Trigonometrical Identities — Exercise 21(B)

(1 + cot² A) + (1 + tan² A) is equal to :

1. $\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$

2. sec² A - cosec² A

3. sin² A - sin⁴ A

4. sec² A.cosec² A

Answer

Solving,

⇒ (1 + cot² A) + (1 + tan² A)

⇒ cosec² A + sec² A

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

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A. \text{cos}^2 A}$

Substituting, sin² A + cos² A = 1, we get :

⇒ $\dfrac{1}{\text{sin}^2 A. \text{cos}^2 A}$

⇒ cosec² A. sec² A

Hence, Option 4 is the correct option.

If a = tan θ and b = sec θ, the relation between a and b is :

1. a × b = 1

2. a² - b² = 1

3. b² - a² = 1

4. a² + b² = 1

Answer

Substituting value of a and b in L.H.S. of option 3, we get :

⇒ b² - a²

⇒ sec² A - tan² A

⇒ 1.

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

Hence, Option 3 is the correct option.

$\dfrac{1}{\text{cot θ + tan θ}}$ is equal to :

1. sin θ + cos θ

2. sin θ. cos θ

3. $\dfrac{1}{\text{sin θ. cos θ}}$

4. $\dfrac{1}{\text{sin θ + cos θ}}$

Answer

Solving,

$\Rightarrow \dfrac{1}{\text{cot θ + tan θ}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos θ}}{\text{sin θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos}^2 θ + \text{sin}^2 θ}{\text{cos θ. sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ. sin θ}}{\text{cos}^2 θ + \text{sin}^2 θ}$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ sin θ. cos θ

Hence, Option 2 is the correct option.

(sec θ - cos θ)² - (sec θ + cos θ)² is equal to :

1. 4

2. 2

3. -2

4. -4

Answer

Solving,

⇒ (sec θ - cos θ)² - (sec θ + cos θ)²

⇒ (sec θ - cos θ + sec θ + cos θ)[(sec θ - cos θ) - (sec θ + cos θ)] [∵ a² - b² = (a + b)(a - b)]

⇒ (sec θ - cos θ + sec θ + cos θ)(sec θ - sec θ - cos θ - cos θ)

⇒ 2 sec θ.(-2 cos θ)

⇒ -4.sec θ.cos θ

⇒ $-4 \times \dfrac{1}{\text{cos θ}} \times \text{cos θ}$

⇒ -4.

Hence, Option 4 is the correct option.

(cot A - cot B)² + (1 + cot A cot B)² is equal to :

1. sec² A - cos² A

2. sec² A - cosec² A

3. (1 + tan A)² - (1 - cot A)²

4. cosec² A . cosec² B

Answer

Solving,

⇒ (cot A - cot B)² + (1 + cot A cot B)²

⇒ cot² A + cot² B - 2 cot A cot B + 1 + cot² A . cot² B + 2 cot A cot B

⇒ cot² A + cot² A . cot² B + 1 + cot² B

⇒ cot² A(1 + cot² B) + (1 + cot² B)

⇒ (1 + cot² B)(1 + cot² A)

⇒ $\Big(1 + \dfrac{\text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(1 + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}\Big)$

⇒ $\Big(\dfrac{\text{sin}^2 B + \text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}\Big)$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ $\Big(\dfrac{1}{{\text{sin}^2 B}}\Big)\Big(\dfrac{1}{\text{sin}^2 A}\Big)$

⇒ cosec² B . cosec² A

Hence, Option 4 is the correct option.

Prove that:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Answer

To prove:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Solving L.H.S. of the above equation,

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

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

Hence, proved that (sec A - tan A)² (1 + sin A) = (1 - sin A).

Prove that :

$\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{(\text{cos}^3 A + \text{sin}^3 A)(\text{cos A - sin A}) + (\text{cos}^3 A - \text{sin}^3 A)(\text{cos A + sin A})}{\text{(cos A + sin A)(cos A - sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^4 A - \text{cos}^3 A\text{ sin A} + \text{cos A sin}^3 A - \text{ sin}^4 A + \text{cos}^4 A + \text{cos}^3 A\text{ sin A} - \text{ sin}^3 A\text{ cos A} - \text{sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos}^4 A - \text{2 sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^4 A - \text{ sin}^4 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^2 A - \text{sin}^2 A)\text{(cos}^2 A + \text{sin}^2 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \text{2 (cos}^2 A + \text{sin}^2 A).$

By formula,

cos² A + sin² A = 1

⇒ 2 × 1 = 2.

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

Hence, proved that $\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2.

Prove that :

$\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1

Answer

By solving L.H.S. of the equation :

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

By formula,

sin² A + cos² A = 1

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

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

Hence, proved that $\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1.

Prove that :

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

Answer

Solving L.H.S. of the equation :

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

By formula,

cos² A = 1 - sin² A

$\Rightarrow 2\Big(\dfrac{1 + \text{sin}^2 A}{1 - \text{sin}^2 A}\Big)$

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

Hence, proved that

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

Prove that :

2 sin² A + cos⁴ A = 1 + sin⁴ A

Answer

Solving L.H.S. of the equation :

⇒ 2 sin² A + cos⁴ A

⇒ 2 sin² A + (cos² A)²

By formula,

cos² A = 1 - sin² A

⇒ 2 sin² A + (1 - sin² A)²

⇒ 2 sin² A + 1 + sin⁴ A - 2 sin² A

⇒ 1 + sin⁴ A.

