Question

prove that cot3theta.sin3theta/cos theta + sin theta whole square + tan3 theta.cos 3 theta/cos theta + sin theta whole square = sec theta cosec theta - 1/cosec theta + sec theta

Answer 1

Tan^3A / Sec^2A + Cot^3A / Cosec^2A

= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)

= Sin^3A/CosA + Cos^3A/SinA

= (Sin^4A + Cos^4A) / SinA.CosA

= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA

= ( 1- 2Sin^A.Cos^A)/ SinA.CosA

RHS = SecA CosecA - 2sinAcosA

= 1/CosA . 1/SinA - 2SinACosA

= (1 - Sin^2A.Cos^2A) / sinAcosA

Hence LHS = RHS (PROVED)

Answer 2

Answer:

★AnSwEr:−

ᴛᴀɴ^3ᴀ / sᴇᴄ^2ᴀ + ᴄᴏᴛ^3ᴀ / ᴄᴏsᴇᴄ^2ᴀ

= (sɪɴ^3ᴀ/ᴄᴏs^3ᴀ) / (1 / ᴄᴏs^2ᴀ) + (ᴄᴏs^3ᴀ/sɪɴ^3ᴀ) / (1 / sɪɴ^2ᴀ)

= sɪɴ^3ᴀ/ᴄᴏsᴀ + ᴄᴏs^3ᴀ/sɪɴᴀ

= (sɪɴ^4ᴀ + ᴄᴏs^4ᴀ) / sɪɴᴀ.ᴄᴏsᴀ

= [ (sɪɴ^2ᴀ + ᴄᴏs^2ᴀ)^2 - 2sɪɴ^2ᴀ.ᴄᴏs^2ᴀ] / sɪɴᴀ.ᴄᴏsᴀ

= ( 1- 2sɪɴ^ᴀ.ᴄᴏs^ᴀ)/ sɪɴᴀ.ᴄᴏsᴀ

ʀʜs = sᴇᴄᴀ ᴄᴏsᴇᴄᴀ - 2sɪɴᴀᴄᴏsᴀ

= 1/ᴄᴏsᴀ . 1/sɪɴᴀ - 2sɪɴᴀᴄᴏsᴀ

= (1 - sɪɴ^2ᴀ.ᴄᴏs^2ᴀ) / sɪɴᴀᴄᴏsᴀ

ʜᴇɴᴄᴇ ,ʟʜs = ʀʜs (ᴘʀᴏᴠᴇᴅ)

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