Question

tan x/1-cot x +cot x/1-tan x =1+sec x.cosec x​

Answer 1

LHS;

tanx / 1-cotx =

• sinx / cosx × sinx / (sinx - cosx). (as tanA = sinA/cosA) (as cotA = cosA/sinA)
• sin²x / cosx (sinx - cosx)

cotx / 1-tanx =

• cosx / sinx × cosx / (cosx - sinx). (as tanA = sinA/cosA) (as cotA = cosA/sinA)
• cos²x / sinx (cosx - sinx)

Thus RHS would be ;

• sin²x / cosx (sinx - cosx) + cos²x / sinx (cosx - sinx)
• sin²x / cosx (sinx - cosx) - cos²x / sinx (-cosx + sinx)
• sin³x / cosx × sinx (sinx - cosx) - cos³x / cosx × sinx (sinx - cosx)
• sin³x - cos³x / cosx × sinx (sinx - cosx)
• (sinx - cosx) ( sin²x + cos²x + sinx•cosx ) / cosx•sinx (sinx - cosx)
• ( sin²x + cos²x + sinx•cosx ) / cosx•sinx
• 1 + sinx•cosx / sinx•cosx. (as sin²A + cos²A = 1)
• 1/ sinx•cosx + sinx•cosx / sinx•cosx
• secx•cosecx + 1 (as sinA = 1/cosecA). (as cosA = 1/secA)
• = RHS

hence proved

#answerwithquality

#BAL