Question

Please solve this sums.​

Answer 1

Given:

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

Solution:

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow \dfrac{1}{cosecA} + \dfrac{cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

We know that:

1/cosecA = sinA

cosecA/cosecA = 1

$\sf \Longrightarrow sinA + 1 = \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow (sinA + 1)(1 - sinA) = sin^2A$

Using the identity (a + b)(a - b) = a² - b² we get:

[Here a = 1, and b = sinA]

$\sf \Longrightarrow (1)^2 - (sinA)^2 = sin^2A$

$\sf \Longrightarrow 1 - sin^2A = sin^2A$

$\sf \Longrightarrow 1 = sin^2A + sin^2A$

$\sf \Longrightarrow 1 = 2sin^2A$

$\sf \Longrightarrow 2sin^2A = 1$

$\sf \Longrightarrow sin^2A = \dfrac{1}{2}$

$\sf \Longrightarrow sinA = \sqrt{\dfrac{1}{2}}$

$\sf \Longrightarrow sinA = \dfrac{1}{\sqrt{2}}$

$\sf \Longrightarrow A = sin^{-1} \ \dfrac{1}{\sqrt{2}}$

We know that:

sin45° = 1/√2

$\sf \Longrightarrow A = 45 ^\circ$

Now, let's check if the LHS = RHS.

(I'm not able to view the whole question, so I'm proving that LHS = RHS)

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

LHS:

$\sf \Longrightarrow \dfrac{1 + cosec45^\circ}{cosec45^\circ}$

Trigonometric ratio used: cosec45° = √2

$\sf \Longrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}}$

$\sf \Longrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf \Longrightarrow \dfrac{\sqrt{2} + (\sqrt{2})^2}{(\sqrt{2})^2}$

$\sf \Longrightarrow \dfrac{\sqrt{2} + 2}{2}$

Now, let's find the value of RHS:

$\sf \Longrightarrow \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow \dfrac{sin^245^\circ}{1 - sin45^\circ}$

Trigonometric ratio used: sin45° = 1/√2

$\sf \Longrightarrow \dfrac{\bigg(\dfrac{1}{\sqrt{2}}\bigg)^2}{1 - \bigg(\dfrac{1}{\sqrt{2}}\bigg)}$

$\sf \Longrightarrow \dfrac{\dfrac{1}{2}}{\bigg(\dfrac{\sqrt{2} - 1}{\sqrt{2}}\bigg)}$

$\sf \Longrightarrow \dfrac{1}{2} \times \dfrac{\sqrt{2}}{\sqrt{2} - 1}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2(\sqrt{2} - 1)}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2} \times \dfrac{2\sqrt{2} + 2}{2\sqrt{2} + 2}$

Identity used: (a + b)(a - b) = a² - b²

$\sf \Longrightarrow \dfrac{\sqrt{2}(2\sqrt{2} + 2)}{(2\sqrt{2})^2 - (2)^2}$

$\sf \Longrightarrow \dfrac{ \sqrt{2}(2\sqrt{2}) + \sqrt{2}(2)}{(4 \times 2) - 4}$

$\sf \Longrightarrow \dfrac{4 + 2\sqrt{2}}{8 - 4}$

$\sf \Longrightarrow \dfrac{2(2 + \sqrt{2})}{4}$

$\sf \Longrightarrow \dfrac{2 + \sqrt{2}}{2}$

LHS = RHS

Hence Proved.

Answer 2

Answer:

Given:

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

Solution:

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow \dfrac{1}{cosecA} + \dfrac{cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

We know that:

1/cosecA = sinA

cosecA/cosecA = 1

$\sf \Longrightarrow sinA + 1 = \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow (sinA + 1)(1 - sinA) = sin^2A$

Using the identity (a + b)(a - b) = a² - b² we get:

[Here a = 1, and b = sinA]

$\sf \Longrightarrow (1)^2 - (sinA)^2 = sin^2A$

$\sf \Longrightarrow 1 - sin^2A = sin^2A$

$\sf \Longrightarrow 1 = sin^2A + sin^2A$

$\sf \Longrightarrow 1 = 2sin^2A$

$\sf \Longrightarrow 2sin^2A = 1$

$\sf \Longrightarrow sin^2A = \dfrac{1}{2}$

$\sf \Longrightarrow sinA = \sqrt{\dfrac{1}{2}}$

$\sf \Longrightarrow sinA = \dfrac{1}{\sqrt{2}}$

$\sf \Longrightarrow A = sin^{-1} \ \dfrac{1}{\sqrt{2}}$

We know that:

sin45° = 1/√2

$\sf \Longrightarrow A = 45 ^\circ$

Now, let's check if the LHS = RHS.

(I'm not able to view the whole question, so I'm proving that LHS = RHS)

$\sf \Longrightarrow \dfrac{1 + cosecA}{cosecA} = \dfrac{sin^2A}{1 - sinA}$

LHS:

$\sf \Longrightarrow \dfrac{1 + cosec45^\circ}{cosec45^\circ}$

Trigonometric ratio used: cosec45° = √2

$\sf \Longrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}}$

$\sf \Longrightarrow \dfrac{1 + \sqrt{2}}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf \Longrightarrow \dfrac{\sqrt{2} + (\sqrt{2})^2}{(\sqrt{2})^2}$

$\sf \Longrightarrow \dfrac{\sqrt{2} + 2}{2}$

Now, let's find the value of RHS:

$\sf \Longrightarrow \dfrac{sin^2A}{1 - sinA}$

$\sf \Longrightarrow \dfrac{sin^245^\circ}{1 - sin45^\circ}$

Trigonometric ratio used: sin45° = 1/√2

$\sf \Longrightarrow \dfrac{\bigg(\dfrac{1}{\sqrt{2}}\bigg)^2}{1 - \bigg(\dfrac{1}{\sqrt{2}}\bigg)}$

$\sf \Longrightarrow \dfrac{\dfrac{1}{2}}{\bigg(\dfrac{\sqrt{2} - 1}{\sqrt{2}}\bigg)}$

$\sf \Longrightarrow \dfrac{1}{2} \times \dfrac{\sqrt{2}}{\sqrt{2} - 1}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2(\sqrt{2} - 1)}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2}$

$\sf \Longrightarrow \dfrac{\sqrt{2}}{2\sqrt{2} - 2} \times \dfrac{2\sqrt{2} + 2}{2\sqrt{2} + 2}$

Identity used: (a + b)(a - b) = a² - b²

$\sf \Longrightarrow \dfrac{\sqrt{2}(2\sqrt{2} + 2)}{(2\sqrt{2})^2 - (2)^2}$

$\sf \Longrightarrow \dfrac{ \sqrt{2}(2\sqrt{2}) + \sqrt{2}(2)}{(4 \times 2) - 4}$

$\sf \Longrightarrow \dfrac{4 + 2\sqrt{2}}{8 - 4}$

$\sf \Longrightarrow \dfrac{2(2 + \sqrt{2})}{4}$

$\sf \Longrightarrow \dfrac{2 + \sqrt{2}}{2}$

LHS = RHS

Hence Proved.