RS Agarwal Solution | Class 10th | Chapter- 12 | Trigonometric Ratios of Some Complemantary Angles | Edugrown

Exercise Ex. 7

Question 1

Without using trigonometric tables, evaluate:

Solution 1

(i)

(ii)              

(iii)

(iv)

(v)

(vi)

Question 2

Without using trigonometric tables, prove that:

(i) 

(ii) 

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)Solution 2

(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)               LHS= 

(v)            

(vi)             

(vii)                 

LHS = RHS

(viii) 

(ix)              LHS = (sin65° + cos25°) (sin65° – cos25°)

Question 3(i)

Without using trigonometric tables, prove that:

sin 53^ᵒ cos37^ᵒ + cos53^ᵒ sin37^ᵒ = 1Solution 3(i)

Question 3(ii)

cos 54^ᵒ cos 36^ᵒ – sin 54^ᵒ sin36^ᵒ = 0Solution 3(ii)

Question 3(iii)

sec 70^ᵒ sin 20^ᵒ + cos 20^ᵒ cosec 70^ᵒ = 2Solution 3(iii)

Question 3(iv)

sin 35^ᵒ sin 55^ᵒ – cos 35^ᵒ cos 55^ᵒ = 0Solution 3(iv)

Question 3(v)

(sin 72^ᵒ + cos 18^ᵒ)(sin72^ᵒ – cos18^ᵒ) = 0Solution 3(v)

Question 3(vi)

tan 48^ᵒ tan 23^ᵒ tan 42^ᵒ tan 67^ᵒ = 1Solution 3(vi)

Question 4(i)

Prove that:

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5(i)

Prove that:

sin θ cos (90^ᵒ – θ) + sin (90^ᵒ – θ) cos θ = 1Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

cos 1^ᵒ cos2^ᵒ cos3^ᵒ … cos 180^ᵒ= 0Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 7

Prove that:

(i)

(ii) 

(iii)

(iv)

(v)Solution 7

(i)                  LHS =

= RHS

(ii)                LHS =

= RHS

(iii)                  LHS =

= RHS

(iv)             LHS = cosec(65° + ) – sec(25°-  ) – tan(55° –  ) + cot(35° +  )

= RHS

(v)        LHS =

             = 0 + 1 = 1

             = RHS
Question 8(i)

Express each of the following in terms of T-ratios of angles lying between 0^ᵒ and 45^ᵒ:

sin 67^ᵒ + cos 75^ᵒSolution 8(i)

Question 8(ii)

cot 65^ᵒ + tan 49^ᵒSolution 8(ii)

Question 8(iii)

sec 78^ᵒ + cosec 56^ᵒSolution 8(iii)

Question 8(iv)

cosec 54^ᵒ + sin 72^ᵒSolution 8(iv)

Question 9

If A, B, C are the angles of a triangle ABC, prove that:Solution 9

A + B + C = 180°

So, B + C= 180° – A

Question 10

If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.Solution 10

Question 11

If sec 2A= cosec (A – 42ᵒ), where 2A is an acute angle, find the value of A.Solution 11

Question 12

If sin3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.Solution 12

Question 13

If tan 2A = cot(A – 12^o), where 2A is an acute angle, find the value of A.Solution 13

Question 14

If sec 4A = cosec(A – 15^o), where 4A is an acute angle, find the value of A.Solution 14

Question 15

Prove that:

Solution 15