Question

If $tan(\frac{\pi }{4}+\theta )+tan(\frac{\pi }{4}-\theta )=a$ then $ta{n}^3(\frac{\pi }{4}+\theta )+ta{n}^3(\frac{\pi }{4}-\theta )$=

• A. $0$
• B. $a$
• C. $3a$
• D. $a^3-3a$

Answer 1

The correct option is D $a^3-3a$

${\tan }^3(\frac{\pi }{4}+\theta )+{\tan }^3(\frac{\pi }{4}-\theta )$ using $a^3+b^3=(a+b)(a^2+b^2-ab)$

$=(\tan (\frac{\pi }{4}+\theta )+\tan (\frac{\pi }{4}-\theta ))({\tan }^2(\frac{\pi }{4}+\theta )+{\tan }^2(\frac{\pi }{4}-\theta )-\tan (\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}-\theta ))$

$=a({(\tan (\frac{\pi }{4}+\theta )+\tan (\frac{\pi }{4}-\theta ))}^2-2\tan (\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}-\theta )-tan(\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}-\theta ))$

$=a(a^2-3tan(\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}-\theta ))$

$=a(a^2-3\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }\times \frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta })$

$=a(a^2-3\frac{1+\tan \theta }{1-\tan \theta }\times \frac{1-\tan \theta }{1+\tan \theta })$

$=a(a^2-3)$

$=a^3-3a$