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

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

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

Question 1.
If A is an acute angle and sin A = \(\\ \frac { 3 }{ 5 } \) find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:

Question 2.
If A is an acute angle and sec A = \(\\ \frac { 17 }{ 8 } \), find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:

Question 3.
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:

Question 4.
If tan A = \(\frac { 1 }{ \sqrt { 3 } } \), find all other trigonometric ratios of angle A.
Solution:

Question 5.
If 12 cosec θ = 13, find the value of \(\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } \)
Solution:

Without using trigonometric tables, evaluate the following (6 to 10):

Question 6.
(i) cos² 26° + cos 64° sin 26° + \(\frac { tan{ 36 }^{ ° } }{ { cot54 }^{ ° } } \)
(ii) \(\frac { sec{ 17 }^{ ° } }{ { cosec73 }^{ ° } } +\frac { tan68^{ ° } }{ cot22^{ ° } } \) + cos² 44° + cos² 46°
Solution:

Question 7.
(i) \(\frac { sin65^{ ° } }{ { cos25 }^{ ° } } +\frac { cos32^{ ° } }{ sin58^{ ° } } \) – sin 28° sec 62° + cosec² 30° (2015)
(ii) \(\frac { sin29^{ ° } }{ { cosec61 }^{ ° } } \) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).
Solution:

Question 8.
(i) \(\frac { { sin }35^{ ° }{ cos55 }^{ ° }+{ cos35 }^{ ° }{ sin }55^{ ° } }{ { cosec }^{ 2 }{ 10 }^{ ° }-{ tan }^{ 2 }{ 80 }^{ ° } } \)
(ii) \({ sin }^{ 2 }{ 34 }^{ ° }+{ sin }^{ 2 }{ 56 }^{ ° }+2tan{ 18 }^{ ° }{ tan72 }^{ ° }-{ cot }^{ 2 }{ 30 }^{ ° }\)
Solution:

Question 9.
(i) \({ \left( \frac { { tan25 }^{ ° } }{ { cosec }65^{ ° } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ ° } }{ { sec65 }^{ ° } } \right) }^{ 2 }+{ 2tan18 }^{ ° }{ tan }45^{ ° }{ tan72 }^{ ° } \)
(ii) \(\left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ ° }-\theta \right) -cot\theta tan\left( { 90 }^{ ° }-\theta \right) \)
Solution:

Question 10.
(i) 2(sec² 35° – cot² 55°) – \(\frac { { cos28 }^{ ° }cosec{ 62 }^{ ° } }{ { tan18 }^{ ° }tan{ 36 }^{ ° }{ tan30 }^{ ° }{ tan54 }^{ ° }{ tan72 }^{ ° } } \)
(ii) \(\frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ ° }+{ cos }^{ 2 }{ 42 }^{ ° }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ ° }{ sec }^{ 2 }{ 52 }^{ ° }{ sin }^{ 2 }{ 38 }^{ ° } }{ { cosec }^{ 2 }{ 70 }^{ ° }-{ tan }^{ 2 }{ 20 }^{ ° } } \)
Solution:

Question 11.
Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) \(\frac { tan\theta }{ tan({ 90 }^{ ° }-\theta ) } +\frac { sin({ 90 }^{ ° }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta \)
(iii) \(\frac { cos({ 90 }^{ ° }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ ° }-\theta )=1\)
(iv) sin (90° – θ) cos (90° – θ) = \(\frac { tan\theta }{ { 1+tan }^{ 2 }\theta } \)
Solution:

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

Question 12.
(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan² A) (1 – sin A) (1 + sin A) = 1.
Solution:

Question 13.
(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.
Solution:

Question 14.
(i) \(\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A\)
(ii) \(\frac { 1 }{ secA+tanA } +\frac { 1 }{ secA-tanA } =2{ sec }A\)
Solution:

Question 15.
(i) \(\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA } \)
(ii) \(\frac { 1-{ tan }^{ 2 }A }{ { cot }^{ 2 }A-1 } ={ tan }^{ 2 }A\)
(iii) \(\frac { sinA }{ 1+cosA } =cosecA-cotA\)
Solution:

Question 16.
(i) \(\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA } \)
(ii) \(\frac { { tan }^{ 2 }\theta }{ { (sec\theta -1) }^{ 2 } } =\frac { 1+cos\theta }{ 1-cos\theta } \)
(iii) \({ (1+tanA) }^{ 2 }+{ (1-tanA) }^{ 2 }=2{ sec }^{ 2 }A\)
(iv) \({ sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A\)
Solution:

Question 17.
(i) \(\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { tanA }{ secA-1 } +\frac { tanA }{ secA+1 } =2cosecA\)
Solution:

Question 18.
(i) \(\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A\)
(ii) \(cotA-tanA=\frac { { 2cos }^{ 2 }A-1 }{ sinA-cosA } \)
(iii) \(\frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA } \)
Solution:

Question 19.
(i) \({ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta \)
(ii) \(\frac { cos\theta }{ 1-tan\theta } -\frac { { sin }^{ 2 }\theta }{ cos\theta -sin\theta } =cos\theta +sin\theta \)
Solution:

Question 20.
(i) cosec⁴ θ – cosec² θ = cot⁴ θ + cot² θ
(ii) 2 sec² θ – sec⁴ θ – 2 cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ.
Solution:

Question 21.
(i) \(\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta \)
(ii) \(\frac { { tan }^{ 3 }\theta -1 }{ tan\theta -1 } ={ sec }^{ 2 }\theta +tan\theta \)
Solution:

Question 22.
(i) \(\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA } \)
(ii) \(\sqrt { \frac { 1-cosA }{ 1+cosA } } =\frac { sinA }{ 1+cosA } \)
Solution:

Question 23.
(i) \(\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA\)
(ii) \(\sqrt { \frac { 1-cosA }{ 1+cosA } } =cosecA-cotA\)
Solution:

Question 24.
(i) \(\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA\)
(ii) \(\frac { cotAcotA }{ 1-sinA } =1+cosecA \)
Solution:

Question 25.
(i) \(\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)\)
(ii) \({ sec }^{ 4 }A-{ tan }^{ 4 }A=1+2{ tan }^{ 2 }A \)
Solution:

Question 26.
(i) cosec⁶ A – cot⁶ A = 3 cot² A cosec² A + 1
(ii) sec⁶ A – tan⁶ A = 1 + 3 tan² A + 3 tan⁴ A
Solution:

Question 27.
(i) \(\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta } \)
(ii) \(\frac { sin\theta }{ cot\theta +cosec\theta } =2+\frac { sin\theta }{ cot\theta -cosec\theta } \)
Solution:

Question 28.
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
(ii) (cosecA – sinA)(secA – cosA) sec²A = tanA
(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1
Solution:

Question 29.
(i) \(\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2\)
(ii) \(\frac { { tan }^{ 2 }A }{ { 1+tan }^{ 2 }A } +\frac { cot^{ 2 }A }{ 1+{ cot }^{ 2 }A } =1\)
Solution:

Question 30.
(i) \(\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA } \)
(ii) \({ (sinA+secA) }^{ 2 }+{ (cosA+cosecA) }^{ 2 }={ (1+secA\quad cosecA) }^{ 2 }\)
(iii) \(\frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 } \)
Solution:

Question 31.
If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1
Solution:

Question 32.
If 7 sin² θ + 3 cos² θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.
Solution:

Question 33.
If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.
Solution:

Question 34.
If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x² – y² = a² – b².
Solution:

Question 35.
If x = h + a cos θ and y = k + a sin θ, prove that (x – h)² + (y – k)² = a².
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths