Question

tan(π/4+¢) + tan(π/4- ¢) = 2sec¢

Answer 1

A/q
=sin(pi/4+x/2) cos (pi/4+x/2)+sin(pi/4-x/2)/cos(pi/4-x/2)
=sin(pi/4+x/2)cos(pi4 - x/2) + sin (pi/4 - x/2) cos (pi/4 + x/2) ] / [cos (pi/4 - x/2)cos (pi/4 + x/2)]

= [ sin (pi/4 + x/2 + pi/4 - x/2) ] / [ cos (pi/4 - x/2) sin (pi/4 - x/2) ]

= 2 sin pi/2 / (2 cos (pi/4 - x/2) sin (pi/4 - x/2))

= 2 / sin (pi/2 - x) 

= 2 / cos x

= 2Secx

Answer 2

Hey !!!

tan(π/4+¢) + tan(π/4-¢)

= tanπ/4+tan¢. .. tanπ/4-tan¢
------------------- + ------------------
1 - tanπ/4×tan¢ 1 + tanπ/4×tan¢

= 1 + tan¢ ....... 1 -tan¢
--------------- + ----------------
1 - tan¢ . .. 1 + tan¢

=( 1 +tan¢)² + ( 1 - tan¢)²
---------------------------------
(1 - tan¢ )( 1 + tan¢ )

= (1 + tan²¢ + 2tan¢ )+( 1 + tan²¢ - 2tan¢)
-----------------------------------------------------
1 - tan²¢

= 2( 1 + tan²¢).
---------------------
1 - tan²¢

•°• 1 + tan²¢ /1 - tan²¢ = cos2¢

= 2/cos2¢ = 2sec¢2¢

⚠ ur question is little bit wrong check it out .

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Hope it helps you !!!

@Rajukumar111