CBSE - Introduction to Trigonometry

In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get

AC² = AB² + BC² = (24)² + 7² = (576+49) cm² = 625 cm²
⇒ AC = 25

(i) sin A = BC/AC = 7/25   
cos A = AB/AC = 24/25

(ii) sin C = AB/AC = 24/25
     cos C = BC/AC = 7/25

By Applying Pythagoras theorem in ΔPQR , we get
PR² = PQ² + QR² = (13)² = (12)² + QR² = 169 = 144  + QR²
⇒  QR² = 25 ⇒  QR = 5 cm

Now,
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
A/q
tan P – cot R = 5/12 - 5/12 = 0

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sin A = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.
 

By Pythagoras theorem we get,
AC² = AB² + BC² 
(4k)² = AB² + (3k)²
16k² - 9k² = AB²
AB² = 7k²
AB = √7 k

cos A = AB/AC = √7 k/4k = √7/4

tan A = BC/AB = 3k/√7 k = 3/√7

Let ΔABC be a right-angled triangle, right-angled at B.

We know that cot A = AB/BC = 8/15   (Given)
Let AB be 8k and BC will be 15k where k is a positive real number. 

By Pythagoras theorem we get,
AC² = AB² + BC² 
AC² = (8k)² + (15k)²
AC² = 64k² + 225k²
AC² = 289k²
AC = 17 k

sin A = BC/AC = 15k/17k = 15/17

sec A = AC/AB = 17k/8 k = 17/8

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sec θ = OP/OM = 13/12   (Given)
Let OP be 13k and OM will be 12k where k is a positive real number. 
By Pythagoras theorem we get,
OP² = OM² + MP² 
(13k)² = (12k)² + MP² 
169k² - 144k² = MP²
MP² = 25k²

MP = 5

Now,

sin θ = MP/OP = 5k/13k = 5/13

cos θ = OM/OP = 12k/13k = 12/13

tan θ = MP/OM = 5k/12k = 5/12

cot θ = OM/MP = 12k/5k = 12/5

cosec θ = OP/MP = 13k/5k = 13/5

Let ΔABC in which CD ⊥ AB.

A/q,

cos A = cos B

⇒ AD/AC = BD/BC

⇒ AD/BD = AC/BC

Let AD/BD = AC/BC = k

⇒ AD = kBD  .... (i)

⇒ AC = kBC  .... (ii)

By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD² = AC² - AD² …. (iii)
and also CD² = BC² - BD² …. (iv)
From equations (iii) and (iv) we get,
AC² - AD² = BC² - BD²
⇒ (kBC)² - (k BD)² = BC² - BD²
⇒ k² (BC² - BD²) = BC² - BD²
⇒ k² = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B  (Angles opposite to equal sides of a triangle are equal-isosceles triangle)

Let ΔABC in which ∠B = 90º and ∠C = θ

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC² 
AC² = (8k)² + (7k)²
AC² = 64k² + 49k²
AC² = 113k²
AC = √113 k

Let ΔABC in which ∠B = 90º,

cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC² 
AC² = (4k)² + (3k)²
AC² = 16k² + 9k²
AC² = 25k²
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5

(i)

If BC is k, then AB will be√3, where k is a positive integer.

In ΔABC,

AC² = AB² + BC²

⇒ tan (A + B) = tan 60°

⇒ tan (A - B) = tan 30°
⇒ (A - B) = 30° ... (ii)

(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

(ii) True.

sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin  90° = 1

Thus the value of sin θ increases as θ increases.

(iii) False.

cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Thus the value of cos θ decreases as θ increases.

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45° 

It is not true for all other values of θ.

Hence, the given statement is false.

(v) True.

cot A = cos A/sin A

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

(i) sin 18°/cos 72°

    = sin (90° - 18°) /cos 72° 

    = cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

    = tan (90° - 36°)/cot 64°

    = cot 64°/cot 64° = 1

(iii) cos 48° - sin 42°

      = cos (90° - 42°) - sin 42°

      = sin 42° - sin 42° = 0

(iv) cosec 31° - sec 59°

     = cosec (90° - 59°) - sec 59°
     = sec 59° - sec 59° = 0

(i) tan 48° tan 23° tan 42° tan 67°
    = tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°
    = cot 42° cot 67° tan 42° tan 67°
    = (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° - sin 38° sin 52°
    = cos (90° - 52°) cos (90°-38°) - sin 38° sin 52°
    = sin 52° sin 38° - sin 38° sin 52° = 0

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

In a triangle, sum of all the interior angles

A + B + C = 180°

⇒ B + C = 180° - A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

⇒ sin (B+C)/2 = sin (90°-A/2)

⇒ sin (B+C)/2 = cos A/2

sin 67° + cos 75°
= sin (90° - 23°) + cos (90° - 15°)
= cos 23° + sin 15°

cosec²A - cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ 1/sin²A = 1 + cot²A
⇒sin²A = 1/(1+cot²A)

Now,
sin²A = 1/(1+cot²A)
⇒ 1 - cos²A = 1/(1+cot²A)
⇒cos²A = 1 - 1/(1+cot²A)
⇒cos²A = (1-1+cot²A)/(1+cot²A)
⇒ 1/sec²A = cot²A/(1+cot²A)
⇒ secA = (1+cot²A)/cot²A

also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A

We know that sec²A−tan²A=1

⇒tan²A=sec²A−1

⇒tan²A=sec²A−1

⇒tanA = √ sec²A−1−−−−−−−−    (1)

We know that cos A = 1 / secA      (2)

We know that sin²A+cos²A=1

⇒sin2A=1−cos2A

Putting (2) in the above equation we get

We know that  cosecA=1/ sinA

Putting (3) in the above equation, we get

cosecA=secA / √ sec2A−1        (4) 

We know that  cotA=1 /tanA

Putting (1) in the above equation, we get

cotA=1 / √sec²A−1                    (5)

 (1), (2), (3), (4) and (5) shows all the other trigonometric ratios of ∠A in terms of sec A.

