IIT JEE , AIEEE Forum - Mathematics , Trignometry

utkarshsingh_91

_[r=1]

^[7]tan² (r *

/16 ) = 35
Plz prove it..
this ques is from fiitjee rankers material

utkarshsingh_91

i am waiting for the answer...!!!

utkarshsingh_91

common every1 ...!!! rates assured for this one...

aishwarykgupta

u sure its 35??....coz i'm getting 7....plz cross-check nd tell....else i'll verify

utkarshsingh_91

it is 35 only as the answer

sachin_gupta1991

As

/16 + 7

/16=

/2

tan ² 7

/16 = cot²pi/16

The required sum reduces to

(tan²pi/16 + cot²pi/16) + (tan²2pi/16+cot²2pi/16)......+ 1 {because tan²4pi/16=1}

= (tan pi/16 + cot pi/16)²+(tan2pi/16+cot2pi/16)²+.............+1-2-2-2

= 4 4

------------------------- + ----------------------------- + .............-5

4sin²pi/16cos²pi/16 4sin²2pi/16cos²2pi/16

= 4 4 4

------------ + ----------------+ --------------------- - 5

sin²pi/8 sin²pi/4 sin²3pi/8

= 4{ 1 1 } + 8 - 5

------------ + -------------

sin²pi/8 sin²3pi/8

On solving this we get

32{sin²pi/8 + sin²3pi/8} + 8 - 5

As sin 3pi/8=cospi/8

sin²pi/8 + sin²3pi/8=1

Therefore

= 32 + 8 - 5

=35

Hence Proved.

Nudge me for any doubt.

aishwarykgupta

nice job....i had deduced the relation that pi/16 + 7pi/16= pi/2
tan^2 (7/16) = cot^2 (pi/16) ....but i was stuck after that...gr8 job dude...