How do you prove # tan^2x / (secx - 1) = secx + 1 #?

2 Answers
Karthik G
Jan 2, 2016

Hint : #tan^2(x) = sec^2(x)-1# and the difference of square rule to prove it. Step by step working is shown below.

Explanation:

To prove #tan^2(x)/(sec(x)-1) = sec(x)+1#

Use the identity #1+tan^2(x) = sec^2(x)#

We can rewrite this as

#tan^2(x) = sec^2(x)-1#

Now back to our problem

LHS
#=tan^2(x)/(sec(x)-1)#

#=(sec^2(x)-1)/(sec(x)-1)#

Recall the difference of square rule

#a^2-b^2 = (a-b)(a+b)#

We need to apply that for #sec^2(x) -1#

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

#=(cancel(sec(x)-1)(sec(x)+1))/cancel(sec(x)-1)#

#=sec(x)+1 = #RHS

Therefore, LHS = RHS thus proved.

Johnson Z.
Jan 2, 2016

Start by deciding on the more difficult side to work on. In this case, it's the left side. Recall the Pythagorean trigonometric identity, #color(red)(tan^2x)=color(green)(sec^2x-1)#. Using this identity, replace #color(red)(tan^2x)# in the equation with #color(green)(sec^2x-1)#.

Left side:

#color(red)(tan^2x)/(secx-1)#

#(color(green)(sec^2x-1))/(secx-1)#

Since "#sec^2x-1#" is a difference of squares, it can be broken down into #color(orange)(secx+1)# and #color(blue)(secx-1)#.

#((color(orange)(secx+1))(color(blue)(secx-1)))/(secx-1)#

You will notice that "#secx-1#" appears both in the numerator and denominator, so they can both be cancelled out.

#((secx+1)color(red)cancelcolor(black)((secx-1)))/color(red)cancelcolor(black)((secx-1))#

#secx+1#

#:.#, LS#=#RS.

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