# Maharashtra Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.
If 2sin A = 1 = $$\sqrt{2}$$ cos B and $$\frac{\pi}{2}$$ < A < π, $$\frac{3 \pi}{2}$$
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos2 A = 1 – sin2 A = 1 – $$\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}$$
∴ cos A = $$\pm \frac{\sqrt{3}}{2}$$
Since $$\frac{\pi}{2}$$ < A < π
A lies in the 2nd quadrant.

We know that,
Sin2 B = 1 – cos2 B = 1 – $$\left(\frac{1}{\sqrt{2}}\right)^{2}$$$$\frac{1}{2}=\frac{1}{2}$$
∴ sin B = $$\pm \frac{1}{\sqrt{2}}$$
Since $$\frac{3 \pi}{2}$$ < B < 2π
B lies in the 4th quadrant,

Question 2.
If  and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, $$\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}$$
∴ sin A = $$\frac{3}{5}$$ and sin B = $$\frac{4}{5}$$
We know that,
cos2 A = 1 – sin2 = 1 – $$\left(\frac{3}{5}\right)^{2}$$ = 1 – $$\frac{9}{25}=\frac{16}{25}$$
∴ Cos A = ± $$[{4}{5}$$
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –$$\frac{4}{5}$$
Sin B = 4/5
We know that,
cos2B = 1 – sin2B = 1 – $$\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}$$
∴ Cos B = ±$$\frac{4}{5}$$
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = $$\frac{1}{2}$$, evaluate $$\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}$$
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = $$\frac{x}{3}$$ and tan θ= $$\frac{y}{4}$$
We know that,
sec2θ – tan2θ = 1

∴ 16x2 – 9y2 = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ =  and cot θ = 
We know that,
cosec2 θ – cot2 θ =

16x2 – 9y2 = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x2 + y2 = (4cos θ – 5sin θ)2 + (4sin θ + 5cos θ)2
= 16cos2θ – 40 sinθ cosθ + 25 sin2θ + 16 sin2 θ + 40sin θ cos θ + 25 cos2 θ
= 16(sin2 θ + cos2 θ) + 25(sin2 θ + cos2 θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = $$\frac{x-5}{6}$$ and cot θ = $$\frac{y-3}{8}$$
We know that,
cosec2 θ – cot2 θ = 1
∴ $$\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}$$ = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = $$\frac{3-2 x}{4}$$ and sec θ = $$\frac{3 y-5}{3}$$θ
We know that, sec2 θ – tan2 θ = 1
∴ $$\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}$$ = 1
∴ $$\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}$$ = 1

Question 5.
If 2sin2 θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin2 θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = $$\frac{-3}{2}$$
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
$$\sqrt{1-\cos ^{2} \theta}$$ = 0 …[ ∵ sin2 θ = 1- cos2 θ]
∴ 1 – cos2 θ = 0
∴ cos2 θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos2 θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos2θ – 11 cos θ + 5 = 0
∴ 2cos2 θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos2 θ = 3sin θ.
Solution:
2cos20 = 3sin θ
∴ 2(1 – sin2 θ) = 3sin θ
∴ 2 – 2sin2 θ = 3sin θ
∴ 2sin2 θ + 3sin 9-2 = θ
∴ 2sin2 θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan2 θ + 3 = 9sec θ
∴ 5(sec2 θ – 1) + 3 = 9sec θ
∴ 5sec2 θ – 5 + 3 = 9sec θ
∴ 5sec2 θ – 9sec θ – 2 = 0
∴ 5sec2 θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos2θ = 16(1 – sin θ)2
∴ 9(1 – sin2 θ) = 16(1 + sin2 θ – 2sin θ)
∴ 9 – 9sin2 θ = 16 + 16sin2 θ – 32sin θ
∴ 25sin2 θ – 32sin θ + 7 = 0
∴ 25sin2 θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = $$\frac{7}{25}$$
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or $$\frac{7}{25}$$
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or $$\frac{7}{25}$$.]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ $$\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5$$
∴ $$\frac{1+\cos \theta}{\sin \theta}=5$$
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos2 θ = 25 sin2 θ
∴ cos2 θ + 2 cos θ + 1 = 25 (1 – cos2 θ)
∴ cos2 θ + 2 cos θ + 1 = 25 – 25 cos2 θ
∴ 26 cos2 θ + 2 cos θ – 24 = 0
∴ 26 cos2 θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = $$\frac{24}{26}=\frac{12}{13}$$
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = $$\frac{12}{13}$$
∴ sec θ = $$\frac{1}{\cos \theta}=\frac{13}{12}$$
[Note: Answer given in the textbook is -1 or $$\frac{13}{12}$$.
However, as per our calculation it is only $$\frac{13}{12}$$.]

