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## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.1

### Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.1 Text Book Back Questions and Answers

Question 1.
Convert the following degree measure into radian measure
(i) 60°
(ii) 150°
(iii) 240°
(iv) -320°
Solutions:
(i) 1°= $$\frac{\pi}{180}$$ radians
∴ 60° = $$\frac{\pi}{180}$$ × 60 radians = $$\frac{\pi}{3}$$ radians.

(ii) 150° = $$\frac{\pi}{180}$$ × 150 radians = $$\frac{5 \pi}{6}$$ radians.

(iii) 240° = $$\frac{\pi}{180}$$ × 240 radians = $$\frac{4 \pi}{3}$$ radians.

(iv) -320° = $$\frac{\pi}{180}$$ × -320 = $$\frac{-16 \pi}{9}$$ radians Question 2.
Find the degree measure corresponding to the following radian measure.
(i) $$\frac{\pi}{8}$$
(ii) $$\frac{9 \pi}{5}$$
(iii) -3
(iv) $$\frac{11 \pi}{18}$$
Solution:
We know that, one radian = $$\frac{180^{\circ}}{\pi}$$
(i) $$\frac{\pi}{8}$$
$$\frac{\pi}{8}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{8}$$ degrees
= $$\frac{45}{2}$$
= 22.5°
= 22°30′ [∵ 0.5° = (0.5 × 60)’ = 30′]

(ii) $$\frac{9 \pi}{5}$$
$$\frac{9 \pi}{5}=\frac{180^{\circ}}{\pi} \times \frac{9 \pi}{5}$$ degrees
= 36 × 9 degrees
= 324°

(iii) -3 = -171.81°
= -171°48′ (∵ 0.8° = (0.8 × 60)’ = 48′)

(iv) $$\frac{11 \pi}{18}$$
$$\frac{11 \pi}{18}=\frac{180}{\pi} \times \frac{11 \pi}{18}$$
= 10 × 11°
= 110°

Question 3.
Determine the quadrants in which the following degree lie.
(i) 380°
(ii) -140°
(iii) 1195°
Solution:
(i) 380° = 360°+ 20°
This is of the form 360° + θ
∴ After one completion of the round, the angle is 20°, 380° lies in the I quadrant. (ii) -140° = -90° + (-50°)
The angle is negative it moves in the anti-clockwise direction.
-140° lies in the III quadrants. (iii) 1195° = (3 × 360°) + 90° + 25°
∴ After three completion round, the angle will lie in the II quadrant.
1195° lies in the II quadrant.  Question 4.
Find the values of each of the following trigonometric ratios.
(i) sin 300°
(ii) cos(-210°)
(iii) sec 390°
(iv) tan(-855°)
(v) cosec 1125°
Solution:
(i) sin 300° = sin(360° – 60°)
[For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative]
= -sin 60°
= $$-\frac{\sqrt{3}}{2}$$

(ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ)
[∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative]
= cos(180° + 30°)
= -cos 30°
= $$-\frac{\sqrt{3}}{2}$$

(iii) sec 390° = sec(360° + 30°)
= sec 30°
= $$\frac{1}{\cos 30^{\circ}}$$
= $$\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}$$
= $$\frac{2}{\sqrt{3}}$$

(iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ)
[∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative]
= -tan(2 × 360° + 135°)
= -tan 135°
= -tan(180° – 45°)
= -(-tan 45°)
= -(-1)
= 1

(v) cosec 1125° = cosec(3 × 360°+ 45°)
= cosec 45°
= $$\frac{1}{\sin 45^{\circ}}$$
= $$\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}$$
= √2

Question 5.
Prove that:
(i) tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0.
(ii) 2 sin2 $$\frac{\pi}{6}$$ + cosec2 $$\frac{7 \pi}{6}$$ cos2 $$\frac{\pi}{3}$$ = $$\frac{3}{2}$$
(iii) $$\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1$$
Solution:
(i) tan(-225°) = -(tan 225°)
= -(tan(180° + 45°))
= – tan 45°
= – 1
cot(-405°) = -(cot 405°)
= – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.]
= -cot 45°
= -1
tan(-765°) = -tan 765°
= -tan(2 × 360° + 45°)
= -tan 45°
= -1
cot 675° = cot (360°+ 315°)
= cot 315°
= cot(360° – 45°)
= -cot 45°
= -1
LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°)
= (-1) (-1) – (-1) (-1)
= 1 – 1
= 0
= RHS.
Hence proved.

