RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles

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RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles

Exercise 8

Question 1.

rs aggarwal class 10 chapter 12 img 1
rs aggarwal class 10 chapter 12 img 2

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:

rs aggarwal class 10 chapter 12 img 4
rs aggarwal class 10 chapter 12 img 5

(v)

rs aggarwal class 10 chapter 12 img 6

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.

rs aggarwal class 10 chapter 12 img 8
rs aggarwal class 10 chapter 12 img 9

(vi)

rs aggarwal class 10 chapter 12 img 10

=(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:

rs aggarwal class 10 chapter 12 img 11
rs aggarwal class 10 chapter 12 img 12

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:

rs aggarwal class 10 chapter 12 img 13

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°)

rs aggarwal class 10 chapter 12 img 14

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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