Trigonometrical Ratios — Exercise 21(D)

If A = 30°, then $\dfrac{\text{sin 2 A}}{\text{1 - cos 2 A}}$ is equal to :

1. cot A

2. tan A

3. sec A

4. cosec A

Answer

$\dfrac{\text{sin 2 A}}{\text{1 - cos 2 A}}\\[1em] = \dfrac{\text{sin (2 x 30°)}}{\text{1 - cos (2 x 30°)}}\\[1em] = \dfrac{\text{sin 60°}}{\text{1 - cos 60°}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{1 - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{2}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{2 - 1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{\cancel{2}}}{\dfrac{1}{\cancel{2}}}\\[1em] = \dfrac{\sqrt3}{1}\\[1em] = \text {cot A}$

Hence, option 1 is the correct option.

If A = 60° and B = 30°; the value of sin A cos B + cos A sin B is equal to :

1. $\dfrac{1}{2}$

2. 1

3. 2

4. ${\sqrt2}$

Answer

sin A cos B + cos A sin B = sin 60°. cos 30° + cos 60°. sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1$

Hence, option 2 is the correct option.

If A = 30° ; 3 sin A - 4 sin³ A is equal to:

1. cos 3A

2. tan 3A

3. sin 3A

4. cot 3A

Answer

3 sin A - 4 sin³ A = 3 sin 30° - 4 sin³ 30°

$= 3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{2} - 4 \times \dfrac{1}{8}\\[1em] = \dfrac{3}{2} - \dfrac{1}{2}\\[1em] = \dfrac{3 - 1}{2} \\[1em] = \dfrac{2}{2} \\[1em] = 1\\[1em] = \text{sin 90°}\\[1em] = \text{sin 3 x 30°}\\[1em] = \text{sin 3A}\\[1em]$

Hence, option 3 is the correct option.

If A = 30°, then cos⁴ A - sin⁴ A is equal to :

1. sin 60°

2. tan 60°

3. cot 60°

4. cos 60°

Answer

cos⁴ A - sin⁴ A = cos⁴ 30° - sin⁴ 30°

$= \Big(\dfrac{\sqrt3}{2}\Big)^4 - \Big(\dfrac{1}{2}\Big)^4\\[1em] = \dfrac{9}{16} - \dfrac{1}{16}\\[1em] = \dfrac{9 - 1}{16}\\[1em] = \dfrac{8}{16}\\[1em] = \dfrac{1}{2}\\[1em] = \text {cos 60°}$

Hence, option 4 is the correct option.

If A = 45° ; then 2 sin A cos A is equal to :

1. 1

2. 0

3. -1

4. 2

Answer

2 sin A cos A = 2 sin 45°. cos 45°

$= 2 \times \dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2}\\[1em] = 2 \times \dfrac{1}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}}\\[1em] = 1$

Hence, option 1 is the correct option.

Given A = 60° and B = 30°, prove that :

sin (A + B) = sin A cos B + cos A sin B

Answer

sin (A + B) = sin A cos B + cos A sin B

L.H.S. = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

∴ L.H.S. = R.H.S.

Hence, sin (A + B) = sin A cos B + cos A sin B.

Given A = 60° and B = 30°, prove that :

cos (A + B) = cos A cos B - sin A sin B

Answer

cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos (A + B) = cos (60° + 30°)

= cos 90° = 0

R.H.S. = cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt3}{2} - \dfrac{\sqrt3}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{\sqrt3}{4} - \dfrac{\sqrt3}{4}\\[1em] = 0$

∴ L.H.S. = R.H.S.

Hence, cos (A + B) = cos A cos B - sin A sin B.

Given A = 60° and B = 30°, prove that :

cos (A - B) = cos A cos B + sin A sin B

Answer

cos (A - B) = cos A cos B + sin A sin B

L.H.S. = cos (A - B) = cos (60° - 30°)

= cos 30° = $\dfrac{\sqrt3}{2}$

R.H.S. = cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{\sqrt3}{4} + \dfrac{\sqrt3}{4}\\[1em] = 2 \times \dfrac{\sqrt3}{4}\\[1em] = \dfrac{\sqrt3}{2}\\[1em]$

∴ L.H.S. = R.H.S.

Hence, cos (A - B) = cos A cos B + sin A sin B.

Given A = 60° and B = 30°, prove that :

$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + \text{ tan A }} . \text{ tan B}}$

Answer

$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$

L.H.S. = tan (A - B) = tan (60° - 30°)

= tan 30° = $\dfrac{1}{\sqrt3}$

$\text{R.H.S.} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}\\[1em] = \dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan} \text{ 60° } . \text{ tan 30°}}\\[1em] = \dfrac{\sqrt3 - \dfrac{1}{\sqrt3}}{1 + \sqrt3 \times \dfrac{1}{\sqrt3}}\\[1em] = \dfrac{\dfrac{\sqrt3 \times \sqrt3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + \cancel{\sqrt3} \times \dfrac{1}{\cancel{\sqrt3}}}\\[1em] = \dfrac{\dfrac{3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{3 - 1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{2}\\[1em] = \dfrac{\dfrac{\cancel{2}}{\sqrt3}}{\cancel{2}}\\[1em] = \dfrac{1}{\sqrt3}$

∴ L.H.S. = R.H.S.

