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Prove that 1/sec x - tan x - 1/cos x =1/cos x - 1/sec x+ tan x |
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Answer» To Prove: [1/(secx – tanx)] – [1/cosx] = [1/cosx] - [1/(secx + tanx)]Proof: LHS = [1/(secx – tanx)] – [1/cosx]But [1/cosx] = sec x = [1 ×(secx + tanx)/ (secx - tanx)(secx + tanx)] – [sec x] = [(sec x + tan x) / (sec2x – tan2x)] – [sec x]But (sec2x – tan2x) = 1LHS = sec x + tan x – [sec x] = tan x ………. (1)RHS = [1/cosx] - [1/(secx + tanx)] = [sec x] - [1 ×(secx - tanx)/ (secx + tanx)(secx - tanx)] = [sec x] - [(sec x - tan x) / (sec2x – tan2x)]But (sec2x – tan2x) = 1RHS = [sec x] - [(sec x - tan x)] = tan x …………… (2)As LHS = RHS , Hence Proved So esy bt likhu kaise |
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