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
   = [sin²(90°-27°) + sin²27°]/[cos²(90°-73°) + cos²73°)]
   = (cos²27° + sin²27°)/(sin²27° + cos²73°)
   = 1/1 =1                       (��� sin²A + cos²A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
   = sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
   = cos 65° cos 65° + sin 65° sin 65°
   = cos²65° + sin²65° = 1

(i) (B) is correct.

9 sec²A - 9 tan²A

= 9 (sec²A - tan²A)
= 9×1 = 9             (��� sec2 A - tan2 A = 1)

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ - cosec θ)   

= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ - 1/sin θ)

= (cos θ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ)²-1²/(cos θ sin θ)

= (cos²θ + sin²θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ)

= (2cos θ sin θ)/(cos θ sin θ) = 2

(iii) (D) is correct.

(secA + tanA) (1 - sinA)

= (1/cos A + sin A/cos A) (1 - sinA)

= (1+sin A/cos A) (1 - sinA)

= (1 - sin²A)/cos A

= cos²A/cos A = cos A

(iv) (D) is correct.

1+tan²A/1+cot²A 

= (1+1/cot²A)/1+cot²A

= (cot²A+1/cot²A)×(1/1+cot²A)

= 1/cot²A = tan²A

(i) (cosec θ - cot θ)² = (1-cos θ)/(1+cos θ)
L.H.S. =  (cosec θ - cot θ)²
           = (cosec²θ + cot²θ - 2cosec θ cot θ)
           = (1/sin²θ + cos²θ/sin²θ - 2cos θ/sin²θ)
           = (1 + cos²θ - 2cos θ)/(1 - cos²θ)
           = (1-cos θ)²/(1 - cosθ)(1+cos θ)
           = (1-cos θ)/(1+cos θ) = R.H.S.

(ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
 L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
            = [cos²A + (1+sin A)²]/(1+sin A)cos A
            = (cos²A + sin²A + 1 + 2sin A)/(1+sin A)cos A
            = (1 + 1 + 2sin A)/(1+sin A)cos A
            = (2+ 2sin A)/(1+sin A)cos A
            = 2(1+sin A)/(1+sin A)cos A
            = 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
           = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
           = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
           = sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]
           = sin²θ/[cos θ(sin θ-cos θ)] - cos²θ/[sin θ(sin θ-cos θ)]
           = 1/(sin θ-cos θ) [(sin²θ/cos θ) - (cos²θ/sin θ)]
           = 1/(sin θ-cos θ) × [(sin³θ - cos³θ)/sin θ cos θ]
           = [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
           = (1 + sin θ cos θ)/sin θ cos θ
           = 1/sin θ cos θ + 1
           = 1 + sec θ cosec θ = R.H.S.

(iv)  (1 + sec A)/sec A = sin²A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
           = (1 + 1/cos A)/1/cos A
           = (cos A + 1)/cos A/1/cos A
           = cos A + 1
R.H.S. = sin²A/(1-cos A)
            = (1 - cos²A)/(1-cos A)
            = (1-cos A)(1+cos A)/(1-cos A)
            = cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
           Dividing Numerator and Denominator by sin A,
           = (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
           = (cot A - 1 + cosec A)/(cot A+ 1 – cosec A)
           = (cot A - cosec²A + cot²A + cosec A)/(cot A+ 1 – cosec A) (using cosec²A - cot²A = 1)
           = [(cot A + cosec A) - (cosec²A - cot²A)]/(cot A+ 1 – cosec A)
           = [(cot A + cosec A) - (cosec A + cot A)(cosec A - cot A)]/(1 – cosec A + cot A)
           =  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
           =  cot A + cosec A = R.H.S.

Dividing Numerator and Denominator of L.H.S. by cos A,

= sec A + tan A = R.H.S.

(vii) (sin θ - 2sin³θ)/(2cos³θ-cos θ) = tan θ
L.H.S. = (sin θ - 2sin³θ)/(2cos³θ - cos θ)
           = [sin θ(1 - 2sin²θ)]/[cos θ(2cos²θ- 1)]
           = sin θ[1 - 2(1-cos²θ)]/[cos θ(2cos²θ -1)]
          = [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]
          = tan θ = R.H.S.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
               = (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
           = (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A
           = 1 + 2 + 2 + 2 + tan²A + cot²A
           = 7+tan²A+cot²A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
           = (1/sin A - sin A)(1/cos A - cos A)
           = [(1-sin²A)/sin A][(1-cos²A)/cos A]
           = (cos²A/sin A)×(sin²A/cos A)
           = cos A sin A
R.H.S. = 1/(tan A+cotA)
            = 1/(sin A/cos A +cos A/sin A)
            = 1/[(sin²A+cos²A)/sin A cos A]
            = cos A sin A
L.H.S. = R.H.S.

(x)  (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A
L.H.S. = (1+tan²A/1+cot²A)
           = (1+tan²A/1+1/tan²A)
           = 1+tan²A/[(1+tan²A)/tan²A]
           = tan²A