Question

Simplify tan($\frac{\pi}{4} + \theta$) . tan($\frac{\pi}{4} - \theta$)

Answer 1

$tan(\pi/4-\theta) tan(\pi/4-\theta)$

we know, tan(A - B) = (tanA - tanB)/(1 + tanA.tanB)

so, $tan(\pi/4-\theta)=\frac{tan\pi/4-tan\theta}{1+tan\pi/4tan\theta}$

$\because tan\frac{\pi}{4}=1$

$\therefore tan(\pi/4-\theta)=\frac{tan\pi/4-tan\theta}{1+tan\pi/4tan\theta}\\\\=\frac{1-tan\theta}{1+tan\theta}$

similarly, $tan(π/4+\theta)=\frac{1+tan\theta}{1-tan\theta}$

now, $tan(\pi/4-\theta) tan(\pi/4-\theta)=\frac{1-tan\theta}{1+tan\theta}\frac{1+tan\theta}{1-tan\theta}$

= $\frac{(1-tan\theta)(1+tan\theta)}{(1+tan\theta)(1-tan\theta)}$

= $\frac{1-tan^2\theta}{1+tan^2\theta}$

we know, cos2x = (1 - tan²x)/(1 + tan²x)

so, $\frac{1-tan^2\theta}{1+tan^2\theta}=cos2\theta$

hence, required answer is $cos2\theta$

Answer 2

Answer:

1

Step-by-step explanation:

Formula used:

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

$tan(\frac{\pi}{4}+\theta)\:tan(\frac{\pi}{4}+\theta)$

$=\frac{tan(\frac{\pi}{4})+tan\theta}{1-tan(\frac{\pi}{4})tan\theta}\:.\frac{tan(\frac{\pi}{4})-tan\theta}{1+tan(\frac{\pi}{4})tan\theta}$

$=\frac{1+tan\theta}{1-tan\theta}\:\frac{1-tan\theta}{1+tan\theta}=1$