KnowledgeBoat Logo
|
OPEN IN APP

Chapter 21

Trigonometrical Ratios — Exercise 21(F)

Class - 9 Concise Mathematics Selina

Exercise 21(F)

Question 1(a)

If sin A = cos A, the measurement of angle A is :

  1. 0°

  2. 30°

  3. 45°

  4. 60°

Answer

Given:

sin A = cos A

⇒ sin A = sin (90° - A)

So, A = 90° - A

⇒ A + A = 90°

⇒ 2A = 90°

⇒ A = 90°2\dfrac{90°}{2}

⇒ A = 45°

Hence, option 3 is the correct option.

Question 1(b)

If sin A = cos B and A ≠ B then the relation between angles A and B is :

  1. A + B = 180°

  2. A - B = 90°

  3. B - A = 90°

  4. A + B = 90°

Answer

Given:

sin A = cos B

⇒ sin A = sin (90° - B)

So, A = 90° - B

⇒ A + B = 90°

Hence, option 4 is the correct option.

Question 1(c)

If A + B = 90°, the value of

cos Asin B×tan Bcot A\dfrac{\text{cos A}}{\text{sin B}}\times \dfrac{\text{tan B}}{\text{cot A}} is :

  1. 1

  2. 2

  3. sin A

  4. cos B

Answer

Given:

cos Asin B×tan Bcot A⇒cos Asin (90° - A)×tan (90° - A)cot A⇒cos Acos A×cot Acot A⇒cosAcosA×cotAcotA⇒1\dfrac{\text{cos A}}{\text{sin B}}\times \dfrac{\text{tan B}}{\text{cot A}}\\[1em] ⇒ \dfrac{\text{cos A}}{\text{sin (90° - A)}}\times \dfrac{\text{tan (90° - A)}}{\text{cot A}}\\[1em] ⇒ \dfrac{\text{cos A}}{\text{cos A}}\times \dfrac{\text{cot A}}{\text{cot A}}\\[1em] ⇒ \dfrac{\cancel{cos A}}{\cancel{cos A}}\times \dfrac{\cancel{cot A}}{\cancel{cot A}}\\[1em] ⇒ 1

Hence, option 1 is the correct option.

Question 1(d)

The value of :

cosec 40° cos 50° + sin 50° sec 40° is:

  1. 1

  2. 2

  3. 3

  4. 0

Answer

Given:

cosec 40° cos 50° + sin 50° sec 40°

= cosec 40° cos (90° - 40°) + sin (90° - 40°) sec 40°

= cosec 40° sin 40° + cos 40° sec 40°

= 1sin 40°×sin 40°+cos 40°×1cos 40°\dfrac{1}{\text{sin 40°}} \times \text{sin 40°} + \text{cos 40°} \times \dfrac{1}{\text{cos 40°}}

= 1 + 1

= 2

Hence, option 2 is the correct option.

Question 1(e)

In a triangle ABC, secA+C2\text{sec}\dfrac{A + C}{2} is equal to:

  1. 0

  2. secB2\text{sec}\dfrac{B}{2}

  3. cosec B

  4. cosecB2\text{cosec}\dfrac{B}{2}

Answer

Given:

In Δ ABC,

⇒∠A+∠B+∠C=180°⇒∠A+∠C=180°−∠B⇒A+C2=180°−B2⇒A+C2=90°−B2⇒secA+C2=sec(90°−B2)⇒secA+C2=cosecB2⇒ ∠ A + ∠ B + ∠ C = 180°\\[1em] ⇒ ∠ A + ∠ C = 180° - ∠ B\\[1em] ⇒ \dfrac{A + C}{2} = \dfrac{180° - B}{2}\\[1em] ⇒ \dfrac{A + C}{2} = 90° - \dfrac{B}{2}\\[1em] ⇒ \text{sec}\dfrac{A + C}{2} = \text{sec}{\Big(90° - \dfrac{B}{2}\Big)}\\[1em] ⇒ \text{sec}\dfrac{A + C}{2} = \text{cosec}\dfrac{B}{2}\\[1em]

Hence, option 4 is the correct option.

Question 2(i)

Evaluate:

cos 22°sin 68°\dfrac{\text{cos 22°}}{\text{sin 68°}}

Answer

cos 22°sin 68°=cos (90° - 68°)sin 68°=sin 68°sin 68°=sin68°sin68°=1\dfrac{\text{cos 22°}}{\text{sin 68°}} = \dfrac{\text{cos (90° - 68°)}}{\text{sin 68°}}\\[1em] = \dfrac{\text{sin 68°}}{\text{sin 68°}}\\[1em] = \dfrac{\cancel{sin 68°}}{\cancel{sin 68°}}\\[1em] = 1

Hence, cos 22°sin 68°\dfrac{\text{cos 22°}}{\text{sin 68°}} = 1.

Question 2(ii)

Evaluate:

tan 47°cot 43°\dfrac{\text{tan 47°}}{\text{cot 43°}}

Answer

tan 47°cot 43°=tan (90° - 43°)cot 43°=cot 43°cot 43°=cot43°cot43°=1\dfrac{\text{tan 47°}}{\text{cot 43°}} = \dfrac{\text{tan (90° - 43°)}}{\text{cot 43°}}\\[1em] = \dfrac{\text{cot 43°}}{\text{cot 43°}}\\[1em] = \dfrac{\cancel{cot 43°}}{\cancel{cot 43°}}\\[1em] = 1

Hence, tan 47°cot 43°\dfrac{\text{tan 47°}}{\text{cot 43°}} = 1.

Question 2(iii)

Evaluate:

sec 75°cosec 15°\dfrac{\text{sec 75°}}{\text{cosec 15°}}

Answer

sec 75°cosec 15°=sec (90° - 15°)cosec 15°=cosec 15°cosec 15°=cosec15°cosec15°=1\dfrac{\text{sec 75°}}{\text{cosec 15°}} = \dfrac{\text{sec (90° - 15°)}}{\text{cosec 15°}}\\[1em] = \dfrac{\text{cosec 15°}}{\text{cosec 15°}}\\[1em] = \dfrac{\cancel{cosec 15°}}{\cancel{cosec 15°}}\\[1em] = 1

Hence, sec 75°cosec 15°\dfrac{\text{sec 75°}}{\text{cosec 15°}} = 1.

Question 2(iv)

Evaluate:

cos 55°sin 35°\dfrac{\text{cos 55°}}{\text{sin 35°}} + cot 35°tan 55°\dfrac{\text{cot 35°}}{\text{tan 55°}}

Answer

cos 55°sin 35°+cot 35°tan 55°=cos (90° - 35°)sin 35°+cot (90° - 55°)tan 55°=sin 35°sin 35°+tan 55°tan 55°=sin⁡35°sin⁡35°+tan⁡55°tan⁡55°=1+1=2\dfrac{\text{cos 55°}}{\text{sin 35°}} + \dfrac{\text{cot 35°}}{\text{tan 55°}}\\[1em] = \dfrac{\text{cos (90° - 35°)}}{\text{sin 35°}} + \dfrac{\text{cot (90° - 55°)}}{\text{tan 55°}}\\[1em] = \dfrac{\text{sin 35°}}{\text{sin 35°}} + \dfrac{\text{tan 55°}}{\text{tan 55°}}\\[1em] = \dfrac{\cancel{\sin 35°}}{\cancel{\sin 35°}} + \dfrac{\cancel{\tan 55°}}{\cancel{\tan 55°}}\\[1em] = 1 + 1\\[1em] = 2

Hence, cos 55°sin 35°+cot 35°tan 55°\dfrac{\text{cos 55°}}{\text{sin 35°}} + \dfrac{\text{cot 35°}}{\text{tan 55°}} = 2.

Question 2(v)

Evaluate:

sin2 40° - cos2 50°

Answer

sin2 40° - cos2 50°

= sin2 (90° - 50°) - cos2 50°

= cos2 50° - cos2 50°

= 0

Hence, sin2 40° - cos2 50° = 0.

Question 2(vi)

Evaluate:

sec2 18° - cosec2 72°

Answer

sec2 18° - cosec2 72°

= sec2 (90° - 72°) - cosec2 72°

= cosec2 72° - cosec2 72°

= 0

Hence, sec2 18° - cosec2 72° = 0.

Question 2(vii)

Evaluate:

sin 15° cos 15° - cos 75° sin 75°

Answer

sin 15° cos 15° - cos 75° sin 75°

= sin (90° - 75°) cos (90° - 75°) - cos (90° - 15°) sin 75°

= cos 75° sin 75° - cos 75° sin 75°

= 0

Hence, sin 15° cos 15° - cos 75° sin 75° = 0.

