Question

Answer the above question​

Answer 1

LHS=(cot^3∅.sin^3∅+tan^3∅.cos^3∅)/cos^2∅+sin^∅+2sin∅cos∅)

=(cot^3∅.sin^3∅+tan^3∅.cos^3∅)/(1+sin∅cos∅)

now,convert cot∅ and tan∅ in terms of cos and sin

={(cos^3∅/sin^3∅)×sin^3∅ +( sin^3∅/cos^3∅)×cos^3∅}/(1+ sin∅cos∅)

=(cos^3∅+sin^3∅)/(1+sin∅cos∅)

a3+b3=(a+b)(a^2+b^2-ab)

also,1+sin∅cos∅=(cos∅+sin∅)^2

(cos∅+sin∅)(cos^2∅+sin^∅-cos∅sin∅)/(cos∅+sin∅)^2

=(1-cos∅sin∅)/(sin∅+cos∅)

=(1-1/sec∅cosec∅)/{(sec∅+cosec∅)/sec∅cosec∅

sec∅cosec∅ will be cancelled out from both numerator. and denominator

we get

LHS=(sec∅cosec∅-1)/(sec∅+cosec∅)=RHS

hence proved