -cos45° = -1/√2

tan 225° = 1 cot 225° = 1

cosec 225° = – √2

sec 225° = – √2

 

240°: sin 240° = sin (180° + 60°) = -sin60° = √3/2

cos 240°= cos (180° + 60°) = – cos 60° = – ½

tan 240° = √3 sec 240° = -2

cosec 240° = -2/√3 cot 240° = 1/√3

 

270°: sin 270° = sin (360° – 90°) = – sin 90° = -1

cos 270° = cos (360° – 90°) = – cos90° = 0

tan 270° = undefined sec 270° = undefined

cosec 270° = -1 cot 270 = 0

 

315°: sin 315° = sin (360° – 45°) = – sin45° = -1/√2

cos 315° = cos (360° – 45°) = cos 45° = 1/√2

tan 315° = -1 sec 315° = √2

cosec 315° = – √2 cot 315° = -1

 

-120°: sin (-120°) = – sin (180° – 60°) = – sin 60° = – √3/2

cos (-120°) = cos (180° – 60°) = – cos 60° = – ½

tan (-120°) = √3 sec (-120°) = -2

cosec (-120°) = -2/√3  cot (-120°) = 1/√3

 

 

-150°: sin (-150°) = – sin (180° – 30°) = – sin 30° = – 1/2

cos (-150°) = cos (180° – 30°) = – cos 30° = – √3/2

tan (-150°) = 1/√3 sec (-150°) = -2/√3

cosec (-150°) = -2 cot (-150°) = √3

 

 

-180°: sin (-180°) = – sin (180° – 0°) = 0

cos (-180°) = cos (180° – 0°) = – cos 0° = – 1

tan (-180°) = 0 sec (-180°) = -1

cosec (-180°) = undefined cot (-180°) = undefined

 

 

-210°: sin (-210°) = – sin (180° + 30°) = sin 30° = 1/2

cos (-210°) = cos (180° – 30°) = – cos 30° = – √3/2

tan (-210°) = – 1/√3  sec (-210°) = -2/√3

cosec (-210°) = 2 cot (-210°) = – √3

 

 

-300°: sin (-300°) = – sin (360° – 60°) = sin 60° = √3/2

cos (-300°) = cos (360° – 60°) = – cos 60° = – 1/2

tan (-300°) = √3  sec (-300°) = 2

cosec (-300°) = 2/√3 cot (-300°) = 1/√3

 

 

-330°: sin (-330°) = – sin (360° – 30°) = sin 30° = 1/2

cos (-330°) = cos (360° – 30°) = – cos 30° = √3/2

tan (-330°) = 1/√3  sec (-330°) = 2/√3

cosec (-330°) = 2 cot (-330°) = √3

 

 

(2) State the signs of

(i) cosec 520°

(ii) cot 1899°

(iii) sin 986°

Solution:

(i) Cosec 520°

520° = 3×180° – 20° lies in 2^nd quadrant

∴ cosec 520° is positive

 

(ii) cot 1899°

1899° = 11×180° – 81° lies in 2^nd quadrant

∴ cot 1899° is negative

 

(iii) Sin 986°

986° = 5×180° + 86° lies in 3^rd quadrant

∴ Sin 986° is negative

 

(3) State the quadrant in which θ lies if

(i) tanθ < 0 and secθ >0

(ii) sinθ < 0 and cosθ <0

(iii) sinθ > 0 and tanθ <0

Solution:

(i) tanθ < 0 & secθ > 0

Fourth quadrant

 

(ii) sinθ < 0 & cosθ < 0

Third quadrant

 

(iii) sinθ > 0 & tanθ < 0

Second quadrant

 

(4) Which is greater sin (1856°) or sin (2006°)?

Solution:

Sin (1856°) = sin (10×180°+56°)

= sin 56° (Lies in 1^st quadrant)

sin (2006°) = sin (11×180°×20°)

= -sin 20° (∵ It lies in 3^rd quadrant)

∴ Sin (1856°) > sin (206°)

 

(5) Which of the following is positive? sin(−310°) or sin(310°)

Solution:

Sin (-310°) is positive because it lies in 1^st quadrant

sin (310°) is negative because it lies in 4^th quadrant

 

(6) Show that 1− 2sinθ cosθ ≥ 0 for all θ ∈ R

Solution:

1 – 2 sinθ cosθ = sin²θ + cos²θ – 2sinθ cosθ

= (sinθ – cosθ)² ≥ 0, θ ∈ R

 

(7) Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R

Solution:

tan² θ + cot² θ = tan² θ + (1/tanθ)²

= (tanθ – 1/tanθ)² + 2tanθ . 1/tanθ

= (tanθ – 1/tanθ)² + 2 ≥ 2

 

(8) If sinθ = x² – y²/x² + y² then find the values of cosθ, tanθ in terms of x and y.

Solution:

(9) If secθ = √2 and 3π/2 < θ < 2π then evaluate

Solution:

Secθ = √2

tanθ = √sec²θ – 1 = √2-1 = ± 1

∵ 3π/2 <θ <2π, tanθ = -1

cotθ = -1

∴ cosecθ = √cot²θ + 1 = ± √1+1 = ± √2

∴ cosecθ = – √2

∴ 1+tanθ+cosecθ/1+cotθ-cosecθ = 1-1- √2/1-1+√2 = -1

 

(10) Prove the following:

(i) sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1

Solution:

sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B

sin²A (sin²B + cos²B + cos²B) + cos²A (sin²B + cos²B)

= sin²A + cos²A = 1 R.H.S

 

(ii) (1+cotθ+tanθ) (sinθ – cosθ)/sec³θ – cosec³θ)

= sin²θ cos²θ

Solution:

(iii) (tanθ + 1/cosθ)² + (tanθ – 1/cosθ)²

= 2 (1+sin²θ/1-sin²θ)

Solution:

(tanθ + 1/cosθ)² + (tanθ – 1/cosθ)²

= tan²θ + sec²θ + tan²θ + sec²θ + 2tanθ secθ – 2tanθ secθ

= 2 (tan²θ + sec²θ)

= 2 (sin²θ/cos²θ + 1/cos²θ)

= 2 (1 + sin²θ/1-sin²θ) = RHS

 

(iv) 2sec²θ – sec⁴θ – 2cosec²θ + cosec⁴θ = cot⁴θ – tan⁴θ

Solution: 

2sec²θ – sec⁴θ – 2cosec²θ + cosec⁴θ

= 2 (1 + tan²θ) – (1 + tan²θ)² – 2 (1 + cot²θ) + (1 + cot²θ)²

= 2 + 2tan²θ – (1 + tan⁴θ + 2tan²θ) – 2 – 2cot²θ + (1 + cot⁴θ + 2cot²θ)

= cot⁴θ + tan⁴θ = RHS

 

 

(v) sin⁴ θ + cos⁴ θ = 1 – 2 sin² θ cos² θ

Solution:

sin⁴ θ + cos⁴ θ

= (sin²θ + cos²θ)² – 2sin²θ cos²θ

= 1 – 2sin²θ cos²θ = RHS

 

 

(vi) 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0

Solution:

2 (sin²θ + cos⁶θ) – 3 (sin⁴θ + cos⁴θ) + 1

= 2 (sin²θ + cos⁴θ – sin²θ cos²θ) (sin²θ + cos²θ) – 3 {(sin²θ + cos²θ)² – 2sin²θ cos²θ} + 1

= 2 (sin⁴ + cos⁴θ – 2sin²θ cos²θ) – 3 (1 – 2sin²θ cos²θ) + 1

= 2 (sin⁴θ + cos⁴θ) – 2sin²θ cos²θ – 3 + 6 sin²θ cos²θ + 1

= 2 (1 – 2sin²θ cos²θ) + 4sin²θ cos²θ – 2

= 2 – 4sin²θ cos²θ + 4sin²θ cos²θ – 2

= 0 = RHS

 

 