Question 11.
If cot θ = $$\frac{3}{4}$$ and π < θ < $$\frac{3 \pi}{2}$$, then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec2θ = 1 + cot2 θ = $$\left(\frac{3}{4}\right)^{2}$$ = 1 + $$\frac{9}{16}$$
∴ cosec2 θ = $$\frac{25}{16}$$
∴ cosec θ = $$\pm \frac{5}{4}$$
Since π < θ < $$\frac{3 \pi}{2}$$
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –$$\frac{5}{4}$$
cot θ = $$\frac{3}{4}$$
tan θ = $$\frac{1}{\cot \theta}=\frac{4}{3}$$
We know that,
sec2 θ = 1 + tan2 θ = 1 + $$\left(\frac{4}{3}\right)^{2}$$
= 1 + $$\frac{16}{9}=\frac{25}{9}$$
∴ sec θ = ±$$\frac{5}{3}$$
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –$$\frac{5}{3}$$
cos θ = $$\frac{1}{\sec \theta}=\frac{-3}{5}$$
∴ 4cosec θ + 5cos θ
= $$4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)$$
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, $$\sqrt{3}$$)
ii. (-1, -1) iv. (-$$\sqrt{3}$$, 1)
Solution:
i. (x, y) = (5, 5)
∴ r = $$\sqrt{x^{2}+y^{2}}$$ = $$\sqrt{25+25}$$
$$=\sqrt{50}=5 \sqrt{2}$$
tan θ = $$\frac{y}{x}=\frac{5}{5}$$ = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are ($$5 \sqrt{2}$$, 45°).

ii. (x, y) = ( 1, $$\sqrt{3}$$)
∴ r = $$\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2$$
tan θ = $$\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}$$
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = $$\sqrt{3}$$]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = $$\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}$$
tan θ = $$\frac{y}{x}=\frac{-1}{-1}=1$$
∴ tan θ = tan $$\frac{\pi}{4}$$
Since the given point lies in the 3rd quadrant,
tan θ = tan $$\left(\pi+\frac{\pi}{4}\right)$$ …[∵ tan (n + x) = tanx]
∴ tan θ = tan $$\left(\frac{5 \pi}{4}\right)$$
∴ θ = $$\frac{5 \pi}{4}$$ = 225°
∴ the required polar co-ordinates are ($$\sqrt{2}$$, 225°).

iv. (x, y) = (-$$\sqrt{3}$$ , 1)
∴ r = $$\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2$$
tan θ = $$\frac{y}{x}=\frac{1}{-\sqrt{3}}$$ = -tan $$\frac{\pi}{6}$$
Since the given point lies in the 2nd quadrant,
tan θ = tan $$\left(\pi-\frac{\pi}{6}\right)$$ …[∵ tan (π – x) = – tanx]
∴ tan θ = tan $$\left(\frac{5 \pi}{6}\right)$$
∴ θ = $$\frac{5 \pi}{6}$$ = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin$$\frac{19 \pi^{e}}{3}$$
ii. cos 1140°
iii. cot $$\frac{25 \pi^{e}}{3}$$
Solution:
i. We know that sine function is periodic with period 2π.
sin $$\frac{19 \pi}{3}$$ = sin $$\left(6 \pi+\frac{\pi}{3}\right)$$ = sin $$\frac{\pi}{3}=\frac{\sqrt{3}}{2}$$

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = $$\frac {1}{2}$$

iii. We know that cotangent function is periodic with period π.
cot $$\frac{25 \pi}{3}$$ = cot $$\left(8 \pi+\frac{\pi}{3}\right)$$ = cot $$\frac{\pi}{3}$$ = $$\frac{1}{\sqrt{3}}$$
dhana work.txt
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