Question

prove that tan(π/4+x)tan(3π/4+x)=-1

Answer 1

Use identity: tan()=cot(2−)

tan (π/4 + x) . tan (3 π/4 +x)
tan ( π/4+x) . cot ( π/2 ( π/4+x)
tan (π/4+x) . cot ( -π/4 - x )
tan ( π/4+x) . - cot (π/4-x)
= - 1

Answer 2

$\tan = ( \frac{ \pi}{4} + a) \times \tan( \frac{3 \pi + A}{4} )$

$using \: tangent \: sum \: of \: angle \: formula \ratio -$

$tan(A+B) = \frac{ \tan A+ \tan B}{1 - \tan A \times \tan B }$

$in \: conjuction \: with \ratio$

$\tan( \frac{ \tan \pi }{4} ) = 1 \: and \: \tan( \frac{ \tan 3\pi }{4} ) = - 1$

$now$

$\tan( \frac{\pi}{4} + A) \times \tan( \frac{3 \pi }{4} + A)$

$= \frac{ \tan (\frac{ \pi }{4} ) + \tan A}{1 - \tan( \frac{ \pi }{4}) \times \tan A }$

$= > \frac{ \tan (\frac{3 \pi }{4} ) + \tan A}{1 - \tan( \frac{ 3\pi }{4}) \times \tan A }$

$= \frac{1 + \tan A }{1 - \tan A} \times \frac{ - 1 + \tan A }{1 + \tan A}$

$= \frac{ - 1 + \tan A }{1 - \tan A} \times \frac{ 1 - \tan A }{1 + \tan A}$

$\boxed{= - 1}$

HENCE R.H.S = L.H.S