Maharashtra Board 11th Maths Solutions Chapter 2 Trigonometry – I Miscellaneous Exercise 2

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

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Miscellaneous Exercise 2

I. Select the correct option from the given alternatives.

Question 1.
The value of the expression
cos1°. cos2°. cos3° … cos 179° =
(A) -1
(B) 0
(C) \(\frac{1}{\sqrt{2}}\)
(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.
\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to
(A) 2cosec A
(B) 2 sec A
(C) 2 sin A
(D) 2 cos A
Answer:
(A) 2cosec A

Explanation:

Question 3.
If α is a root of 25cos² θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to
(A) \(-\frac{24}{25}\)
(B) \(-\frac{13}{18}\)
(C) \(\frac{13}{18}\)
(D) \(\frac{24}{25}\)
Answer:
(A) \(-\frac{24}{25}\)

Explanation:

25 cos² θ + 5 cos θ – 12 = 0
∴ (5cos θ + 4) (5 cos θ – 3) = 0
∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π,
cos α < 0
∴ cos α = \(-\frac{4}{5}\)
sin² α = 1 – cos² α = 1 – \(\frac{16}{25}=\frac{9}{25}\)
∴ sin α = \(\pm \frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π sin α > 0
∴ sin α = 3/5
sin 2 α = 2 sin α cos α
= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.
If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to
(A) \(\frac{\sqrt{3}}{2}\)
(B) \(\frac{2}{\sqrt{3}}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt{3}\)
Answer:
(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.
If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to
(A) 2
(B) mn
(C) 2m
(D) 2n
Answer:
(A) 2
Explanation:

Question 6.
If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is
(A) \(\frac{14}{25}\)
(B) \(\frac{20}{21}\)
(C) \(\frac{21}{20}\)
(D) \(\frac{15}{16}\)
Answer:
(B) \(\frac{20}{21}\)

Explanation:
cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1
∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)
Subtracting (ii) from (i), we get
2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)
∴ cot θ = \(\frac{21}{20}\)
∴ tan θ = \(\frac{20}{21}\)

Question 7.
\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals
(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
(A) \(\frac{2 q}{1+q^{2}}\)
(B) \(\frac{2 q}{1-q^{2}}\)
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)
(D) \(\frac{1+q^{2}}{2 q}\)
Answer:
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ (cosec θ + cot θ)q = 1
∴ cosec θ + cot θ = 1/q …….(ii)
Subtracting (i) from (ii), we get
2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)
∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.
The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in
(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
(C) \(\frac{\pi}{2}\)
(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
tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined
cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1
sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined
cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 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

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

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° = \(\frac{y}{x}\)

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

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

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 if
i. tan θ < 0 and sec θ > 0
ii. sin θ < 0 and cos θ < 0
iii. sin θ > 0 and tan θ < 0
Solution:
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
= sin² θ + cos² θ – 2sin θ cos θ
= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

Question 7.
Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R.
Solution:

Question 8.
If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.
Solution:
Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
we know that
cos²θ = 1 – sin² θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.
If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)
Solution:
Given sec θ = \(\sqrt{2}\)
We know that,
tan² θ = sec² θ – 1
= (\(\sqrt{2}\)) – 1
= 2 – 1 = 1
∴ tan θ = ±1
Since \(\frac{3 \pi}{2}\) < θ < 2π
θ lies in the 4th quadrant.
∴ tan θ < 0
∴ tan θ = -1

Question 10.
Prove the following:

i. sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1
Solution:
L.H.S. = sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A (cos² B + sin² B) + cos² A (sin² B + cos² B)
= sin²A(1) + cos²A(1)
= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)
Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)
Solution:
L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)
= (tanθ + secθ)² + (tanθ – secθ)²
= tan² θ + 2 tan θ sec θ + sec² θ
+ tan² θ – 2 tan θ sec θ +.sec² θ
= 2(tan² θ + sec² θ)

iv. 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ
Solution:
LHS.
= 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  = 2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²
= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ)
+ (1+ cot² θ)²
= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ)
– 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ
= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2
– 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ
= cot⁴ θ – tan⁴ θ = R.H.S.