(ii) 2 sin2 $$\frac{\pi}{6}$$ + cosec2 $$\frac{7 \pi}{6}$$ cos2 $$\frac{\pi}{3}$$ = $$\frac{3}{2}$$
LHS = 2 sin2 $$\frac{\pi}{6}$$ + cosec2 $$\frac{7 \pi}{6}$$ cos2 $$\frac{\pi}{3}$$
[∵ $$\frac{7 \pi}{6}$$ = 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio.
210° lies in 3rd quadrant, cosec θ is negative.]
= 2$$\left(\sin \frac{\pi}{6}\right)^{2}$$ + (cosec (180° + 30°))2 $$\left(\cos \frac{\pi}{3}\right)^{2}$$
= 2 $$\left(\frac{1}{2}\right)^{2}$$ + (-cosec 30°)2 . $$\left(\frac{1}{2}\right)^{2}$$
= $$2 \times \frac{1}{4}+(-2)^{2} \frac{1}{4}$$
= $$\frac{2}{4}+\frac{4}{4}=\frac{6}{4}$$
= $$\frac{6}{4}$$
= $$\frac{3}{2}$$
= RHS

(iii) sec($$\frac{3 \pi}{2}$$ – θ) = sec (270° – θ) = -cosec θ
[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’]
sec(θ – $$\frac{5 \pi}{2}$$) = $$\sec \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)$$
= sec($$\frac{5 \pi}{2}$$ – θ) [∵ sec(-θ) = θ]
= sec(450° – θ)
= sec (360° + (90° – θ))
= sec (90° – θ)
= cosec θ
[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’]
tan($$\frac{5 \pi}{2}$$ + θ) = tan(450° + θ)
[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’]
= tan (360° + (90° + θ))
= tan (90° + θ)
= -cot θ
$$\tan \left(\theta-\frac{5 \pi}{2}\right)=\tan \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)$$
= $$-\tan \left(\frac{5 \pi}{2}-\theta\right)$$ [∵ tan(-θ) = -tan θ]
= -tan(450° – θ)
= -tan(360° + (90° – θ))
= -tan(90° – θ)
= -cot θ
LHS = $$\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)$$
= -cosec θ (cosec θ) + (-cot θ) (-cot θ)
= -cosec2 θ + cot2 θ
= -(1 + cot2 θ) + cot2 θ [∵ 1 + cot2 θ = cosec2 θ]
= -1
= RHS Question 6.
If A, B, C, D are angles of a cyclic quadrilateral, prove that: cos A + cos B + cos C + cos D = 0.
Solution:
Note: If the vertices of a quadrilateral lie on the circle then the quadrilateral is called a cyclic quadrilateral.
In a cyclic quadrilateral sum of opposite angles are 180°. Since A, B, C, D are angles of cyclic quadrilateral
A + C = 180° and B + D = 180°
LHS = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
= RHS

Question 7.
Prove that (ii) sin θ . cos{sin($$\frac{\pi}{2}$$ – θ) . cosec θ + cos($$\frac{\pi}{2}$$ – θ) . sec θ} = 1
Solution:  (ii) sin θ . cos{sin($$\frac{\pi}{2}$$ – θ) . cosec θ + cos($$\frac{\pi}{2}$$ – θ) . sec θ} = 1  Question 8.
Prove that: cos 510° cos 330° + sin 390° cos 120° = -1.
Solution:
LHS = cos 510° cos 330° + sin 390° cos 120°
= cos(360° + 150°) cos(360° – 30°) + sin(360° + 30°) × cos(180° – 60°)
= cos 150° cos 30° + sin 30° (-cos 60°)
= cos(180° – 30°) cos 30° + sin 30° cos 60°
= -cos 30° cos 30° + $$\frac{1}{2} \times\left(\frac{-1}{2}\right)$$
= $$-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$$
= $$-\frac{3}{4}-\frac{1}{4}$$
= $$\frac{-3-1}{4}$$
= -1

Question 9.
Prove that:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = sin2 x. Solution:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = (tan x) (-cot(π – x) – cos x cos x
[∵ cot(x – π) = cot(-(π – x)) = -cot(π – x) = cot x]
= tan x cot x – cos2 x
= 1 – cos2 x
= sin2 x [∵ sin2 x + cos2 x = 1 ⇒ sin2 x = (1 – cos2 x)]   Question 10.
If sin θ = $$\frac{3}{5}$$, tan φ = $$\frac{1}{2}$$ and $$\frac{\pi}{2}$$ < θ < π < φ < $$\frac{3 \pi}{2}$$, then find the value of 8 tan θ – √5 sec φ.
Solution: Given that sin θ = $$\frac{3}{5}=\frac{\text { Opposite side }}{\text { Hypotenuse }}$$
∵ AB = $$\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}$$ = 4
Here θ lies in second quadrant [∵ $$\frac{\pi}{2}$$ < θ < π]
∵ tan θ is negative.
tan θ = $$-\frac{3}{4}$$
Also given that tan Φ = $$\frac{1}{2}=\frac{\text { Opposite side }}{\text { Adjacent side }}$$
∴ PR = $$\sqrt{\mathrm{PQ}^{2}+\mathrm{QP}^{2}}=\sqrt{4+1}=\sqrt{5}$$
Here Φ lies in third quadrant (∵ π < Φ < $$\frac{3 \pi}{2}$$)
∴ sec Φ is negative.
$$\sec \phi=\frac{1}{\cos \phi}=-\frac{1}{\left(\frac{2}{\sqrt{5}}\right)}=-\frac{\sqrt{5}}{2}$$
Now 8 tan θ – √5 sec Φ = $$8\left(-\frac{3}{4}\right)-\sqrt{5}\left(\frac{-\sqrt{5}}{2}\right)$$
= 2 × (-3) + $$\frac{5}{2}$$
= -6 + $$\frac{5}{2}$$
= $$\frac{-12+5}{2}$$
= $$\frac{-7}{2}$$