Contents
GSEB Solutions for Class 10 mathematics – Trigonometry (English Medium)
Exercise-9.1
Question 1:
In ∆ABC, m∠A = 90. If AB = 5, AC = 12 and BC = 13, find sinC, cosC, tanB, cosB, sinB.
Solution :
Question 2:
In ∆ABC, m∠B = 90. If BC = 3 and AC = 5, find all the six trigonometric ratios of ∠A.
Solution :
Question 3:
If cosA = \(\frac{4}{5}\) find sinA and tanA.
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Question 4:
If coseceθ = \(\frac{13}{5}\), find tanθ and cosθ.
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Question 5:
If cosB = \(\frac{1}{3}\), find the other five trigonometric ratios.
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Question 6:
In ∆ABC, m∠A = 90 and if AB : BC = 1 : 2 find sinB, cosC, tanC.
Solution :
Question 7:
If tanθ = \(\frac{4}{3}\), find the value of \(\frac{5\sin \theta +2\cos \theta }{3\sin \theta -\cos \theta }\).
Solution :
Question 8:
If secθ = \(\frac{13}{5}\), find the value of \(\frac{2\sin \theta +3\cos \theta }{5\cos \theta -4\sin \theta }\).
Solution :
Question 9:
If sinB = \(\frac{1}{2}\), prove that 3cosB – 4cos3B = 0.
Solution :
Question 10:
If tanA = \(\sqrt{3}\) verify that
- sin2A + cos2A = 1
- sec2A – tan2A = 1
- 1 + cot2A = cosec2A
Solution :
Question 11:
If cosθ = \(\frac{2\sqrt{2}}{3}\), verify that tan2θ – sin2θ = tan2θ . sin2θ.
Solution :
Question 12:
In ∆ABC, m∠B = 90, AC + BC = 25 and AB = 5, determine the value of sinA, cosA and tanA.
Solution :
Question 13:
In ∆ABC, m∠c = 90 and m∠A = m∠B,
- Is cosA = cosB ?
- Is tanA = tanB ?
- Will the other trigonometric ratios of ∠A and ∠B be equal ?
Solution :
Question 14:
If 3cotA = 4, examine whether \(\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\) = cos2A – sin2A.
Solution :
Question 15:
If pcotθ = q, examine whether \(\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }\) = \(\frac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}\)
Solution :
Question 16:
State whether the following are true or false. Justify your answer :
- Sinθ = \(\frac{3}{2}\), for some angle having measure θ.
- Cosθ = \(\frac{2}{3}\), for some angle having measure θ.
- cosecA = \(\frac{5}{2}\) for some measure of angle A.
- The value of tanA is always less than 1.
- secB = \(\frac{3}{5}\) for some ∠B.
- cosθ = 100 for some angle having measure θ.
Solution :
Exercise-9.2
Question 1:
Verify :
- cos60 = 1 – 2sin230 = 2cos230 – 1 = cos230 – sin230
- sin60 = 2sin30 cos30
- sin60 = \(\frac{2\tan 30}{1+{{\tan }^{2}}30}\)
- cos60 = \(\frac{1-{{\tan }^{2}}30}{1+{{\tan }^{2}}30}\)
- cos90 = 4cos330 — 3cos30
Question 1(1):
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Question 1(2):
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Question 1(3):
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Question 1(4):
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Question 1(5):
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Question 2:
Evaluate :
- \(\frac{\sin 30+\tan 45-\cos ec60}{\sec 30+\cos 60+\cot 45}\)
- \(\frac{5{{\cos }^{2}}60+4{{\sec }^{2}}30-{{\tan }^{2}}45}{{{\sin }^{2}}30+{{\cos }^{2}}30}\)
- 2sin230 cot30 – 3cos260 sec230
- 3cos230 + sec230 + 2cos0 + 3sin90 – tan260
Solution :
Question 3:
In ∆ABC, m∠B = 90, find the measure of the parts of the triangle other than the ones which are given below :
- m∠C = 45, AB = 5
- m∠A= 30, AC = 10
- AC = 6\(\sqrt{2}\), BC = 3\(\sqrt{6}\)
- AB = 4, BC = 4
Question 3(1):
Solution :
For ∆ABC, ∠B is a right angle.
m∠A + m∠B + m∠C = 180°
∴ m∠A + 90° + 45° = 180°
∴ m∠A = 45°
Now, m∠A = m∠C = 45°
∴ BC = AB
(∵ The sides opposite to congruent angles are equal.)