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

Hence, proved that 2 sin² A + cos⁴ A = 1 + sin⁴ A.

Prove that :

$\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{(sin A - sin B)(sin A + sin B) + (cos A - cos B)(cos A + cos B)}}{\text{(cos A + cos B)(sin A + sin B)}} = 0 \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 B + \text{cos}^2 A - \text{cos}^2 B}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A -\text{(sin}^2 B + \text{cos}^2 B)}{\text{(cos A + cos B)(sin A + sin B)}}$

By formula,

sin² θ + cos² θ = 1

$\therefore \dfrac{1 - 1}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow 0.$

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

Hence, proved that $\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0.

Prove that :

(cosec A - sin A)(sec A - cos A) = $\dfrac{1}{\text{tan A + cot A}}$

Answer

Solving L.H.S. of the equation :

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

By formula,

1 - sin² A = cos² A

1 - cos² A = sin² A

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

Solving R.H.S. of the equation :

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

By formula,

sin² θ + cos² θ = 1

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

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

Hence, proved that (cosec A - sin A)(sec A - cos A) = $\dfrac{1}{\text{tan A + cot A}}$.

Prove that :

(1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B

Answer

By formula,

sec² θ = 1 + tan² θ

Solving L.H.S. of the equation :

⇒ (1 + tan A. tan B)² + (tan A - tan B)²

⇒ 1 + tan² A tan² B + 2 tan A tan B + tan² A + tan² B - 2 tan A tan B

⇒ 1 + tan² A tan² B + tan² A + tan² B

⇒ 1 + tan² A + tan² A tan² B + tan² B

⇒ sec² A + tan² B(tan² A + 1)

⇒ sec² A + tan² B sec² A

⇒ sec² A(1 + tan² B)

⇒ sec² A sec² B.

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

Hence, proved that (1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B.

Prove that :

$\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A

Answer

Solving L.H.S. of the equation :

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

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

Hence, proved that $\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A.

If x cos A + y sin A = m and x sin A - y cos A = n, then prove that :

x² + y² = m² + n²

Answer

To prove:

x² + y² = m² + n²

Substituting value of m and n in R.H.S. of the equation :

= (x cos A + y sin A)² + (x sin A - y cos A)²

= x² cos² A + y² sin² A + 2xy cos A sin A + x² sin² A + y² cos² A - 2xy sin A cos A

= x² cos² A + x² sin² A + y² cos² A + y² sin² A

= x²(sin² A + cos² A) + y²(sin² A + cos² A)

By formula,

sin² A + cos² A = 1

⇒ x² × 1 + y² × 1

⇒ x² + y².

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

Hence, proved that x² + y² = m² + n².

If m = a sec A + b tan A and n = a tan A + b sec A, show that :

m² - n² = a² - b²

Answer

To prove:

m² - n² = a² - b²

Substituting value of m and n in L.H.S. of the above equation :

= (a sec A + b tan A)² - (a tan A + b sec A)²

= a² sec² A + b² tan² A + 2ab sec A tan A - (a² tan² A + b² sec² A + 2ab sec A tan A)

= a² sec² A - a² tan² A + b²tan² A - b² sec² A + 2ab sec A tan A - 2ab sec A tan A

= a² (sec² A - tan² A) + b² (tan² A - sec² A)

= a² (sec² A - tan² A) - b² (sec² A - tan² A)

By formula,

sec² A - tan² A = 1

⇒ a² × 1 - b² × 1

⇒ a² - b²

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

Hence, proved that m² - n² = a² - b².

If x = r cos A cos B, y = r cos A sin B and z = r sin A, then prove that :

x² + y² + z² = r²

Answer

To prove:

⇒ x² + y² + z² = r²

Substituting value of x, y and z in L.H.S. of the equation :

= (r cos A cos B)² + (r cos A sin B)² + (r sin A)²

= r² cos² A cos² B + r² cos² A sin² B + r² sin² A

= r²cos² A(cos² B + sin² B) + r² sin² A

⇒ r²cos² A + r²sin² A [∵ sin² θ + cos² θ = 1]

⇒ r²(cos² A + sin² A)

⇒ r² × 1 [∵ sin² θ + cos² θ = 1]

⇒ r².

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

Hence, proved that x² + y² + z² = r².

If $\dfrac{\text{cos A}}{\text{cos B}} = m \text{ and } \dfrac{\text{cos A}}{\text{sin B}}$= n,

show that:

(m² + n²) cos² B = n²

Answer

To prove:

(m² + n²) cos² B = n²

Substituting value of m and n in L.H.S. of the above equation :

$\Rightarrow \Big[\Big(\dfrac{\text{cos A}}{\text{cos B}}\Big)^2 + \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2\Big]\text{cos}^2 B \\[1em] \Rightarrow \Big[\dfrac{\text{cos}^2 A}{\text{cos}^2 B} + \dfrac{\text{cos}^2 A}{\text{sin}^2 B}\Big]\text{cos}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{ cos}^2 B \text{ sin}^2 B} \times \text{cos}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A(\text{sin}^2 B + \text{cos}^2 B)}{\text{sin}^2 B}$

By formula,

⇒ sin² B + cos² B = 1

$\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 B} \\[1em] \Rightarrow \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2 \\[1em] \Rightarrow n^2.$

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

Hence, proved that (m² + n²) cos² B = n².