Question 2(viii)

Evaluate:

sin 42° sin 48° - cos 42° cos 48°

Answer

sin 42° sin 48° - cos 42° cos 48°

= sin (90° - 48°) sin 48° - cos (90° - 48°) cos 48°

= cos 48° sin 48° - sin 48° cos 48°

= 0

Hence, sin 42° sin 48° - cos 42° cos 48° = 0.

Question 3(i)

Evaluate :

sin (90° - A) sin A - cos (90° - A) cos A

Answer

sin (90° - A) sin A - cos (90° - A) cos A

= cos A sin A - sin A cos A

= 0

Hence, sin (90° - A) sin A - cos (90° - A) cos A = 0.

Question 3(ii)

Evaluate :

sin2 35° - cos2 55°

Answer

sin2 35° - cos2 55°

= sin2 (90° - 55°) - cos2 55°

= cos2 55° - cos2 55°

= 0

Hence, sin2 35° - cos2 55° = 0.

Question 3(iii)

Evaluate :

cot 54°tan 36°\dfrac{\text{cot 54°}}{\text{tan 36°}} + tan 20°cot 70°\dfrac{\text{tan 20°}}{\text{cot 70°}} - 2

Answer

cot 54°tan 36°+tan 20°cot 70°−2=cot (90° - 36°)tan 36°+tan (90° - 70°)cot 70°−2=tan 36°tan 36°+cot 70°cot 70°−2=tan36°tan36°+cot70°cot70°−2=1+1−2=2−2=0\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\text{cot (90° - 36°)}}{\text{tan 36°}} + \dfrac{\text{tan (90° - 70°)}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\text{tan 36°}}{\text{tan 36°}} + \dfrac{\text{cot 70°}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\cancel{tan 36°}}{\cancel{tan 36°}} + \dfrac{\cancel{cot 70°}}{\cancel{cot 70°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 2 - 2\\[1em] = 0

Hence, cot 54°tan 36°+tan 20°cot 70°−2=0\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0.

Question 3(iv)

Evaluate :

cos2 25° - sin2 65° - tan2 45°

Answer

cos2 25° - sin2 65° - tan2 45°

= cos2 (90° - 65°) - sin2 65° - tan2 45°

= sin2 65° - sin2 65° - tan2 45°

= - tan2 45°

= - 1

Hence, cos2 25° - sin2 65° - tan2 45° = -1.

Question 3(v)

Evaluate :

(sin 77°cos 13°)2\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + (cos 77°sin 13°)2\Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - 2 cos2 45°

Answer

(sin 77°cos 13°)2+(cos 77°sin 13°)2−2 cos245°=(sin (90° - 13°)cos 13°)2+(cos (90° - 13°)sin 13°)2−2 cos245°=(cos 13°cos 13°)2+(sin 13°sin 13°)2−2 cos245°=(cos13°cos13°)2+(sin13°sin13°)2−2 cos245°=12+12−2×(12)2=1+1−2×(12)=2−1=1\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{sin (90° - 13°)}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos (90° - 13°)}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{cos 13°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{sin 13°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\cancel{cos 13°}}{\cancel{cos 13°}}\Big)^2 + \Big(\dfrac{\cancel{sin 13°}}{\cancel{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = 1^2 + 1^2 - 2 \times \Big(\dfrac{1}{\sqrt2}\Big)^2\\[1em] = 1 + 1 - 2 \times \Big(\dfrac{1}{2}\Big)\\[1em] = 2 - 1\\[1em] = 1

Hence, (sin 77°cos 13°)2+(cos 77°sin 13°)−2 cos245°=1\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big) - \text{2 cos}^2 45° = 1.

Question 4(i)

Show that :

tan 10° tan 15° tan 75° tan 80° = 1

Answer

tan 10° tan 15° tan 75° tan 80° = 1

L.H.S. = tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= 1tan 80°×1tan 75°×tan 75°×tan 80°\dfrac{1}{\text{tan 80°}} \times \dfrac{1}{\text{tan 75°}} \times \text{tan 75°} \times \text{tan 80°}

= 1tan80°×1tan75°×tan75°×tan80°\dfrac{1}{\cancel{tan 80°}} \times \dfrac{1}{\cancel{tan 75°}} \times \cancel{tan 75°} \times \cancel{tan 80°}

= 1

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, tan 10° tan 15° tan 75° tan 80° = 1.

