Question

Can someone PLEASE help me solve this equation ? due soon

Answer 1

$\sf{Given : 3tanx + 7 = \dfrac{2}{(1 - sinx)(1 + sinx)}}$

We know that : (a - b)(a + b) = a² - b²

$\implies \sf{3tanx + 7 = \dfrac{2}{1 - sin^2x}}$

We know that : 1 - sin²x = cos²x

$\implies \sf{3tanx + 7 = \dfrac{2}{cos^2x}}$

$\sf{\bigstar \ \ We \ know \ that : \boxed{\sf{\dfrac{1}{cos^2x} = sec^2x}}}$

$\implies \sf{3tanx + 7 = 2sec^2x}$

We know that : sec²x = 1 + tan²x

$\implies \sf{3tanx + 7 =2(1 + tan^2x)}$

$\implies \sf{2 + 2tan^2x - 3 tanx - 7 = 0}$

$\implies \sf{2tan^2x - 3 tanx - 5 = 0}$

$\implies \sf{2tan^2x - 5tanx + 2tanx - 5 = 0}$

$\implies \sf{2tanx(tanx + 1) - 5(tanx + 1) = 0}$

$\implies \sf{(tanx + 1)(2tanx - 5) = 0}$

$\implies \sf{tanx = -1 \ (or) \ tanx = \dfrac{5}{2} }$