Question

pls solve this with the easiest method known

Answer 1

taking 
tan x = sin x / cos x 
cot x = cos x / sin x 

we have 

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

Taking LCM and subtracting we get 

[ sin^3 x - cos^3 x ]/[(sin x)(cos x)(sin x - cos x)] 

now a^3 - b^3 = (a-b)^3 + 3ab(a-b) 

using the above expansion for (sin^3 x - cos^3 x) 
and cancelling out (sin x - cos x) from all the terms we have 

[ ( sin x - cos x)^2 + 3(sin x)(cos x) ] / [ (sin x)(cos x) ] 

(a-b)^2 = a^2 + b^2 - 2ab 

so ( sin x - cos x)^2 = 1 - 2(sin x)(cos x) 

since sin^2 x + cos^2 x = 1 

so we have 

[ 1 + (sin x)(cos x) ] / [ (sin x)(cos x) ] 

= [(cosec x)(sec x)] + 1

Answer 2

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