Question

if Cosecø - Sinø = A³ , Secø-Cosø = b³ prove that a²b²(a²+b²)=1 ..... i need the answer in copy and specially in the step how Cos4/3ø×Cos2/3ø+Sin4/3ø×Sin2/3ø = Cos²ø+Sin²ø... i need a step by step cancellation process don't copy it from google its an request
;-) ​

Answer 1

Answer:

Step-by-step explanation:

cosec∅-sin∅=a³

(1/sin∅)-sin∅=a³

(1-sin²∅)/sin∅=a³

cos²∅/sin∅=a³

cos^(2/3)∅/sin^(1/3)∅=a

whole squaring both sides, we get

cos^(4/3)∅/sin^(2/3)∅=a²

sec∅-cos∅=b²

(1/cos∅)-cos∅=b³

(1-cos²∅)/cos∅=b³

sin²∅/cos∅=b³

sin^(2/3)∅/cos^(1/3)∅=b

whole squaring both sides, we get

sin^(4/3)∅/cos^(2/3)∅=b²

now,

a²b²= (cos^(4/3)∅/sin^(2/3)∅).(sin^(4/3)∅/cos^(2/3)∅)

      = cos^(2/3)∅.sin^(2/3)∅

a²+b² = cos^(4/3)∅/sin^(2/3)∅  +   sin^(4/3)∅/cos^(2/3)∅

        take LCM

          = (cos^(4/3+2/3)∅ + sin^(4/3+2/3)∅)/(sin^(2/3)∅)(cos^(2/3)∅)

          = (cos²∅+sin²∅)/ (sin^(2/3)∅)(cos^(2/3)∅)

          = 1/(sin^(2/3)∅)(cos^(2/3)∅)

a²b²(a²+b²) = (sin^(2/3)∅)(cos^(2/3)∅) ×  1/(sin^(2/3)∅)(cos^(2/3)∅)

                  =1

hence proved