Question 4(ii)

Show that :

sin 42° sec 48° + cos 42° cosec 48° = 2

Answer

sin 42° sec 48° + cos 42° cosec 48° = 2

L.H.S. = sin 42° sec 48° + cos 42° cosec 48°

= sin (90° - 48°) sec 48° + cos (90° - 48°) cosec 48°

= cos 48° sec 48° + sin 48° cosec 48°

= cos 48°1cos 48°+sin 48°1sin 48°\text{cos 48°} \dfrac{1}{\text{cos 48°}} + \text{sin 48°} \dfrac{1}{\text{sin 48°}}

= cos48°1cos48°+sin48°1sin48°\cancel{cos 48°} \dfrac{1}{\cancel{cos 48°}} + \cancel{sin 48°} \dfrac{1}{\cancel{sin 48°}}

= 1 + 1

= 2

R.H.S. = 2

∴ L.H.S. = R.H.S.

Hence, sin 42° sec 48° + cos 42° cosec 48° = 2.

Question 5(i)

Express the following in terms of angles between 0° and 45° :

sin 59° + tan 63°

Answer

sin 59° + tan 63°

= sin (90° - 31°) + tan (90° - 27°)

= cos 31° + cot 27°

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

Question 5(ii)

Express the following in terms of angles between 0° and 45° :

cosec 68° + cot 72°

Answer

cosec 68°+ cot 72°

= cosec (90° - 22°) + cot (90° - 18°)

= sec 22° + tan 18°

Hence, cosec 68°+ cot 72° = sec 22° + tan 18°.

Question 5(iii)

Express the following in terms of angles between 0° and 45° :

cos 74° + sec 67°

Answer

cos 74° + sec 67°

= cos (90° - 16°) + sec (90° - 23°)

= sin 16° + cosec 23°

Hence, cos 74° + sec 67° = sin 16° + cosec 23°.

Question 6

For triangle ABC, show that :

(i) sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}

(ii) tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}

Answer

(i) sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}

According to angle sum property,

∠A+∠B+∠C=180°∠A+∠B=180°−∠C⇒A+B2=180°−C2∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ A + ∠ B = 180° - ∠ C\\[1em] ⇒ \dfrac{A + B}{2} = \dfrac{180° - C}{2}\\[1em]

L.H.S.=sinA+B2=sin180°−C2=sin(90°−C2)=cosC2\text{L.H.S.} = \text{sin}\dfrac{A+B}{2}\\[1em] = \text{sin}\dfrac{180° - C}{2}\\[1em] = \text{sin}\Big(90° - \dfrac{C}{2}\Big)\\[1em] = \text{cos}\dfrac{C}{2}

R.H.S. = cosC2\text{cos}\dfrac{C}{2}

∴ L.H.S. = R.H.S.

Hence, sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}.

(ii) tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}

According to angle sum property,

∠A+∠B+∠C=180°∠B+∠C=180°−∠A⇒B+C2=180°−A2∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ B + ∠ C = 180° - ∠ A\\[1em] ⇒ \dfrac{B + C}{2} = \dfrac{180° - A}{2}\\[1em]

L.H.S.=tanB+C2=tan(180°−A2)=tan(90°−A2)=cotA2\text{L.H.S.} = \text{tan}\dfrac{B+C}{2}\\[1em] = \text{tan}\Big(\dfrac{180° - A}{2}\Big)\\[1em] = \text{tan}\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{cot}\dfrac{A}{2}

R.H.S. = cotA2\text{cot}\dfrac{A}{2}

∴ L.H.S. = R.H.S.

Hence, tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}.

Question 7(i)

Evaluate :

3 sin 72° cos 18°\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}} - sec 32°cosec 58°\dfrac{\text{sec 32°}}{\text{cosec 58°}}

Answer

3 sin 72° cos 18°−sec 32°cosec 58°=3 sin (90° - 18°) cos 18°−sec (90° - 58°)cosec 58°=3 cos 18° cos 18°−cosec 58°cosec 58°=3cos18°cos18°−cosec58°cosec58°=3−1=2\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\text{ sin (90° - 18°)}}{\text{ cos 18°}} - \dfrac{\text{sec (90° - 58°)}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\text{ cos 18°}}{\text{ cos 18°}} - \dfrac{\text{cosec 58°}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\cancel{ cos 18°}}{\cancel{ cos 18°}} - \dfrac{\cancel{cosec 58°}}{\cancel{cosec 58°}}\\[1em] = 3 - 1\\[1em] = 2

Hence, 3 sin 72° cos 18°−sec 32°cosec 58°=2\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} = 2.

