Question

tanA/secA-1=secA+1/tanA​

Answer 1

Answer:

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Step-by-step explanation:

tanA + secA - 1 / tanA - secA + 1 = 1 + sinA / cosA. LHS = tanA + secA -1 / tanA - secA + 1. = tanA + secA + tan2A - sec2A / tanA - secA + 1. as tan2A + 1 = sec2A.

Answer 2

Step-by-step explanation:

tan x /( sec x - 1) = (sec x + 1) / tan x

LHS: multiply both numerator and the denominator by the conjugate of the denominator

sec x + 1

=> [tan x(sec x+1)] / [(sec x+ 1) (sec x-1)]

=> tan x (sec x- 1) / sec²x - 1

=> tan x (sec x+1) / tan²x [ sec²x - 1 = tan²x ]

=> (sec x + 1) / tan x

[tan x cancels out in the numerator ]