Question

prove that 1/sec x + tan x = sec x - tan x

Answer 1

Answer:

$\frac{1}{secx+tanx}=secx-tanx$

Step-by-step explanation:

$LHS=\frac{1}{secx+tanx}\\=\frac{secx-tanx}{(secx+tanx)(secx-tanx)}\\=\frac{secx-tanx}{sec^{2}x-tan^{2}x}$

/* By Trigonometric identity:

sec²x-tan²x=1 */

$=\frac{secx-tanx}{1}\\=secx-tanx=RHS$

Therefore,

$\frac{1}{secx+tanx}=secx-tanx$

•••♪

Answer 2

Answer:

Multiply the left hand side by secx - tanx to obtain (secx - tanx) / (sec2x - tan2x)

Now, recall that 1 + tan2x = sec2x.  So, sec2x - tan2x = 1.

So, the left hand side simplifies to (secx - tanx) / 1 = secx - tanx = right hand side

Step-by-step explanation: