Math, asked by sejal4855, 11 months ago

pls solve this que fast (82.que)

Attachments:

Answers

Answered by arjun6068
1

HERE IS UR ANSWER

⬇️⬇️⬇️⬇️⬇️

LHS

= (tanA + cosecB)^2 - (cotB-secA)^2

= (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A)

= tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step:

(sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

= 2tanAcosecB + 2cotBsecA

= 2(tanAcosecB + cotBsecA)

RHS

= 2tanAcotB(cosecA + secB)

= 2tanAcotB((1 / sinA) + (1 / cosB))

= 2(tanA / sinA)cotB + 2tanA(cot B / cosB)

= 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB))

= 2(1 / cosA)cotB + 2 tanA(1 / sinB)

= 2secAcotB + 2tanAcosecB

= 2(secAcotB + tanAcosecB)

ꃅꂦᖘꍟ ꀤ꓄ ꃅꍟ꒒ᖘꌗ ꌩꂦꀎ 


sejal4855: i did not understand RHS explaination
sejal4855: hey
Similar questions