Question 7(ii)

Evaluate :

3 cos 80° cosec 10° + 2 sin 59° sec 31°

Answer

3 cos 80° cosec 10° + 2 sin 59° sec 31°

= 3 cos (90° - 10°) cosec 10° + 2 sin (90° - 31°) sec 31°

= 3 sin 10° cosec 10° + 2 cos 31° sec 31°

= 3sin 10°×1sin 10°+2cos 31°×1cos 31°3 \text{sin 10°} \times \dfrac{1}{\text{sin 10°}} + 2 \text{cos 31°} \times \dfrac{1}{\text{cos 31°}}

= 3sin10°×1sin10°+2cos31°×1cos31°3 \cancel{sin 10°} \times \dfrac{1}{\cancel{sin 10°}} + 2 \cancel{cos 31°} \times \dfrac{1}{\cancel{cos 31°}}

= 3 + 2

= 5

Hence, 3 cos 80° cosec 10° + 2 sin 59° sec 31° = 5.

Question 7(iii)

Evaluate :

sin 80°cos 10°\dfrac{\text{sin 80°}}{\text{cos 10°}} + sin 59° sec 31°

Answer

sin 80°cos 10°+sin 59° sec 31°=sin (90° - 10°)cos 10°+sin (90° - 31°) sec 31°=cos 10°cos 10°+cos 31° sec 31°=cos10°cos10°+cos 31°1cos 31°=1+cos31°1cos31°=1+1=2\dfrac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°}\\[1em] = \dfrac{\text{sin (90° - 10°)}}{\text{cos 10°}} + \text{sin (90° - 31°) sec 31°}\\[1em] = \dfrac{\text{cos 10°}}{\text{cos 10°}} + \text{cos 31° sec 31°}\\[1em] = \dfrac{\cancel{cos 10°}}{\cancel{cos 10°}} + \text{cos 31°}\dfrac{1}{\text{cos 31°}}\\[1em] = 1 + \cancel{cos 31°}\dfrac{1}{\cancel{cos 31°}}\\[1em] = 1 + 1\\[1em] = 2

Hence, sin 80°cos 10°+sin 59° sec 31°=2\dfrac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°} = 2.

Question 7(iv)

Evaluate :

tan (55° - A) - cot (35° + A)

Answer

tan (55° - A) - cot (35° + A)

= tan [(90° - 35°) - A] - cot (35° + A)

= tan [90° - (35° + A)] - cot (35° + A)

= cot (35° + A) - cot (35° + A)

= 0

Hence, tan (55° - A) - cot (35° + A) = 0.

Question 7(v)

Evaluate :

cosec (65° + A) - sec (25° - A)

Answer

cosec (65° + A) - sec (25° - A)

= cosec [(90° - 25°) + A] - sec (25° - A)

= cosec [90° - (25° - A)] - sec (25° - A)

= sec (25° - A) - sec (25° - A)

= 0

Hence, cosec (65° + A) - sec (25° - A) = 0.

Question 7(vi)

Evaluate :

2tan 57°cot 33°2\dfrac{\text{tan 57°}}{\text{cot 33°}} - cot 70°tan 20°\dfrac{\text{cot 70°}}{\text{tan 20°}} - 2cos 45°{\sqrt2} \text{cos 45°}

Answer

2tan 57°cot 33°−cot 70°tan 20°−2cos45°=2tan (90° - 33°)cot 33°−cot (90° - 20°)tan 20°−2cos45°=2cot 33°cot 33°−tan 20°tan 20°−212=2cot33°cot33°−tan20°tan20°−212=2−1−1=02\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\dfrac{\text{tan (90° - 33°)}}{\text{cot 33°}} - \dfrac{\text{cot (90° - 20°)}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\dfrac{\text{cot 33°}}{\text{cot 33°}} - \dfrac{\text{tan 20°}}{\text{tan 20°}} - {\sqrt2} \dfrac{1}{\sqrt2}\\[1em] = 2\dfrac{\cancel{cot 33°}}{\cancel{cot 33°}} - \dfrac{\cancel{tan 20°}}{\cancel{tan 20°}} - \cancel{\sqrt2} \dfrac{1}{\cancel{\sqrt2}}\\[1em] = 2 - 1 - 1\\[1em] = 0

Hence, 2tan 57°cot 33°−cot 70°tan 20°−2cos45°=02\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°} = 0.

