Question

8. tan (45° + x) - tan (45° - x) = 2 tan 2x.​

Answer 1

Answer:

To prove:

$tan(45^0+x) - tan(45^0 - x) = 2tan2x$

Solution:

$\implies tan(45^0 + x) - tan(45^0 - x)$

$\implies \dfrac{tan45^0 + tanx}{1 - tan45^0 tanx} - \dfrac{tan45^0 - tanx}{1+tan45^0 tanx}$

$\implies \dfrac{1+tanx}{-tanx} - \dfrac{1-tanx}{1+tanx}$

$\implies \dfrac{(1+tanx)^2 - (1-tanx^2)}{1-tan^2x}$

$\implies \dfrac{1 + tan^2x + 2tanx - 1 - tan^2x + 2tanx}{1-tan^2x}$

$\implies \dfrac{4tanx}{1-tan^2x}$

$\implies 2 \dfrac{2tanx}{1-tan^2x}$

$\boxed{\implies{\boxed{2tan2x = RHS}}}$

Identities used:

1️⃣ $tan(A+B) = \dfrac{tanA + tanB}{1-tanA tanB}$

2️⃣ $tan2 x = \dfrac{2tanx}{1 - tan^2x}$

Additional information:

$\begin{lgathered}\boxed{\begin{array}{l}\sf Fundamental \ trigonometric \ identities: \\</p><p>sin^2 \theta + cos^2 \theta = 1 \\ 1 + tan^2\theta = sec^2 \theta \\ 1 + cot^2 \theta = cosec^2\theta \\\end{array}}\end{lgathered}$