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Answer 1

If the question istan2θ+csc2θtan2θ+csc2θtan2⁡θ+csc2⁡θtan2⁡θ+csc2⁡θ,then the solution will be2−(32)2+32=72−34=14−34=1142−(32)2+32=72−34=14−34=114If the question istan2θ−csc2θtan2θ+csc2θtan2⁡θ−csc2⁡θtan2⁡θ+csc2⁡θ,then the solution is2−322+32=4−324+42=172−322+32=4−324+42=17296 ViewsUpvoteShare RecommendedAll
Amit Singh, studied at Army Public School, LucknowAnswered May 13, 2017Let x=thetaSec x=(3) ^(1/2)Sec x= (H) /BTherfore H=(3) ^(1/2)B=1Using Pythagoras H^2=B^2+P^2P=(2) ^(1/2)Tan x=P/B=(2) ^(1/2)Cosec x=H/P=(3/2) ^(1/2)[tan^2 x-cosec^2 x]/[tan^2+cosec^2]=[(2) -(3/2) ] /[(2) +(3/2) ]=(1/3)/(7/3)=1/7 is The Answer!=1/7