Introduction to Trigonometry

Introduction to Trigonometry

1. In D ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) SinA , CosA
(ii) SinC , CosC.

Solution:

2. In the figure, find tan P – cot R.

Solution:In a right angled D PQR
PR² = PQ² + QR²13² = 12² + QR²QR² = 169 – 144 = 25
QR = 5
\tan P - cot R = = 0

3. sin A =,calculate cos A and tan A.

Solution:Let the right angled D ABC be right angled at B.

In a right angled D ABC

AC² = AB² + BC²16 = AB² + BC
AB² = 16 – 9 = 7
AB =
sin A =
cos A = =

tan A = .

4. Given 15 cot A = 8, find sin A and sec A.

Solution:15 cot A = 8
Þ cot A =

Let ABC be the D right angled at B
Cot A = 8/15
Þ AB = 8, BC = 15
AC² = AB² + BC²
      = 8² + 15²
      = 64 + 225
      = 289
AC = = 17
sin A = =
sec A =  =

5. Given secq = calculate all other trigonometric Ratios.

Solution:

In a right angled D ABC,
AC² = AB² + BC²13² - 12² = AB²AB² = 169 - 144
AB = 5

6. If Ð A and Ð B are acute angles such that cos A = cos B,then show that Ð A = Ð B.

Solution:If cosA = cosB
Þcos A – cos B = 0
only when = 45⁰.

7. If cot q = ,evaluate : (i) (ii) cot²q.

Solution:If cot q =
(i) = =

(ii) cot²q = .

8. If 3 cotA = 4, Check whether =cos²A – sin²A or not.

Solution:If 3 cot A = 4
=  =  =  ………………… (1)
cos²A – sin²A = sin²A(cot²A - 1) Dividing and multiplying by sin²A
                   = sin²A
                   = sin²A= sin²A
                   =  =  …………………………. (2)
Since (1) = (2)
LHS = RHS.

9. In triangle ABC, right-angled at B, if tan A =find the value of (i) sin A cos C +cos A sin C (ii)cos A cos C – sin A sin C.

Solution:If tan A =
In a right angled D ABC, right angled at B
AC² = AB² + BC²
\AC = = 2
sin A = , cos A = sin C = cos C = 1/2

(i) sin A cos C + cos A sin C
= = = 1

(ii) cos A cos C – sin A sin C
= 0.

10. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution:
PR + QR = 25
PQ = 5 cm
In a right angled D PQR, PQ² + QR² = PR² …………………… (1)
Let PR = x
QR = (25 – x)
From 1 x² = 5² + (25 – x)²
x² = 25 + (25)² – 2 ´ 25x + x²
    = 625 + 25 – 50x
50x = 650
x =  = 13
PR = 13 cm
QR = 12 cm
PQ = 5 cm
sin P =  =
cos P =  =
- tan P = = .
- 11. State whether the following are true or false. Justify your answer.
  (i) The value of tan A is always less than 1.
  (ii) sec A = for some value of angle A.
  (iii) cos A is the abbreviation used for the cosecant of angle A.
  (iv) cot A is the product of cot and A.
  (v) sin q =.
  Solution:(i) False. The value of tan 90^◦is grater than 1
  
  (ii) True. sec A = Þ cos A = as 12 the hypotenuse is the largest side of triangle.
  
  (iii) False. cos A is the abbreviation for cosine of angle A
  
  (iv) No,False. cos A is not the product of cot and A
  
  (v) False, Because the hypotenuse is the longest side whereas in sin q = , 3 which is the denominator is cannot be the hypothenuse.
  12. Evaluate the following:
  (i) sin 60° cos 30° + sin 30° cos 60°
  
  (ii) 2 tan2 45° + cos2 30° – sin2 60°
  (iii)
  (iv)
  (v) .
  Solution:
  
  =
  13. Choose the correct option and justify your choice:(i) =(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
  
  (ii) =(A) tan 90° (B) 1 (C) sin 45° (D) 0
  
  (iii) sin 2A = 2 sin A is true when A =
  (A) 0° (B) 30° (C) 45° (D) 60°
  
  (iv)=(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°.
  Solution:(i) =  =
                        =
                        =  =
                        = sin 60⁰
  Answer: A
  
  (ii)  =  = 0
  Answer: D
  (iii) sin 2A = 2 sin A is true when A = 45⁰
  Answer: A
  
  (iv)  =  =
                       =
                       =
                       =
                       =´
                       =  =  = tan 60⁰.
  
