Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles download pdf.
RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles
Exercise 8
Question 1.
Question 2:
Solution:
(i) LHS = cos81° – sin9°
= cos(90° -9°)- sin9°
= sin9° – sin9°
= 0
= RHS
(ii) LHS = tan71° – cot19°
=tan(90° – 19°) – cot19°
=cot19° – cot19°
=0
= RHS
(iii) LHS = cosec80° – sec10°
= cosec(90° – 10°) – sec(10°)
= sec10° – sec10°
= 0
= RHS
(iv) cosec^2 72° − tan^2 18° = 1
LHS = cosec^2 72° – tan^2 18°
= cosec^2 72° – tan2 (90 – 72)°
= cosec^2 72° – cot^2 72°
= 1 = RHS
(v) cos^2 75° + cos^2 15° = 1
LHS = cos^2 75° + cos^2 15°
= cos^2 75° + cos^2 (90 – 75)°
= cos^2 75° + sin^2 75°
= 1= RHS
(vi) tan^2 66° − cot^2 24° = 0
LHS = tan^2 66° − cot^2 24°
= tan^2 66° – cot^2 (90 – 66)°
= tan^2 66° – tan^2 66°
= 0
= RHS
(vii) sin^2 48° + sin^2 42° = 1
LHS = sin^2 48° + sin^2 42°
= sin^2 48° + sin^2 (90 – 48)°
= sin^2 48° + cos^2 48°
= 1
= RHS
(viii) cos^2 57° − sin^2 33° = 0
LHS = cos^2 57° − sin^2 33°
= cos^2 57° – sin^2 (90 – 57)°
= cos^2 57° – cos^2 57°
= 0
=RHS
(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0
LHS = (sin 65° + cos 25°)(sin 65° − cos 25°)
= sin^2 65° – cos^2 25°
= sin^2 65° – cos^2 (90 – 65)°
= sin^2 65° – sin^2 65°
= 0
=RHS
Question 3.
Solution:
(i) LHS = sin53° cos37° + cos53° sin37°
= sin53° cos(90° – 53°) + cos53° sin (90° – 53°)
= sin53° x sin53° + cos53° x cos53°
= sin^2 53° + cos^2 53°
= 1
= RHS
(ii) LHS = cos54° cos36° − sin54° sin36°
= cos54° cos36° − sin(90° -36°) sin(90° -54°)
= cos54° cos36° – cos36°cos54°
= 0
=RHS
(iii) LHS = sec70° sin20° + cos20° cosec70°
= sec(90° – 20°) sin20° + cos20° cosec(90° – 20°)
= cosec 20° sin20° + cos20° sec 20°
= 1 +1
= 2
=RHS
(iv) LHS = sin35° sin55° − cos35° cos55°
= sin(90° – 55°) sin(90° – 35°) − cos35° cos55°
= cos55° cos35° – cos35° cos55°
= 0
=RHS
(v) LHS = (sin72° + cos18°)(sin72° − cos18°)
=(sin^2 72° – cos^2 18°)
= (sin^2 72° – cos^2 (90° – 72°))
= sin^2 72° – sin^2 72°
= 0
=RHS
(vi) LHS = tan48° tan23° tan42° tan67°
=tan48° tan23° tan (90° – 48°) tan (90° – 23°)
= tan48° tan23° cot48° cot23°
= 1×1
= 1
=RHS
Question 4.
Solution:
(v)
Question 5.
Solution:
(i)
LHS = sin θ cos (90° – θ) + sin (90° – θ) cos θ
= sin θ sin θ + cos θ cos θ
= sin^2 θ + cos^2 θ
= 1
= R.H.S.
Hence proved.
(vi)
=(1+1)/(3×1)
= 2/3 = RHS
Hence proved.
(vii)
LHS = cot θ tan (90° -θ) – sec (90° – θ)cosec θ +√3tan 12°tan 60°tan 78°
= cot θ cot θ – cosec θ cosec θ +√3 tan 60° tan 12° tan 78°
= cot^2 θ – cosec^2 θ +√3 tan 60° tan 12° tan(90-12)°
= – (cosec^2 θ – cot^2 θ) +√3 tan 60° tan 12° cot 12°
= – 1 + √3(√3 × 1)
= – 1 + 3
= 2
= R.H.S.
