**I. Select the correct option from the given alternatives.**

**Question 1.The value of the expressioncos1°. cos2°. cos3° … cos 179° =**

(A) -1

(B) 0

(D) 1**Answer:**

(B) 0

Explanation:

cos 1° cos 2° cos 3° … cos 179°

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

= 0 …[∵ cos 90° = 0]

**Question 2.**

(A) 2cosec A

(B) 2 sec A

(C) 2 sin A

(D) 2 cos A

**Answer:**

(A) 2cosec A

Explanation:

**Question 3.**

Explanation:

**Question 4.**

Explanation:

**Question 5.**(A) 2

(B) mn

(C) 2m

(D) 2n

**Answer:**

(A) 2

Explanation:

**Question 6.**

Explanation:

**Question 7.**

(A) 0

(B) 1

(C) sin θ

(D) cos θ

**Answer:**

(D) cos θ

Explanation:

**Question 8.If cosec θ – cot θ = q, then the value of cot θ is**

**Question 9.**

(A) A.P.

(B) G.P.

(C) H.P.

(D) Not in progression**Answer:**

(B) G.P.

Explanation:

**Question 10.The value of tan 1°.tan 2° tan 3° equal to**

(A) -1

(B) 1

(D) 2

**Answer:**

(B) 1

Explanation:

tan1° tan2° tan3° … tan89°

= (tan 1° tan 89°) (tan 2° tan 88°)

…(tan 44° tan 46°) tan 45°

= (tan 1 ° cot 1 °) (tan 2° cot 2°)

…(tan 44° cot 44°) . tan 45°

…tan(∵ 90° – θ) = cot θ]

= 1 x 1 x 1 x … x 1 x tan 45° =1

**II. Answer the following:**

**Question 1.Find the trigonometric functions of:90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°Solution:**

Angle of measure 90° :

Let m∠XOA = 90°

Its terminal arm (ray OA)

intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1

sin 90° = y = 1

cos 90° = x = 0

Angle of measure 120° :

Let m∠XOA =120°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 225° :

Let m∠XOA = 225°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 45° – 45° – 90° triangle.

OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :

Let m∠XOA = 240°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :

Let m∠XOA = 270°

Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).

x = 0 andy = – 1

sin 270° = y = -1

cos 270° = x = 0

tan 270° =

Angle of measure 315° :

Let m∠XOA = 315°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 45° – 45° – 90° triangle.

OP = 1

[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):

Let m∠XOA = – 120°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :

Let m∠XOA = – 150°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Angle of measure (-180°):

Let m∠XOA = – 180°

Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).

∴ x = – 1 andy = 0

sin (-180°) = y = 0

cos (-180°) = x

= -1

Angle of measure (- 210°):

Let m∠XOA = -210°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Angle of measure (- 300°):

Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 1st quadrant, x>0,y>0

Angle of measure (- 330°):

Let m∠XOA = – 330°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0

**Question 2.State the signs of:i. cosec 520°ii. cot 1899°iii. sin 986°Solution:**

i. 520° =360° + 160°

∴ 520° and 160° are co-terminal angles.

Since 90° < 160° < 180°,

160° lies in the 2nd quadrant.

∴ 520° lies in the 2nd quadrant,

∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°

∴ 1899° and 99° are co-terminal angles.

Since 90° < 99° < 180°,

99° lies in the 2nd quadrant.

∴ 1899° lies in the 2nd quadrant.

∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°

∴ 986° and 266° are co-terminal angles.

Since 180° < 266° < 270°,

266° lies in the 3rd quadrant.

∴ 986° lies in the 3rd quadrant.

∴ sin 986° is negative.

**Question 3.State the quadrant in which 6 lies ifi. tan θ < 0 and sec θ > 0ii. sin θ < 0 and cos θ < 0iii. sin θ > 0 and tan θ < 0Solution:**

i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0

sec θ is positive in 1st and 4th quadrants.

∴ θ lies in the 4th quadrant.

ii. sin θ < 0

sin θ is negative in 3rd and 4th quadrants, cos θ < 0

cos θ is negative in 2nd and 3rd quadrants.

.’. θ lies in the 3rd quadrant.

iii. sin θ > 0

sin θ is positive in 1st and 2nd quadrants, tan θ < 0

tan θ is negative in 2nd and 4th quadrants.

∴ θ lies in the 2nd quadrant.

**Question 4.Which is greater?sin (1856°) or sin (2006°)Solution:**

1856° = 5 x 360° + 56°

∴ 1856° and 56° are co-terminal angles.

Since 0° < 56° < 90°, 56° lies in the 1st quadrant.

∴ 1856° lies in the 1st quadrant,

∴ sin 1856° >0 …(i)

2006° = 5 x 360° + 206°

∴ 2006° and 206° are co-terminal angles.

Since 180° < 206° < 270°,

206° lies in the 3rd quadrant.

∴ 2006° lies in the 3rd quadrant,

∴ sin 2006° <0 …(ii)

From (i) and (ii),

sin 1856° is greater.

**Question 5.Which of the following is positive?sin(-310°) or sin(310°)Solution:**

Since 270° <310° <360°,

310° lies in the 4th quadrant.

∴ sin (310°) < 0

-310° = -360°+ 50°

∴ 50° and – 310° are co-terminal angles.

Since 0° < 50° < 90°, 50° lies in the 1st quadrant.

∴ – 310° lies in the 1st quadrant.

∴ sin (- 310°) > 0

∴ sin (- 310°) is positive.

**Question 6.Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.Solution:**

1 – 2 sin θ cos θ

= si θ + co θ – 2sin θ cos θ

= (sin θ – cos θ ≥ 0 for all θ ∈ R

**Question 7.Show that ta θ + co θ ≥ 2 for all θ ∈ R.Solution:**

**Question 8.**

**Question 9.**

θ lies in the 4th quadrant.

∴ tan θ < 0

∴ tan θ = -1

**Question 10.Prove the following:**

Solution:

Solution:

= 2(1)

= 2 = R.H.S.

∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)