ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

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Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \(\\ \frac { 8 }{ 15 } \), find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q1.4

Question 2.
Evaluate the following:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) \(\frac { { sec }29^{ O } }{ { cosec }61^{ O } } \) + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) \(\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } } \) + sin2 63° + cos 63° sin 27°
Solution:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q2.3

Question 3.
If \(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \), then find the value of x.
Solution:
Given
\(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q3.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q3.2

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4.
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA \)
Solution:
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
L.H.S = \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q4.2

Question 5.
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
L.H.S = \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q5.2

Question 6.
(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) \(\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta \)
Solution:
L.H.S. = sin2 θ + cos4 θ
= (1 – cos2 θ + cos4 θ
= 1 – cos2 θ + cos4 θ
= 1 – cos2 θ (1 – cos2 θ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q6.2

Question 7.
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) \(\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA\)
Solution:
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q7.3

Question 8.
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.
Solution:
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
L.H.S = \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q8.2

Question 9.
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \(\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta } \)
Solution:
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
L.H.S = \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.3
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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q9.5

Question 10.
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
Solution:
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q10.2

Question 11.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
Solution:
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q11.1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q12.2

Question 13.
If sec θ + tan θ = p, prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Solution:
sec θ + tan θ = p,
prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
\(\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q13.2

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos2 A = \(\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)
Solution:
m = \(\\ \frac { sinA }{ sinB } \)
n = \(\\ \frac { tanA }{ tanB } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q14.1

Question 15.
If sec A = \(x+ \frac { 1 }{ 4x } \), then prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
Solution:
sec A = \(x+ \frac { 1 }{ 4x } \)
To prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q15.2

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test Q16.3

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