Help with trigonometry !?

I am going to take a guess that the left side is: (tanA) / (secA - 1), and the right side is (tanA + secA + 1) / (secA - 1 + tanA).

And it looks like maybe I was right. I tried several test values for A, and the two sides do equal.

Next, what do we have in terms of trig identities. I include a link to a handy one, below.

So what about the one tan²A + 1 = sec²A ?

Rearrange this to be tan²A = sec²A - 1.

So this can factor out to tan²A = sec²A - 1 = (secA + 1)(secA - 1). Maybe that can help.

please make ur ques more clear!!

Would like to help but , unfortunately , presentation is unclear and open to doubt.

........... tanA

LHS = ---------

.......... secA-1

mulriply top & bottom by secA+1

.. tanA.(secA+1)

= ---------------------

....... sec²A-1

.....tanA(1+cosA)/cosA

= -----------------------------

............ tan²A

.... 1+cosA

= -------------

..... sinA

..........tanA + secA + 1

RHS =---------------------

.......... tanA + secA - 1

multiply top & bottom by cosA

.... sinA + 1 + cosA

= ------------------------

.... sinA +1 - cosA

multiply top & bottom by (sinA + 1 + cosA)

... sin²A + 1 + 2sinA + cos²A + 2sinA.cosA +2cosA

= ---------------------------------------------------------------

............. sin²A + 1 + 2sinA - cos²A

.... 2 + 2sinA + 2sinAcosA + 2cosA

= -------------------------------------------

............. 2sinA(1+sinA)

...... 2(1+cosA) +2sinA(1+cosA)

= --------------------------------------------

.... 2sinA(1+cosA)

......(1+cosA)

= ------------------

...... sinA

so proved LHS = RHS

what have u written..canniot even understand use brackets