Math, asked by kings07, 11 months ago

Cot^3theta. Sin^3 theta/(cos theta+sin theta)^2+tan^3theta .cos^3 theta/(cos theta+sin theta ) ^2= sec theta. Cosec theta-1/cosec theta +sec theta

Answers

Answered by XxMissPaglixX

${\huge{\pink{\underline{\underline{\red{\mathbb{\bigstar{\red{AnSwEr:-}}}}}}}}}$

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)

(¯`◕‿◕´¯) ᴛʜᴀɴᴋ ʏᴏᴜ (¯`◕‿◕´¯)

Answered by Anonymous

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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