Question

Solve : tan^-1(x-5/x-6)+tan^-1(x+5/x+6)=pi/4

Answer 1

Answer:

± 7 / √2

Explanation:

tan⁻ ¹ ( x - 5 / x - 6 ) + tan⁻ ¹ ( x + 5 / x + 6 ) = π / 4

Using tan⁻ ¹ a + tan⁻ ¹ b = tan⁻ ¹ ( a + b / 1 - ab ) we get

$\Rightarrow \sf tan^{ - 1} \dfrac{ \dfrac{x - 5}{x - 6} + \dfrac{x + 5}{x + 6} }{1 - \bigg(\dfrac{x + 5}{x + 6} \bigg) \bigg( \dfrac{x - 5}{x - 6} \bigg)} = \dfrac{\pi}{4}$

$\Rightarrow \sf tan^{ - 1} \bigg( \dfrac{2x^2 - 60}{-11} \bigg) = \dfrac{\pi}{4}$

Taking tan on both sides

$\Rightarrow \sf \dfrac{2x^2 - 60 }{-11} = tan\dfrac{\pi}{4}$

⇒ ( 2x² - 60 ) / - 11 = 1

⇒ 2x² - 60 = - 11

⇒ 2x² = - 11 + 60

⇒ 2x² = 49

⇒ x² = 49 / 2

⇒ x = ± √49/2

⇒ x = ± √49 / √2

⇒ x = ± 7 / √2

Therefore the value of x is ± 7 / √2.