Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.1

Question 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:

Angle of measure 0°:

Let m∠XOA = 0° = 0^{c}

Its terminal arm (ray OA) intersects the standard

unit circle in P(1, 0).

Hence,x = 1 and y = 0

sin 0° = y = 0,

cos 0° = x = 1,

tan 0° = \(\frac{y}{x}=\frac{0}{1}\) = 0

cot 0° = \(\frac{x}{y}=\frac{1}{0}\) which is not defined

sec 0° = \(\frac{1}{x}=\frac{1}{1}\) = 1

cot 0° = \(\frac{1}{y}=\frac{1}{0}\) which is not defined,

Angle of measure 30°:

Let m∠XOA = 30°

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 1st quadrant, x > 0, y > 0

∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Angle of measure 45°:

Let m∠XOA = 45°

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 1st quadrant, x > 0, y > 0

∴ x = OM = \(\frac{1}{\sqrt{2}}\) and

y = PM = \(\frac{1}{\sqrt{2}}\)

∴ P = (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

Angle of measure 60°:

Let m∠XOA = 60°

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 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,

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

Angle of measure 180°:

Let m∠XOA = 180°

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

∴ x = – 1 and y = 0

sin 180° =y = 0

cos 180° = x = -1

tan 180° = \(\frac{y}{x}\)

= \(\frac{0}{-1}\) = 0

Cosec 180° = \(\frac{1}{y}\)

= \(\frac{1}{0}\)

which is not defined.

sec 180°= \(\frac{1}{x}=\frac{1}{-1}\) = -1

cot 180° = \(\frac{x}{y}=\frac{-1}{0}\) , which is not defined.

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,

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

∴ x = -OM = \(\frac{-\sqrt{3}}{2}\) and y = -PM = \(\frac{-1}{2}\)

∴ P ≡( \(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\) )

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

x = OM = \(\frac{1}{2}\) = and y = -PM = \(\frac{-\sqrt{3}}{2}\)

sin 300° = y = \(\frac{-\sqrt{3}}{2}\)

cos 300° = x = \(\frac{1}{2}\)

tan 300° = \(\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\)

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 4th quadrant, x > 0, y < 0

Angle of measure 30°

Let m∠XOA = -30°

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 4th quadrant x > 0, y < 0

Angle of measure 45°:

Let m∠XOA = 45°

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 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our calculation it is \(-\frac{1}{\sqrt{2}}\) ]

Angle of measure (-60°):

Let m∠XOA = -60°

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 4’ quadrant,

x > 0, y < 0

x = OM =\(\frac{1}{2}\) and y = -PM = \(-\frac{\sqrt{3}}{2}\)

Angle of measure (-90°):

Let m∠XOA = -90°

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

∴ x = 0 and y = -1

sin (-90°) = y = -1

cos (-90°) = s = 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 3rd 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 2nd quadrant, x < 0, y > 0

Angle of measure 2400):

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° – 900 triangle.

Since point P lies in the 2nd 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 and y = 1

sin (- 270°) = y = 1

cos (- 270°) = x = 0

tan(-270°)= \(\frac{y}{x}=\frac{1}{0}\)

which is not defined.

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,

Question 2.

State the signs of:

i. tan 380°

ii. cot 230°

iii 468°

Solution:

1. 380° = 360° + 20°

∴ 380° and 20° are co-terminal angles.

Since 0° < 20° <90°0,

20° lies in the l quadrant.

∴ 380° lies in the 1st quadrant,

∴ tan 380° is positive.

ii. Since, 180° <230° <270°

∴ 230° lies in the 3rd quadrant.

∴ cot 230° is positive.

iii. 468° = 360°+108°

∴ 468° and 108° are co-terminal angles.

Since 90° < 108° < 180°,

108° lies in the 2nd quadrant.

∴ 468° lies in the 2nd quadrant.

∴ sec 468° is negative.

Question 3.

State the signs of cos 4^{c} and cos 4°. Which of these two functions is greater?

