Question

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Answer 1

To prove:

$\Longrightarrow \sf \ \dfrac{2 - cosec^2A}{cosec^2A + 2cotA} = \dfrac{sinA - cosA}{sinA + cosA}$

Solution:

Considering the LHS of the given equation we get;

$\Longrightarrow \sf \ \dfrac{2 - cosec^2A}{cosec^2A + 2cotA}$

The RHS of the given equation contains sins and cosA in it. Therefore, we'll try to express the LHS in terms of sinA and cosA as well.

For the same purpose, we'll substitute the following:

⇒ cosec²A = 1/sin²A

⇒ cotA = cosA/sinA

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

Taking LCM in the numerator we get;

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

Taking LCM in the denominator we get;

$\Longrightarrow \sf \ \dfrac{\bigg\{\dfrac{2sin^2A - 1}{sin^2A}\bigg\}}{\bigg\{\dfrac{1 + 2cosAsinA}{sin^2A}\bigg\}}$

$\Longrightarrow \sf \ \bigg\{\dfrac{2sin^2A - 1}{sin^2A}\bigg\} \times \bigg\{\dfrac{sin^2A}{1 + 2cosAsinA}\bigg\}$

On cancelling sin²A we get;

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

Substitute 1 = sin²A + cos²A in the denominator and numerator.

$\Longrightarrow \sf \ \dfrac{2sin^2A - \Big\{sin^2A + cos^2A\Big\}}{\Big\{sin^2A + cos^2A\Big\} + 2cosAsinA}$

$\Longrightarrow \sf \ \dfrac{2sin^2A - sin^2A - cos^2A}{sin^2A + cos^2A + 2cosAsinA}$

The denominator is of the form a² + b² + 2ab = (a + b)². Therefore, we can substitute it with (sinA + cosA)².

[a = sinA; b = cosA]

$\Longrightarrow \sf \ \dfrac{sin^2A - cos^2A}{\big\{sinA + cos\big\}^2}$

The numerator is of the form a² - b² = (a + b)(a - b), and therefore can be written in the form of (a + b)(a - b).

The denominator is of the form (a + b)² = (a + b)(a + b), and therefore can be written in the form of (a + b)(a + b).

$\Longrightarrow \sf \ \dfrac{\big\{sinA + cosA\big\}\big\{sinA - cosA\big\}}{\big\{sinA + cos\big\}\big\{sinA + cosA\big\}}$

Cancelling (sinA + cosA) in both the numerator and denominator we get:

$\Longrightarrow \sf \ \dfrac{sinA - cosA}{sinA + cosA}$

LHS = RHS

Hence proved.

Answer 2

Hope it helps..

More to know ,

• Sin² A + Cos²A = 1

• Sec²A - Tan²A = 1

• Cos²A - 1 = Cosec²A

• Sin (90 - A) = Cos A
• Cos (90 - A) = Sin A

• Tan (90 - A) = Cot A
• Cot (90 - A) = Tan A

• Cosec (90 - A) = Sec A
• Sec (90 - A) = Cosec A

Sin A = 1/Cosec A

CosA = 1/Sec A

TanA = 1/Cot A

Cosec A = 1/Sin A

Sec A = 1/Cos A

Cot A = 1/ Tan A

Tan A = SinA/Cos A

Cot A = CosA/Tan A

Hope it helps!