(vii) cos⁴ θ − sin⁴ θ +1= 2cos² θ

Solution:

cos⁴θ – sin⁴θ + 1

= cos²θ + sin²θ + sin⁴θ + cos⁴θ

= cos²θ (1 + cos²θ) + sin²θ + sin⁴θ

= (1 – sin²θ) (2 – sin²θ) + sin²θ + sin⁴θ

= 2 – 3 sin²θ – sin⁴θ + sin²θ + sin⁴θ

= 2 (1 – sin²θ) = 2cos²θ = RHS

 

(viii) sin⁴θ +2sin²θ ·cos²θ = 1 − cos⁴θ

Solution:

sin⁴θ + 2sin²θ ·cos²θ

= sin⁴θ + 2sin²θ cos²θ + cos⁴θ – cos⁴θ

= (sin²θ + cos²θ)² – cos⁴θ

= 1 – cos⁴θ = RHS

(ix) sin³θ + cos³θ/sinθ + cosθ + sin²θ – cos³θ/sinθ – cosθ = 2

Solution:

(x) tan² θ − sin² θ = sin⁴ θ sec² θ

Solution:

tan²θ – sin²θ

= sin²θ/cos²θ – sin²θ

= sin²θ – sin²θ cos²θ/cos²θ

= sin²θ (1 – cos²θ)/cos²θ

= sin⁴θ sec²θ = RHS

 

(xi) (sinθ + cosecθ)² + (cosθ + secθ)² = tan²θ + cot²θ + 7

Solution:

(sinθ + cosecθ)² + (cosθ + secθ)²

= sin²θ + 1/sin²θ + 2sinθ 1/sinθ + cos²θ + 1/cos²θ + 2cosθ 1/cosθ

= (sin²θ + cos²θ) + 2 + 2 + (cosec²θ + sec²θ)

= 5 + (1 + cot²θ + 1 + tan²θ)

= tan²θ + cot²θ + 7 = RHS

 

(xii) sin⁸θ − cos⁸θ = (sin²θ − cos²θ) (1 − 2 sin²θ cos²θ)

Solution:

= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)

= (sin²θ – cos²θ) (sin²θ + cos²θ) {(sin²θ + cos²θ)² – 2sin²θ cos²θ}

= (sin²θ – cos²θ) (1 – 2sin²θ cos²) = RHS

 

(xiii) sin⁶A + cos⁶A = 1 − 3sin²A +3 sin⁴A

Solution:

sin⁶A + cos⁶A

= (sin²A + cos²A)³ – 3sin²A cos²A (sin²A + cos²A)

= 1 – 3 sin²A (1 – sin²A)

= 1 – 3sin²A (1 – sin²A)

= 1 – 3 sin²A + 3sin⁴A = RHS

 

 

(xiv) (1+ tanA·tanB)² + (tanA−tanB)² = sec² A·sec² B

Solution:

(1 + tanA tanB)² + (TanA – tanB)²

= 1 + tan²A tan²B + 2tanA tanB + tan²A + tan²B – 2tanA tanB

= (1 + tan²B) (1 + tan²A)

= sec²A sec²B = RHS

 

(xv) 1 + cotθ + cosecθ/1 – cotθ + cosecθ = cosecθ + cotθ – 1/cotθ – cosecθ + 1

Solution:

1 + cotθ + cosecθ/1 – cotθ + cosecθ

cosec²θ – cot²θ = 1

= (cosecθ + cotθ) (cosecθ – cotθ) = 1

= cosecθ + cotθ/1 = 1/cosecθ – cotθ

= cosecθ + cotθ + 1/coecθ + cotθ – 1 = 1 + cosecθ – cotθ/1 – cosecθ + cotθ (By componendo & dividendo)

= cosecθ + cotθ + 1/1 – cotθ + cosecθ

= cosecθ + cotθ – 1/cotθ – cosecθ + 1

 

(xvi) tanθ + secθ – 1/tanθ + secθ + 1 = tanθ/secθ + 1

(xvii) cosecθ + cotθ – 1/cosecθ + cotθ + 1 = 1 – sinθ/cosθ

(xviii) Cosecθ + cotθ + 1/cotθ + cosecθ – 1 = cotθ/cosecθ – 1

Solution: