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 cos37ᵒ + cos53 sin37ᵒ = 1Solution 3(i)

Question 3(ii)

cos 54ᵒ cos 36ᵒ – sin 54 sin36ᵒ = 0Solution 3(ii)

Question 3(iii)

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

Question 3(iv)

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

Question 3(v)

(sin 72ᵒ + cos 18)(sin72ᵒ – cos18) = 0Solution 3(v)

Question 3(vi)

tan 48ᵒ tan 23 tan 42 tan 67ᵒ = 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 – θ) + sin (90ᵒ – 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 cos2 cos3 … cos 180= 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 and 45:

sin 67 + cos 75Solution 8(i)

Question 8(ii)

cot 65ᵒ + tan 49Solution 8(ii)

Question 8(iii)

sec 78 + cosec 56Solution 8(iii)

Question 8(iv)

cosec 54 + sin 72Solution 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 – 26o), where 3A is an acute angle, find the value of A.Solution 12

Question 13

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

Question 14

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

Question 15

Prove that:

Solution 15


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