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] Share with your friends Share 0 Shruti Tyagi answered this 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 0 View Full Answer