Question

tan+Sec-1 by tan-Sec+1=1+Sin by Cos

Answer 1

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Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$\frac{tanA+secA-1}{tanA-secA+1} =\frac{1+sinA}{cosA}$

LHS:

$\frac{tanA+secA-1}{tanA-secA+1}$

$=\frac{tanA+secA-(sec^{2}A-tan^{2}A)  }{tanA-secA+1}$  ( ∵$sec^{2}A-tan^{2}A=1$)

$=\frac{(tanA+secA)-(secA-tanA) (secA+tanA) }{tanA-secA+1}$

$=\frac{(tanA+secA)(1-secA+tanA)}{tanA-secA+1}$

$=tanA+secA$

$=\frac{sinA}{cosA} +\frac{1}{cosA}$

$=\frac{1+sinA}{cosA}$

$=<strong>RHS</strong>$

Hence Proved.