Question 7(vii)

Evaluate :

cot2 41°tan2 49°\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2 sin2 75° cos2 15°2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}

Answer

cot2 41°tan2 49°−2 sin2 75° cos2 15°=cot2 (90° - 49°)tan2 49°−2 sin2 (90° - 15°) cos2 15°=tan249°tan2 49°−2 cos215° cos2 15°=tan249°tan249°−2cos215°cos215°=1−2=−1\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\text{cot}^2 \text{ (90° - 49°)}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ (90° - 15°)}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\text{tan}^2 \text{49°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ cos}^2 \text{15°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\cancel{tan^2 49°}}{\cancel{tan^2 49°}} - 2\dfrac{\cancel{ cos^2 15°}}{\cancel{ cos^2 15°}}\\[1em] = 1 - 2\\[1em] = - 1

Hence, cot2 41°tan2 49°−2 sin2 75° cos2 15°=−1\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}} = -1.

Question 7(viii)

Evaluate :

cos 70°sin 20°\dfrac{\text{cos 70°}}{\text{sin 20°}} + cos 59°sin 31°\dfrac{\text{cos 59°}}{\text{sin 31°}} - 8 sin2 30°

Answer

cos 70°sin 20°+cos 59°sin 31°−8 sin230°=cos (90° - 70°)sin 20°+cos (90° - 31°)sin 31°−8×(12)2=sin 70°sin 20°+sin 31°sin 31°−8×(14)=sin70°sin20°+sin31°sin31°−2=1+1−2=0\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°}\\[1em] =\dfrac{\text{cos (90° - 70°)}}{\text{sin 20°}} + \dfrac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] =\dfrac{\text{sin 70°}}{\text{sin 20°}} + \dfrac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{4}\Big)\\[1em] =\dfrac{\cancel{sin 70°}}{\cancel{sin 20°}} + \dfrac{\cancel{sin 31°}}{\cancel{sin 31°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 0

Hence, cos 70°sin 20°+cos 59°sin 31°−8 sin230°=0\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°} = 0.

Question 7(ix)

Evaluate :

14 sin 30° + 6 cos 60° - 5 tan 45°.

Answer

14 sin 30° + 6 cos 60° - 5 tan 45°

=14×12+6×12−5×1= 14 \times \dfrac{1}{2} + 6 \times \dfrac{1}{2} - 5 \times 1

= 7 + 3 - 5

= 10 - 5

= 5

Hence, 14 sin 30° + 6 cos 60° - 5 tan 45° = 5.

Question 8

A triangle ABC is right-angled at B; find the value of sec A . sin C - tan A . tan Csin B\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}.

Answer

Given:

ABC is right-angled triangle at B.

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 90° + ∠ C = 180°

⇒ ∠ A + ∠ C = 180° - 90°

⇒ ∠ A + ∠ C = 90°

⇒ ∠ A = 90° - ∠ C

Now,

sec A . sin C - tan A . tan Csin B=sec (90° - C) . sin C - tan (90° - C). tan Csin B=cosec C . sin C - cot C. tan Csin B=1sin C×sin C−1tan C×tan Csin B=1sinC×sinC−1tanC×tanCsin B=1−1sin B=0\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{sec (90° - C) . sin C - tan (90° - C). tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{cosec C . sin C - cot C. tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\text{sin C}} \times \text{sin C} - \dfrac{1}{\text{tan C}} \times \text{tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\cancel{sin C}} \times \cancel{sin C} - \dfrac{1}{\cancel{tan C}} \times \cancel{tan C}}{\text{sin B}}\\[1em] = \dfrac{1 - 1}{\text{sin B}}\\[1em] = 0

Hence, sec A . sin C - tan A . tan Csin B=0\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}} = 0.

Question 9

In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.

(i) sin (90° - 3A) . cosec 42° = 1

(ii) cos (90° - A) . sec 77° = 1

Answer

(i) sin (90° - 3A) . cosec 42° = 1

⇒ cos 3A . cosec 42° = 1

⇒ cos 3A x 1sin 42°\dfrac{1}{\text{sin 42°}} = 1

⇒ cos 3A = sin 42°

⇒ cos 3A = sin (90° - 48°)

⇒ cos 3A = cos 48°

So, 3A = 48°

⇒ A = 48°3\dfrac{48°}{3}

⇒ A = 16°

Hence, A = 16°.

(ii) cos (90° - A) . sec 77° = 1

⇒ sin A . sec 77° = 1

⇒ sin A x 1cos 77°\dfrac{1}{\text{cos 77°}} = 1

⇒ sin A = cos 77°

⇒ sin A = cos (90° - 13°)

⇒ sin A = sin 13°

So, A = 13°

Hence, A = 13°.

PrevNext