Question

√(secA-1)/(secA+1) + √(secA+1)/(secA-1) = 2cosecA

Answer 1

Refer attachment for your answer

Answer 2

To Prove:

$\sf \sqrt{\dfrac{secA - 1}{secA + 1}} \ + \ \sqrt{\dfrac{secA + 1}{secA - 1}} = 2cosecA$

Solution:

Taking the LHS we get:

$\Longrightarrow \sf \sqrt{\dfrac{secA - 1}{secA + 1}} \ + \ \sqrt{\dfrac{secA + 1}{secA - 1}}$

Rationalizing the denominator we get;

$\Longrightarrow \sf \sqrt{\dfrac{secA - 1}{secA + 1}} \times \sqrt{\dfrac{secA - 1}{secA - 1}} + \ \sqrt{\dfrac{secA + 1}{secA - 1}} \times \sqrt{\dfrac{secA + 1}{secA + 1}}$

$\Longrightarrow \sf \sqrt{\dfrac{secA - 1}{secA + 1} \times \dfrac{secA - 1}{secA - 1}} + \ \sqrt{\dfrac{secA + 1}{secA - 1} \times \dfrac{secA + 1}{secA + 1}}$

Using the below algebraic identities we get:

• (a - b)(a - b) = (a - b)²

• (a + b)(a + b) = (a + b)²

• (a + b)(a - b) = a² - b²

$\Longrightarrow \sf \sqrt{\dfrac{\big(secA - 1\big)^2}{\big(secA\big)^2 - \big(1\big)^2}} + \ \sqrt{\dfrac{\big(secA + 1\big)^2}{\big(secA\big)^2 - \big(1\big)^2}}$

Using sec²A - 1 = tan²A we get;

$\Longrightarrow \sf \sqrt{\dfrac{\big(secA - 1\big)^2}{tan^2A}} + \ \sqrt{\dfrac{\big(secA + 1\big)^2}{tan^2A}}$

Roots and squares get cancelled.

$\Longrightarrow \ \sf \dfrac{secA - 1}{tanA} + \dfrac{secA + 1}{tanA}$

$\Longrightarrow \ \sf \dfrac{secA - 1 + secA + 1}{tanA}$

$\Longrightarrow \ \sf \dfrac{2secA}{tanA}$

We know that;

• secA = 1/cosA
• tanA = sinA/cosA

$\Longrightarrow \ \sf \dfrac{2 \times \bigg[\dfrac{1}{cosA}\bigg]}{\bigg[\dfrac{sinA}{cosA}\bigg]}$

$\Longrightarrow \ \sf \dfrac{2}{sinA}$

We know that;

• 1/sinA = cosecA

$\Longrightarrow \ \sf 2 cosecA$

LHS = RHS

Hence proved.