Please answer question 29 .Please don't send any links please.

29 .         E v a l u t e   w i t h o u t   u sin g   t r i g o n o m e t r i c   t a b l e s   : sin 2   θ   +   sin 2   ( 90   -   θ ) 3 ( s e c 2   61 °   -   c o t 2   29 ° )   -   3   c o t 2   30 °   sin 2   54 °   s e c 2   36 ° 2 ( cos e c 2   65 °   -   tan 2   25 ° )                                      [Foreign 2006]

Dear student,

We have,sin2θ+sin290-θ3sec261°-cot229°-3cot230°sin254°sec236°2cosec265°-tan225°=sin2θ+cos2θ3sec261°-cot290-61°-332sin254°sec290-54°2cosec265°-tan290-65°     --[sin90-θ=cosθ]=sin2θ+cos2θ3sec261°-tan261°-3×3×sin254°cosec254°2cosec265°-cot265°        --[sec90-θ=cosecθ, cot90-θ=tanθ]=13sec261°-tan261°-9×sin254°×cosec254°2×1         --[sin2θ+cos2θ=1 and cosec2θ-cot2θ=1]=13×1-9×sin254°×1sin254°2×1                                            --[sec2θ-tan2θ=1]=13-92=2-276=-256

Regards

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