  Answer: C
  14. tan (A + B) = and tan (A – B) =; 0° < A + B 90°; A > B, find A and B.
  Solution:If tan(A + B) =
  tan(A - B) = 1/
  If tan A+B =
  Þ A + B = 60⁰ ………………………(1)
  tan (A – B) =
  A – B = 30⁰ ……………………..(2)
  Adding (1) ´ (2) 2A = 90⁰A = 45⁰
  \B = 15^0.
  15.  State whether the following are true or false. Justify your answer.
  (i) sin (A + B) = sin A + sin B.
  
  (ii) The value of sin  increases as  increases.
  
  (iii) The value of cos increases as  increases.
  
  (iv) sin  = cos  for all values of .
  
  (v) cot A is not defined for A = 0°.
  Solution:(i) sin(A+B) = sin A + sin B
  False
  
  (ii) Value of sin q as q increase
  True
  
  (iii) Value of cos q as q ncreases
  False
  
  (iv) sinq = cosq for all values of q
  False
  
  (v) cot A is not defined for A = 0⁰
  True.
  16. Evaluate:
  (i)
  (ii)
  
  (iii) cos48⁰ – sin 42⁰(iv)cosec 31⁰ – sec 59⁰.
  Solution:(i)
      =
      =  = 1
  
  (ii)  =  =  = 1
  
  (iii) cos48⁰ – sin 42⁰cos48⁰ = cos(90 - 42⁰)
            = sin 42⁰
  \cos 48⁰ – sin 42⁰ = sin 42⁰ – sin 42⁰                            = 0
  
  (iv) cosec 31⁰ – sec 59⁰
  cosec 31⁰ = cosec(90⁰ - 59⁰)
                = sec 59⁰
  LHS = sec 59⁰ – sec 59⁰ = 0.
  
  21.  A, B and C are interior angles of a triangle ABC, then show that sin = cos .
  
  Solution:
  
  22.  Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
  
  Solution:
  
  23.  Express the trigonometric ratios sin A, cos A and tan A in terms of sec A.
  
  Solution:
  
  24.  Write all the other trigonometric ratios of Ð A in terms of sec A.
  
  Solution:sin A =
             =
  cos A =
  tan A =  =  ¸
          =  ´ sec A.
              =
  
  25.  Evaluate:(i) (ii) sin 25⁰. cos 65⁰ + cos 25⁰ sin 65^0.
  
  Solution:
  
  26.  Choose the correct option. Justify your choice.
  (i) 9 sec² A – 9 tan² A =
  (A) 1 (B) 9 (C) 8 (D) 0
  
  (ii) (1 + tan q + secq ) (1 + cot q – cosec q ) =
  (A) 0 (B) 1 (C) 2 (D) –1
  
  (iii) (sec A + tan A) (1 – sin A) =
  (A)sec A (B) sin A (C) cosec A (D) cos A
  
  (iv)
  (A) sec² A (B) –1 (C) cot² A (D) tan² A.
  
  Solution:(i) 9 sec²A - 9 tan²A
     = 9 (1 + tan²A) – 9 tan²A
     = 9 + 9 tan²A - 9 tan²A
     = 9
  Answer (B)
  
  (ii)
  
  (iii) (sec A + tan A) (1 – sin A)
  sec A – sin A sec A + tan A – tan A sin A
  +  - .sinA
  
  (iv)  =  =  = tan²A.
  