∴ BC = 5
Question 3(2):
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Question 3(3):
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Question 3(4):
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Question 4:
In a rectangle ABCD, AB = 20, m∠BAC = 60, calculate the length of side \(\overline{BC}\) and diagonals \(\overline{AC}\) and \(\overline{BD}\).
Solution :
Question 5:
If θ is measure of an acute angle and cosθ = sinθ, find the value of 2tan2θ + sin2θ + 1
Solution :
Question 6:
If α is measure of an acute angle and 3sinα = 2cosα, prove that \({{\left( \frac{1-{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha } \right)}^{2}}+{{\left( \frac{2\tan \alpha }{1+{{\tan }^{2}}\alpha } \right)}^{2}}=1\)
Solution :
Question 7:
If A = 30 and B = 60, verify that
- sin(A + B) = sinA cosB + cosA sinB,
- cos(A + B) = cosA cosB – sinA sinB
Solution :
Question 8:
If sin(A – B) = sinA cosB – cosA sinB and cos(A – B) = cosA cosB + sinA sinB, find the values of sin15 and cos15.
Solution :
Question 9:
State whether the following are true or false. Justify your answer :
- The value of sinθ increases as θ increases from 0 to 90.
- sinθ = cosθ for all value of θ.
- cos(A + B) = cosA + cosB
- tanA is not defined for A = 90.
- The value of cotθ increases as θ increases from 0 to 90.
Solution :
Exercise-9.3
Question 1:
Evaluate :
- \(\frac{\cos 18}{\sin 72}\)
- tan48 – cot42
- cosec32 – sec58
- \(\frac{\cos 70}{\sin 20}\)+cos59 ∙ cosec31
- Sec70 sin20 – cos20 cosec70
- Cos(40 – θ) – sin(50 + θ) + \(\frac{{{\cos }^{2}}40+{{\cos }^{2}}50}{{{\sin }^{2}}40+{{\sin }^{2}}50}\)
- \(\frac{\cos 70}{\sin 20}\)+ \(\frac{\cos 55\cos ec35}{\tan 5\tan 25\tan 45\tan 65\tan 85}\)
- Cot12 ∙ cot38 ∙ cot52 ∙ cot60 ∙ cot78
- \(\frac{\sin 18}{\cos 72}\)+ \(\sqrt{3}\)(tan10 tan30 tan45 tan50 tan80)
- \(\frac{\cos 58}{\sin 32}\)+ \(\frac{\sin 22}{\cos 68}\)– \(\frac{\cos 38\cos ec52}{\tan 18\tan 35\tan 60\tan 72\tan 55}\)
Question 1(1):
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Question 1(2):
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Question 1(3):
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Question 1(7):
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Question 1(8):
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Question 1(9):
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Question 1(10):
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Question 2:
Prove the following :
Question 2(1):
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Question 2(2):
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Question 2(3):
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Question 2(4):
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Question 3:
Express the following in terms of trigonometric ratios of angles having measure between 0 and 45:
- sin85 + cosec85
- cos89 + cosec87
- sec81 + cosec54
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Question 4:
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Question 5:
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Question 6:
If 3θ is the measure of an acute angle and sin3θ = cos(θ – 26), then find the value of θ.
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Question 7:
If 0 < θ < 90, θ, sinθ = cos30, then obtain the value of 2tan2θ – 1.
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Question 8:
If tanA = cotB, prove that A + B = 90, where A and B are measures of acute angles.