Hence, $\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$.

If A = 30°, then prove that :

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

Answer

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{sin 2 A} = \text{sin (2 x 30°)} = \text{sin 60°} = \dfrac{\sqrt3}{2}$

$\text{2nd term} = \text{2 sin A cos A}\\[1em] = 2 \times \text{sin 30°} \times \text{cos 30°}\\[1em] = 2 \times \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{2}$

$\text{3rd term} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{2 \times \dfrac{1}{\sqrt3}}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{4}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times \sqrt3}\\[1em] = \dfrac{\sqrt3}{2}$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$.

If A = 30°, then prove that :

$\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

Answer

$\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{cos 2 A} = \text{cos (2 x 30°)} = \text{cos 60°} = \dfrac{1}{2}$

$\text{2nd term} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A}\\[1em] = \text{cos}^2 \text{ 30°} - \text{sin}^2 \text{ 30°}\\[1em] = \Big(\dfrac{\sqrt3}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} - \dfrac{1}{4}\\[1em] = \dfrac{3 - 1}{4}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

$\text{3rd term} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{3}{3} - \dfrac{1}{3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3 - 1}{3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{\dfrac{2}{\cancel{3}}}{\dfrac{4}{\cancel{3}}}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

If A = 30°, then prove that :

2 cos² A - 1 = 1 - 2 sin² A

Answer

2 cos² A - 1 = 1 - 2 sin² A

$\text{L.H.S.} = 2 \text{cos}^2 A - 1\\[1em] = 2 \text{cos}^2 30° - 1\\[1em] = 2 \times \Big(\dfrac{\sqrt3}{2}\Big)^2 - 1\\[1em] = 2 \times \dfrac{3}{4} - 1\\[1em] = \dfrac{3}{2} - \dfrac{2}{2}\\[1em] = \dfrac{3 - 2}{2}\\[1em] = \dfrac{1}{2}$

$\text{R.H.S.} = 1 - 2 \text{sin}^2 A \\[1em] = 1 - 2 \text{sin}^2 30°\\[1em] = 1 - 2 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = 1 - 2 \times \dfrac{1}{4}\\[1em] = \dfrac{2}{2} - \dfrac{1}{2}\\[1em] = \dfrac{2 - 1}{2}\\[1em] = \dfrac{1}{2}$

∴ L.H.S. = R.H.S.

Hence, 2 cos² A - 1 = 1 - 2 sin² A.

If A = 30°, then prove that :

sin 3A = 3 sin A - 4 sin³ A

Answer

sin 3A = 3 sin A - 4 sin³ A

L.H.S. = sin 3A

= sin (3 x 30°)

= sin 90°

= 1

$\text{R.H.S.} = 3 \text{sin A} - 4 \text{sin}^3 A\\[1em] = 3 \text{sin 30°} - 4 \text{sin}^3 30°\\[1em] = 3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3\\[1em] = \dfrac{3}{2} - 4 \times \dfrac{1}{8}\\[1em] = \dfrac{3}{2} - \dfrac{1}{2}\\[1em] = \dfrac{3 - 1}{2}\\[1em] = \dfrac{2}{2}\\[1em] = 1$

∴ L.H.S. = R.H.S.

Hence, sin 3A = 3 sin A - 4 sin³ A.

If A = B = 45°, show that :

sin (A - B) = sin A cos B - cos A sin B

Answer

sin (A - B) = sin A cos B - cos A sin B

L.H.S. = sin (A - B)

= sin (45° - 45°)

= sin 0°

= 0

R.H.S. = sin A cos B - cos A sin B

= sin 45° . cos 45° - cos 45°. sin 45°

$= \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2} - \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2}\\[1em] = \dfrac {1}{2} - \dfrac {1}{2}\\[1em] = 0$

∴ L.H.S. = R.H.S.

Hence, sin (A - B) = sin A cos B - cos A sin B.

If A = B = 45°, show that :

cos (A + B) = cos A cos B - sin A sin B

Answer

cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos (A + B)

= cos (45° + 45°)

= cos 90°

= 0

R.H.S. = cos A cos B - sin A sin B

= cos 45° . cos 45° - sin 45°. sin 45°

$= \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2} - \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2}\\[1em] = \dfrac {1}{2} - \dfrac {1}{2}\\[1em] = 0$

∴ L.H.S. = R.H.S.

Hence, cos (A + B) = cos A cos B - sin A sin B.