  27.  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
  (i) (cosecq - cot q )² =
  
  (ii)= 2 secA
  
  (iii)= 1 +secq cosecq
  
  (iv) =
  (v)= Cosec A +cot A
  
  Solution:(i) (cosecq - cot q )² =
  RHS  =
  
                    =
                    = cosec^2q- 2cosecq . cotq + cot^2q                  = (cosec q - cotq )²
  Hence Proved.
  (ii)
                    =  =
                    =  =
                    =  = 2 sec A
  
  (iii)
  
  (iv)  =
  LHS
  RHS  =
              = LHS
  
  (v)
  Using the identity cosec²A = 1 + cot²A.
  Taking the LHS of the given equation
  
  Dividing the numerator and denominator by sin A
  
  = (Changing = cot A ´
  =
  
  (Replacing cosec²A - cot²A = (cosec A + cot A) (cosec A - cot A) using the identity cosec²A - cot²A = 1
  
  =
  =
  = cosec A + cotA.
  
  28.  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
  (i)  = secA + tanA
  (ii) (sin A + cosec A)² + (cos A + sec A)²= 7 +tan²A + Cot²A
  (iii) (cosecA –sinA)(secA- cosA) =
  
  (iv)= .
  
  Solution:(i)
        =  Taking the conjugate of (1-sinA) + multiplying the numerator and denominator by it.
        =
        =  (Since 1 - sin²A = cos²A)
        =  =
        =  = sec A + tan A (As  = sec A,  = tan A)
  
  (ii) (sin A + cosec A)² + (cos A + sec A)²= sin²A + cosec²A + 2sinA.cosec A + cos²A + sec²A + 2cos A. sec A
        = sin²A + cos²A + cosec²A + sec²A + 2sinA .  + 2cos A.
        = 1 + 2 + 2 + cosec²A + sec²A (using cosec A =  secA = )
        = 5 + 1 + tan²A + 1 + cot²A
        = 7 + tan²A + cot²A (1 + tan²A = sec²A) 1 + cot²A = cosec²A
  
  (iii) (cosec A - sin A) (secA - cosA)
  LHS
  (cosec A - sin A) (sec A - cos A)
  
  (since 1 - sin²A = cos²A and 1 - cos²A = sin²A)
  =  = sin A cosA
  RHS  =
                         =  (Taking LCM as cosA. sinA)
                         =  (since sin²A + cos²A = 1)
  LHS = RHS
  Hence the result.
  
  (iv)
  LHS
  
  RHS
  
  Thus, LHS = RHS.
  17. Show that tan 48⁰ tan 23⁰ tan 42⁰ tan 67⁰ = 1.
  Solution:(i) tan 48⁰ tan23⁰ tan 42⁰ tan67⁰ = 1
  LHS = tan 48⁰. tan 42⁰ tan 23⁰ tan67⁰= tan 48⁰. tan(90⁰ - 48⁰). tan(90⁰ - 67⁰) tan 67⁰
  = tan 48⁰. cot 48⁰. cot67⁰tan 67⁰ .= 1 ´ 1 = 1.
  18. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
  Solution:tan 2A = cot (A – 18) ………………. (1)
  where 2A is an acute angle
  tan 2A = cot(90 – 2A) = cot(A – 18)
  cot 90 – 2A and A - 18 are acute angles.
  90 – 2A = A – 18
  3A = 108
  A = 36⁰.
  19. If tan A = cot B, prove that A + B = 90°.
  Solution:If tan A = cot B …………… (1)
  tan A = cot (90 – A)
  From (1)
  cot (90 - A) = cot B
  Þ 90 – A = B
  or A + B = 90⁰.
  20. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
  Solution:If sec 4A = cosec (A – 20⁰)
  sec 4A = cosec (90 – 4A)
  \cosec(90 – 4A) = cosec (A – 20⁰)
  Since (90 – 4A) and A – 20⁰ are acute angles
  90 – 4A = A – 20⁰
  90 + 20 = 5A
  A =
  = 22⁰.