Solution:

Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)

Since 1^{c} = 57° nearly,

180° < 4^{c} < 270°

∴ 4^{c} lies in the third quadrant.

∴ cos 4^{c} < 0 ………(ii)

From (i) and (ii),

cos 4° is greater.

Question 4.

State the quadrant in which 6 lies if

i. sin θ < 0 and tan θ > 0

ii. cos θ < 0 and tan θ > 0

Solution:

i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0

tan θ is positive in 1st and 3rd quadrants.

∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0

tan θ is positive in 1st and 3rd quadrants.

∴ θ lies in the 3rd quadrant.

Question 5.

Evaluate each of the following:

i. sin 30° + cos 45° + tan 180°

ii. cosec 45° + cot 45° + tan 0°

iii. sin 30° x cos 45° x lies tan 360°

Solution:

i. We know that,

sin30° = 1/2, cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 180° = 0

sin30° + cos 45° +tan 180°

= \(\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2}\)

ii. We know that,

cosec 45° = \(\sqrt{2}\) , cot 45° = 1, tan 0° = 0

cosec 45° + cot 45° + tan 0°

= \(\sqrt{2}\) + 1 + 0 = \(\sqrt{2}\) + 1

iii. We know that,

sin 30° = \(\frac{1}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 360° = 0

sin 30° x cos 45° x tan 360°

= \(\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\) = 0

Question 6.

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

Solution:

Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).

∴ x = 3 and y = -4

r = OP

Question 7.

If cos θ = \(\frac{12}{13}, 0<\theta<\frac{\pi}{2}\) find the value of \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^{2} \theta}\)

Solution:

cos θ = \(\frac{12}{13}\)

We know that,

sin^{2} θ = 1 – cos^{2}θ

∴ sin θ = ± \(\frac{5}{13}\)

Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, ∴ sin θ > 0

Question 8.

Using tables evaluate the following:

i. 4 cot 45° – sec2 60° + sin 30°

ii.\(\cos ^{2} 0+\cos ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{2}\)

Solution:

i. We know that,

cot 45° = 1, sec 60° = 2, sin 30° = 1/2

4 cot 45° – sec^{2} 60° + sin 30°

= 4(1) – (2)^{2} + \(\frac{1}{2}\)

= 4 – 4 + \(\frac{1}{2}=\frac{1}{2}\)

ii. We know that,

Question 9.

Find the other trigonometric functions if

i. cot θ = \(-\frac{3}{5}\), and 180 < θ < 270

ii. Sec A = \(-\frac{25}{7}\) and A lies in the second quadrant.

iii cot x = \(\frac{3}{4}\), x lies in the third quadrant.

iv. tan x = \(\frac{-5}{12}\) x lies in the fourth quadrant.

Solution:

i. cot θ = \(-\frac{3}{5}\)

we know that,

sin^{2}θ = 1 – cos^{2}θ

= 1 – \(\left(-\frac{3}{5}\right)^{2}\)

= 1 – \(\frac{9}{25}=\frac{16}{25}\)

∴ sin θ = ± \(\frac{4}{5}\)

Since 180° < 0 < 270°,

θ lies in the 3rd quadrant.

∴ sin θ < 0

Since A lies in the 2nd quadrant,

tan A < 0

iii. Given, cot x = \(\frac{3}{4}\)

We know that,

cosec^{2} x = 1 + cot^{2} x

= 1 + \(\left(\frac{3}{4}\right)^{2}=1+\frac{9}{16}=\frac{25}{16}\)

∴ cosec x = ± \(\frac{5}{4}\)

Since x lies in the 3rd quadrant, cosec x < 0

iv. Given, tan x = \(-\frac{5}{12}\)

sec^{2} x = 1 + tan^{2}

= 1 + \(\left(-\frac{5}{12}\right)^{2}\)

= 1 + \(\frac{25}{144}=\frac{169}{144}\)

∴ sec x = ± \(\frac{13}{12}\)

Since x lies in the 4th quadrant,

sec x > 0