Question

tan[π/4+A]+tan[π/4-A] divide by tan[π/4+A]-tan[π/4-A] =cosec2A​

Answer 1

Answer:

Step-by-step explanation:

tan[π/4+A]+tan[π/4-A]/tan[π/4+A]-tan[π/4-A]

//Solve numerator and denominator separately

tan[π/4+A]+tan[π/4-A]

//Tan (A+B) = TanA+ TanB / 1- TanATanB;

//Tan (A-B) = TanA - TanB / 1 + TanATanB

=> Tan(π/4) + TanA/ 1 - Tan{π/4}TanA + Tan(π/4) -  TanA/ 1 + Tan{π/4}TanA

=> 1 + TanA/ 1 - TanA + 1 - TanA/1+TanA

= 2(1+Tan²A) / 1 - Tan²A  

//Now for denominator

tan[π/4+A]-tan[π/4-A]

=>  Tan(π/4) + TanA/ 1 - Tan{π/4}TanA -  Tan(π/4) -  TanA/ 1 + Tan{π/4}TanA

=> 1 + TanA/ 1 - TanA -  1 - TanA/1+TanA

= 2(2TanA) / 1 - Tan²A

//Substitute back the values in main equation

2(1+Tan²A) / 1 - Tan²A /  2(2TanA) / 1 - Tan²A

= 2Sec²A /4TanA

=1/2sincosa

= 1/Sin2A

= Coesc2a

= R.H.S

Hence Proved

Answer 2

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

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

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

$in \: conjuction \: with \ratio$

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

$now$

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

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

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

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

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

$\boxed{= - 1}$