Trigonometrical Identities — Exercise 22(B)

Eliminate θ between the given equations:

x = a cosec θ, y = b cot θ

Answer

Given,

x = a cosec θ, y = b cot θ

⇒ cosec θ = $\dfrac{x}{a}$

⇒ cot θ = $\dfrac{y}{b}$

Using the identity

cosec² θ - cot² θ = 1

Substitute the values:

$\Rightarrow \Big(\dfrac{x}{a}\Big)^2 - \Big(\dfrac{y}{b}\Big)^2 = 1 \\[1em] \Rightarrow \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

Hence, the required relation is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.

Eliminate θ between the given equations:

x = a cot θ + b cosec θ, y = a cosec θ + b cot θ

Answer

x = a cot θ + b cosec θ .....(1)

y = a cosec θ + b cot θ .....(2)

Subtract equation (2) from (1):

⇒ x - y = a cot θ + b cosec θ - (a cosec θ + b cot θ)

⇒ x - y = a cot θ + b cosec θ - a cosec θ - b cot θ

⇒ x - y = a (cot θ - cosec θ) - b (cot θ - cosec θ)

⇒ x - y = (a - b) (cot θ - cosec θ)

Add equation (2) from (1):

⇒ x + y = a cot θ + b cosec θ + (a cosec θ + b cot θ)

⇒ x + y = a (cot θ + cosec θ) + b (cot θ + cosec θ)

⇒ x + y = (a + b) (cot θ + cosec θ)

Now multiply:

⇒ (x + y)(x - y) = (a + b)(a - b)(cot θ + cosec θ)(cot θ - cosec θ)

⇒ x² - y² = (a² - b²)(cot² θ - cosec² θ)

By identity: cot² θ − cosec ² θ = −1

⇒ x² - y² = - (a² - b²)

⇒ x² - y² = b² - a²

Hence, the required relation is x² - y² = b² - a² .

Eliminate θ between the given equations:

x = a sec³ θ, y = b tan³ θ

Answer

Given,

⇒ x = a sec³ θ

⇒ sec³ θ = $\dfrac{x}{a}$

⇒ sec² θ = $\Big(\dfrac{x}{a}\Big)^\dfrac{2}{3}$

⇒ y = b tan³ θ

⇒ tan³ θ = $\dfrac{y}{b}$

⇒ tan² θ = $\Big(\dfrac{y}{b}\Big)^\dfrac{2}{3}$

Using the identity

sec² θ - tan² θ = 1

Substitute,

$\Big(\dfrac{x}{a}\Big)^\dfrac{2}{3} - \Big(\dfrac{y}{b}\Big)^\dfrac{2}{3} = 1$

Hence,the required relation is $\Big(\dfrac{x}{a}\Big)^\dfrac{2}{3} - \Big(\dfrac{y}{b}\Big)^\dfrac{2}{3} = 1$.

Eliminate θ between the given equations:

$\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1, \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1$

Answer

$\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1.....(1)\\[1em] \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1.....(2)$

Square and add equations (1) and (2):

$\Rightarrow \Big[\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta\Big]^2 + \Big[\Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta\Big]^2 = 1^2 + (-1)^2 \\[1em] \Rightarrow \Big[\Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + 2\dfrac{xy}{ab} \sin \theta \cos \theta \Big] + \Big[\Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta - 2\dfrac{xy}{ab} \sin \theta \cos \theta\Big] = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + \Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 (\cos^2 \theta + \sin^2 \theta) + \Big(\dfrac{y}{b}\Big)^2 (\sin^2 \theta + \cos^2 \theta) = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2 .$

Hence, the required relation is $\Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2$.

Eliminate θ between the given equations:

x = h + a cos θ, y = k + b sin θ

Answer

⇒ x = h + a cos θ

⇒ cos θ = $\dfrac{x - h}{a}$.....(1)

⇒ y = k + b sin θ

⇒ sin θ = $\dfrac{y - k}{b}$.....(2)

Square and add equations (1) and (2):

$\Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = \cos^2 \theta + \sin^2 \theta \\[1em] \Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = 1 \\[1em] \Rightarrow \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 .$

Hence,the required relation is $\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$.

If $\dfrac{\cos A}{\cos B}$ = m and $\dfrac{\cos A}{\sin B}$ = n, prove that : (m² + n²)cos²B = n²

Answer

To prove:

(m² + n²)cos²B = n²

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

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

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

Hence, proved that (m² + n²)cos²B = n².

If x = a sec A cos B, y = b sec A sin B and z = c tan A, prove that $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$

Answer

⇒ x = a sec A cos B

sec A cos B = $\dfrac{x}{a}$....(1)

⇒ y = b sec A sin B

sec A sin B = $\dfrac{y}{b}$....(2)

Square and add equations (1) and (2) :

$\Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A \cos^2 B + \sec^2 A \sin^2 B \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A (\cos^2 B + \sin^2 B) \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A ...(3)$

Now,

⇒ z = c tan A

tan A = $\dfrac{z}{c}$

tan² A = $\dfrac{z^2}{c^2}$...(4)

Subtract (4) from (3):

$\Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = \sec^2 A - \tan^2 A \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1.$

Hence, proved that $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$.