Question

if (a^2 - b^2 )sinA + 2ab. cosA = a^2 + b^2 ,find the value of tanA ​

Answer 1

Answer

$\sf\frac{a^2 - b^2}{2ab}$

Solution

Given

(a² - b²)sinA + 2ab·cosA = a² + b²

We can write it as

2ab·cosA = a² + b² - (a² - b²)sinA

Squaring both sides, we get

(2ab·cosA)² = (a² + b²)² + (a² - b²)²sin²A - 2(a² + b²)(a² - b²)sinA

Using, sin²A = 1 - cos²A,

→ 4a²b²·cos²A = a⁴ + b⁴ + 2a²b² + (a² - b²)²(1 - cos²A) - 2(a² + b²)(a² - b²)sinA

→ 4a²b²·cos²A = (a² + b²)² + (a² - b²)²- (a² - b²)²cos²A - 2(a² + b²)(a² - b²)sinA

Transporting 4a²b²·cos²A on other side

→ (a² + b²)² + (a² - b²)² - a⁴cos²A - b⁴cos²A + 2a²b²·cos²A - 4a²b²·cos² - 2(a⁴ - b⁴)sinA = 0

It becomes after simplifying

→ (a² + b²)² + (a² - b²)² - (a⁴cos²A + b⁴cos²A + 2a²b²·cos²A) - 2(a⁴ - b⁴)sinA = 0

Taking out negative cos²A common

→ (a² + b²)² + (a² - b²)² - (cos²A)(a² + b²)² - 2(a⁴ - b⁴)sinA = 0

Taking out (a² + b²)² common

→ (a² + b²)²(1 - cos²A) + (a² - b²)² - 2(a² + b²)(a² - b²)sinA = 0

Using (1 - cos²A) = sin²A

→ (a² + b²)²(sin²A) + (a² - b²)² - 2(a² + b²)(a² - b²)sinA = 0

Using a² + b² - 2ab = (a - b)²

→ [(a² + b²)sinA - (a² - b²)]² = 0

→ (a² + b²)sinA - a² + b² = 0

→ sinA = (a² - b²)/(a² + b²)

Now, we know that cos²A = 1 - sin²A

So, cos²A = $\sf 1 - (\frac{a^2 - b^2}{a^2 + b^2})^2$

It becomes, cos²A = $\sf \frac{(a^2 + b^2)^2 - (a^2 - b^2)^2}{(a^2 + b^2)^2}$

Using a² - b² = (a + b)(a - b),

cos²A = $\sf\frac{(a^2 + b^2 - a^2 + b^2)(a^2 + b^2 + a^2 - b^2)}{(a^2 + b^2)^2}$

cos²A = $\sf\frac{4a^2b^2}{(a^2 + b^2)^2}$

Taking under root on both sides,

cosA = $\sf\frac{2ab}{a^2 + b^2}$

We know that tanA = sinA/cosA

So, tanA = $\sf\frac{a^2 - b^2}{a^2 + b^2} \times \frac{a^2 + b^2}{2ab}$

TanA = $\sf\frac{a^2 - b^2}{2ab}$