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Question 9:
If sec2A = cosec(A – 42), where 2A is the measure of an acute angle, find the value of A.
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Question 10:
If 0 < θ < 90 and secθ = cosec60, find the value Of 2cos2θ – 1.
Solution :
Exercise-9
Prove the following by using trigonometric identities : (1 to 19)
Question 1:
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Question 2:
2sin2θ + 4sec2θ + 5cot2θ + 2cos2θ – 4tan2θ – 5cosec2θ = 1
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Question 3:
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Question 4:
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Question 5:
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Question 6:
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Question 7:
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Question 8:
(sinθ + cosecθ)2 + (cosθ + secθ)2 = 7 + tan2θ + cot2θ.
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Question 9:
2sec2θ – sec4θ – 2cosec2θ + cosec4θ = cot4θ – tan4θ.
Solution :
L.H.S.= 2sec2θ – sec4θ – 2cosec2θ + cosec4θ
= 2(1 + tan2θ) – (1 + tan2θ)2 – 2(1 + cot2θ) + (1 + cot2θ)2
(∵ Subs. sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ)
= 2 + 2tan2θ – (1 + 2tan2θ + tan4θ) – 2 – 2cot2 θ + (1 + 2cot2 θ + cot4 θ)
= 2 + 2tan2θ – 1 – 2tan2θ – tan4θ – 2 – 2cot2θ + 1 + 2cot2 θ + cot4 θ
= cot4 θ – tan4 θ
= R.H.S.
Question 10:
(sinθ – secθ)2 + (cosθ – cosecθ)2 = (1 – secθ ∙ cosecθ)2.
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Question 11:
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Question 12:
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Question 13:
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Question 14:
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Question 15:
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Question 16:
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Question 17:
sin4θ – cos4θ = sin2θ – cos2θ = 2sin2θ – 1 = 1 – 2cos2θ.
Solution :
We have, sin4θ – cos4θ
= (sin2θ – cos2θ)(sin2θ + cos2θ)
= sin2θ – cos2θ .. .. (1)
Now, sin2θ – cos2θ
= sin2θ – (1 – sin2θ)
= 2sin2θ – 1 .. .. (2)
Also sin2θ – cos2θ
= 1 – cos2θ – cos2θ
= 1 – 2cos2θ .. .. (3)
Hence, from (1), (2) and (3), we get
sin4θ – cos4θ = sin2θ – cos2θ
= 2sin2θ – 1 = 1 – 2 – cos2θ
Question 18:
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Question 19:
2(Sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1 = 0
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Question 20:
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p.
Solution :
Question 21:
If tanθ + Sinθ = a and tanθ – sinθ = b, then prove that a2 – b2 = 4\(\sqrt{ab}\)
Solution :
Question 22:
acosθ + bsinθ = p and asinθ – bcosθ = q, then prove that a2 + b2 = p2 + q2.
Solution :
We have, acosθ + bsinθ = p ……(i)
asinθ – bcosθ = q ……(ii)
Squaring and adding (i) and (ii), w get
∴ (acosθ + bsinθ)2 + (asinθ – bcosθ)2 = p2 + q2
∴ a2cos2θ + 2abcosθsinθ + b2sin2θ + a2sin2θ – 2absinθcosθ + b2cos2 = p2 + q2
∴ a2(cos2θ + sin2θ) + b2(sin2θ + cos2θ) = p2 + q2
∴a2(1) + b2(1) = p2 + q2
∴ a2 + b2 = p2 + q2
Question 23:
secθ + tanθ = p, then obtain the values Of secθ, tanθ and Sinθ in terms Of p.
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Question 24:
Evaluate the following :
Question 24(1):
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Question 24(2):
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Question 25:
If sinA + cosA = \(\sqrt{2}\)sin(90 – A), then obtain the value of cotA.
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Question 26:
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Question 27:
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Question 28:
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Question 29:
Select a proper option (a), (b), (c) or (d) from given options :
Question 29(1):
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