Trigonometrical Identities — Exercise 21(C)

sin² A + sin² (90° - A) is equal to :

1. -1

2. 1

3. 2

4. -2

Answer

sin (90° - A) = cos A

To prove:

sin² A + sin² (90° - A)

⇒ sin² A + cos² A

⇒ 1.

Hence, Option 2 is the correct option.

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

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

2. sec $\dfrac{B}{2}$

3. cosec $\dfrac{B + A}{2}$

4. none of these

Answer

In triangle ABC,

By angle sum property of triangle,

⇒ A + B + C = 180°

⇒ A + C = 180° - B

⇒ $\dfrac{A + C}{2} = \dfrac{180° - B}{2}$ ........(1)

Substituting value of $\dfrac{A + C}{2}$ in sec $\dfrac{C + A}{2}$, we get :

$\Rightarrow \text{sec } \dfrac{C + A}{2} \\[1em] \Rightarrow \text{sec } \dfrac{180° - B}{2} \\[1em] \Rightarrow \text{sec } \Big(90° - \dfrac{B}{2}\Big) \\[1em] \Rightarrow \text{cosec } \dfrac{B}{2}.$

Hence, Option 1 is the correct option.

$\dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}}$ + 5 is equal to :

1. -3

2. 4

3. 3

4. -4

Answer

Solving,

$\Rightarrow \dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{cot (90° - 44°)}}{\text{tan 44°}} - 3 \dfrac{\text{sec (90° - 70°)}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{tan 44°}}{\text{tan 44°}} - 3 \dfrac{\text{cosec 70°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow 1 - 3 \times 1 + 5 \\[1em] \Rightarrow 1 - 3 + 5 \\[1em] \Rightarrow 3.$

Hence, Option 3 is the correct option.

sin 67° . cos 23° + cos 67° . sin 23° is equal to :

1. -1

2. 2 sin 67°

3. 2 cos 23°

4. 1

Answer

Solving,

⇒ sin 67° . cos 23° + cos 67° . sin 23°

⇒ sin 67° . cos (90° - 67°) + cos 67° . sin (90° - 67°)

⇒ sin 67° . sin 67° + cos 67° . cos 67°

⇒ sin² 67° + cos² 67°

⇒ 1.

Hence, Option 4 is the correct option.

$\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}$ is equivalent to :

1. cos² θ - 1

2. sin² θ

3. sin² θ - cos² θ

4. sin² θ - 1

Answer

Solving,

$\Rightarrow \dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ . sin θ}}{\text{cos θ}} \\[1em] \Rightarrow \text{sin}^2 θ.$

Hence, Option 2 is the correct option.

Show that :

tan 10° tan 15° tan 75° tan 80° = 1

Answer

Solving L.H.S. of the equation :

⇒ tan 10° tan 15° tan 75° tan 80°

⇒ tan 10° tan 15° tan (90 - 15)° tan (90 - 10)°

By formula,

tan (90° - A) = cot A

⇒ tan 10° tan 15° cot 15° cot 10°

⇒ tan 10° × tan 15° $\times \dfrac{1}{\text{tan 15°}} \times \dfrac{1}{\text{tan 10°}}$

⇒ 1.

Since, L.H.S. = R.H.S.

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

Show that :

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

Answer

Solving L.H.S. of the equation :

⇒ sin 42° sec (90 - 42)° + cos 42° cosec (90 - 42)°

By formula,

sec (90° - A) = cosec A and cosec (90° - A) = sec A.

⇒ sin 42° cosec 42° + cos 42° sec 42°

⇒ sin 42° $\times \dfrac{1}{\text{sin 42°}}$ + cos 42° $\times \dfrac{1}{\text{cos 42°}}$

⇒ 1 + 1

⇒ 2.

Since, L.H.S. = R.H.S.

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

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

(i) sin 59° + tan 63°

(ii) cosec 68° + cot 72°

Answer

(i) Solving,

⇒ sin 59° + tan 63°

⇒ sin (90 - 31)° + tan (90 - 27)°

By formula,

sin (90° - A) = cos A and tan (90° - A) = cot A.

⇒ cos 31° + cot 27°.

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

(ii) Solving,

⇒ cosec 68° + cot 72°

⇒ cosec (90 - 22)° + cot (90 - 18)°

By formula,

cosec (90° - A) = sec A and cot (90° - A) = tan A.

⇒ sec 22° + tan 18°.

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

Show that :

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

Answer

To prove:

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

By formula,

sin (90° - A) = cos A and cos (90° - A) = sin A.

Substituting above values in L.H.S. :

$\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} \times \dfrac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A sec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A.

Show that :

sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0

Answer

By formula,

cos (90° - A) = sin A, sec (90° - A) = cosec A, cosec (90° - A) = sec A and sin (90° - A) = cos A.

$\Rightarrow \text{sin A cos A} - \dfrac{\text{sin A sin A cos A}}{\text{cosec A}} - \dfrac{\text{cos A cos A sin A}}{\text{sec A}} \\[1em] \Rightarrow \text{sin A cos A} - \dfrac{\text{sin}^2 A \text{cos A}}{\dfrac{1}{\text{sin A}}} - \dfrac{\text{cos}^2 A \text{sin A}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin}^3 A \text{ cos A} - \text{cos}^3 A \text{sin A} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin A cos A}(\text{sin}^2 A + \text{cos}^2 A)$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \text{sin A cos A} - \text{sin A cos A} \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0.

For triangle ABC, show that :

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

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

Answer

(i) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠A + ∠B = 180° - ∠C .........(1)

Given equation,

sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$

Substituting value of (A + B) from (1) in L.H.S. of above equation :

$\Rightarrow \text{sin} \dfrac{180° - C}{2} \\[1em] \Rightarrow \text{sin} \Big(90° - \dfrac{C}{2}\Big)$

By formula,

sin(90° - θ) = cos θ

$\therefore \text{cos} \dfrac{C}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$.

(ii) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠B + ∠C = 180° - ∠A .........(1)

Given equation,

tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$

Substituting value of (B + C) from (1) in L.H.S. of above equation :

$\Rightarrow \text{tan} \dfrac{180° - A}{2} \\[1em] \Rightarrow \text{tan } \Big(90° - \dfrac{A}{2}\Big)$

By formula,

tan(90° - θ) = cot θ

$\Rightarrow \text{cot} \dfrac{A}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$.

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

Answer

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠A = 90° - ∠C.

Substituting value of A in $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ we get,

$\Rightarrow \dfrac{\text{sec (90° - C). cosec C - tan (90° - C). cot C}}{\text{sin 90°}}$

By formula,

tan (90° - C) = cot c, sec (90° - C) = cosec C and cosec² C - cot² C = 1.

$\Rightarrow \dfrac{\text{cosec C. cosec C - cot C. cot C}}{1} \\[1em] \Rightarrow \text{cosec}^2 C - \text{cot}^2 C \\[1em] \Rightarrow 1.$

Hence, $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ = 1.

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) Given,

⇒ sin (90° - 3A). cosec 42° = 1

⇒ sin (90° - 3A) × $\dfrac{1}{\text{sin 42°}}$ = 1

⇒ sin (90° - 3A) = sin 42°

⇒ 90° - 3A = 42°

⇒ 3A = 90° - 42°

⇒ 3A = 48°

⇒ A = 16°.

Hence, A = 16°.

(ii) Given,

⇒ cos (90° - A). sec 77° = 1

⇒ cos (90° - A) × $\dfrac{1}{\text{cos 77°}}$ = 1

⇒ cos (90° - A) = cos 77°

⇒ 90° - A = 77°

⇒ A = 90° - 77°

⇒ A = 13°.

Hence, A = 13°.

Prove that :

$\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 - cos² θ

Answer

By formula,

cos (90° - θ) = sin θ.

Solving L.H.S. of the equation

$\Rightarrow \dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ cos θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \text{sin}^2 θ \\[1em] \Rightarrow \text{1 - cos}^2 θ.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 - cos² θ.

Evaluate :

sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

Answer

Solving,

⇒ sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

⇒ sin² 34° + sin² (90° - 34°) + 2 tan 18° tan (90° - 18°) - cot² 30°

By formula,

sin(90° - A) = cos A and tan(90° - A) = cot A

⇒ sin² 34° + cos² 34° + 2 tan 18° cot 18° - cot² 30°

By formula,

sin² θ + cos² θ = 1

⇒ 1 + 2 tan 18° $\times \dfrac{1}{\text{tan 18°}} - (\sqrt{3})^2$

⇒ 1 + 2 - 3

⇒ 3 - 3

⇒ 0.

Hence, sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30° = 0.

Evaluate :

cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60°

Answer

Solving,

⇒ cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60°

⇒ cosec² 57° - tan² (90° - 57°) + cos 44° cosec (90° - 44°) - $\sqrt{2} \times \dfrac{1}{\sqrt{2}} - (\sqrt{3})^2$

By formula,

tan(90° - A) = cot A and cosec(90° - A) = sec A

⇒ cosec² 57° - cot² 57° + cos 44° sec 44° - 1 - 3

By formula,

cosec² A - cot² A = 1

⇒ 1 + cos 44° $\times \dfrac{1}{\text{cos 44°}}$ - 1 - 3

⇒ 1 + 1 - 1 - 3

⇒ -2.

Hence, cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60° = -2.