Hence proved.
Question 6:
Solution:
Question 7:
Solution:
L.H.S.
= sin (70°+θ) — cos (20° — θ)
= sin (70°+θ) — cos [90°-(70° + θ)]
= sin (70°+θ) — sin (70° + θ)
= 0
= R.H.S.
Hence Proved.
(ii)
L.H.S.
= tan (55° — θ) — cot (35° + θ)
= tan (90°-(35° +θ)) — cot (35° + θ)
= cot (35° + θ) – cot (35° + θ)
= 0
=RHS
Hence Proved.
(iii)
L.H.S.
= cosec (67° + θ) — sec (23° — θ)
= cosec (67° + θ) – sec (90° —(23° + θ))
= cosec (67° + θ) – cosec (67° + θ)
= 0
=RHS
Hence Proved.
(iv)
L.H.S.
= cosec (65° + θ) — sec (25° — θ) — tan (55° — θ) + cot (35° + θ)
= cosec (65° + θ) — sec (90° – (65° + θ)) — tan (90° – (35° + θ)) + cot (35° + θ)
= cosec (65° + θ) — cosec (65° + θ)) — cot (35° + θ) + cot (35° + θ)
= 0
= R.H.S.
(v)
L.H.S.
= sin (50° +θ) — cos (40° — θ) + tan 1° tan 10° tan 80° tan 89°
= sin ((90°-(40° – θ)) — cos (40° — θ) + (tan 1° tan 89°)(tan10° tan 80°)
= cos (40° – θ) – cos (40° – θ) + {tan 1° tan (90°-1°)}{tan10° tan(90°-10°)}
= 0 + {(tan 1° cot 1°}{tan10° cot 10°}
= 0 + 1 = 1
= R.H.S.
Question 8.
Solution:
Question 9.
Solution:
Given function is : tanC+A2=cotB2
Sum of all the angles of a triangle = 180 degree
So, A + B + C = 180o
Or A + C = 180o – B
And, (A + C)/2 = (180o – B)/2 = 90o – B/2
Now, tan (A + C)/2 = tan(90o – B/2) = cot B/2
Hence Proved.
Question 10.
Solution:
cos 2θ = sin 4θ …(1)
We know that,
sin( 90° – θ) = cos θ
So, equation (1) can be written as
sin (90° – 2θ) = sin 4θ
On comparing both sides
90° – 2θ = 4θ
90° = 4θ + 2θ
6θ = 90°
or θ = 15°
The value of θ is 15°
Question 11:
Solution: Given: sec2A = cosec(A − 42°)
The value of angle A is 44 degrees.
Question 12:
Solution:
sin 3 A = cos (A − 26°) (given)
or cos (90° – 3A) = cos (A – 26°)
On comparing
90° – 3A = A – 26°
A + 3A = 90° + 26°
4A = 116° = 29°
The value of A is 29°.
Question 13:
Solution:
tan 2A = cot (A – 12°)
or cot (90° – 2A) = cot (A – 12°)
On comparing
90° – 2A = A – 12 °
A + 2A = 90° + 12°
3A = 102°
A = 34°
The value of A is 34°
Question 14:
Solution: sec 4A = cosec (A – 15°)
or cosec (90° – 4A) = cosec (A – 15°)
On comparing
90° – 4A = A – 15°
A + 4A = 90° + 15°
5A = 105°
A = 21°
The value of A is 21°.
Question 15:
Solution:
= 2/3 cosec^2 58°- 2/3 cot 58° tan 32° – 5/3 tan 13° tan 37° tan 45° tan 53° tan 77°
= 2/3 cosec^2 58°- 2/3 cot 58° tan (90-58)° – 5/3 tan 45° (tan 13° tan 77°) (tan 37° tan 53°)
= 2/3 cosec^2 58°- 2/3 cot 58° cot 58° – 5/3 × 1 x (tan 13° tan (90-13)°) × (tan 37° tan (90-37)°)
= 2/3 cosec^2 58°- 2/3 cot^2 58° – 5/3 × (tan 13° cot 13°) (tan 37° cot 37°)
= 2/3 [cosec^2 58°- cot^2 58°] – 5/3
= (2/3) – (5/3)
= -1
= R.H.S.
Complete RS Aggarwal Solutions Class 10
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