v. sin⁴ θ + cos⁴ θ = sin⁴ θ + cos⁴ θ
Solution:
L.H.S. = sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ
… [ v a² + b² = (a + b)² – 2ab]
= 1 – 2sin² θ cos² θ
= R.H.S.

vi. 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0
L.H.S =
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0
= sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin2 0 + cos2 0)
…[••• a³ + b³ = (a + b)³ – 3ab(a + b)]
= (1)³ – 3 sin² θ cos² θ(1)
= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos² θ
…[Y a² + b² = (a + b)² – 2ab]
= 1-2 sin² θ cos² θ
L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1
= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c
= R.H.S.

vii. cos⁴ θ – sin⁴ θ + 1 = 2cos²θ
L.H.S. = cos⁴ θ – sin⁴ θ + 1
= (cos² θ)² – (sin² θ)² + 1 = (cos²θ + sin²θ) c(os² θ – sin²θ) +1
= (1) (cos²θ – sin²θ) + 1 = cos² θ + (1 – sin²θ)
= cos² θ + cos²θ = 2cos²θ = R.H.S.

viii. sin⁴θ + 2sin²θ cos²θ = 1 – cos⁴θ
L.H.S. = sin⁴θ + 2sin²θ cos²θ = sin²θ(sin²θ + 2cos²θ)
= (sin²θ) (sin²θ + cos²θ + cos²θ) = (1 – cos²θ) (1 + cos²θ)
= 1 – cos⁴θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)
Solution:

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)
= 2 (sin² θ + cos² θ)
= 2(1)
= 2 = R.H.S.

x. tan² θ – sin² θ = sin⁴ θ sec² θ
Solution:
L.H.S. = tan² θ – sin² θ
= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin²θ
= sin² θ (\(\frac{1}{\cos ^{2} \theta}-1 \))
= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)
= (sin² θ) (sin² θ)sec² θ
= sin⁴ θ sec² θ
= R.H.S

xi. (sinθ + cosecθ)² + (cos θ + see θ)² = tan² θ + cot² θ + 7
Solution:
L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)²
= sin ² θ + cosec² θ + 2sinθ cosec θ
+ cos² θ + sec² θ + 2sec0 cos0
= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2
= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7
= R.H.S.

xii. sin⁸θ – cos⁸θ = (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
Solution:
L.H.S. = sin⁸θ – cos⁸θ
= (sin⁴θ)² – (cos⁴θ)²
= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)
= [(sin² θ)² – (cos² θ)² ]
. [(sin² θ)² + (cos² θ)² ]
= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] …[Y a² + b² = (a + b)² – 2ab]
= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ)
= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
= R.H.S.

xiii. sin⁶A + cos⁶A = 1 – 3 sin²A + 3sin⁴A
Soluiton:
L.H.S. = sin⁶A + cos⁶A
= (sin² A)³ + (cos² A)³
= (sin² A + cos² A)³
– 3sin²A cos²A(sin² A + cos² A)
…[ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3sin²A cos²A (1)
= 1 – 3sin²A cos²A
= 1 – 3 sin²A (1 – sin²A)
= 1 – 3 sin²A + 3sin⁴A
= R.H.S.

xiv. (1 + tanA tanB)² + (tanA – tanB)² = sec ²A sec²B
Solution:
L.H.S. = (1 + tanA tanB)² + (tanA – tanB)²
= 1 + 2tanA tanB + tan²A tan² + tan² A- 2tanA tanB + tan²B
= 1 + tan²A + tan² B + tan²A tan²B
= 1(1+ tan²A) + tan² B(1 + tan²A)
= (1 + tan²A) (1 + tan²B)
= sec²A sec²B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)
Solution:
We know that cosec²θ – cot² θ = 1
∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)
Solution:
We know that
tan²θ = sec² θ – 1
∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)
Solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)
solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)