Please Help Me With This Trigonometry Question?

Question

Please Help Me With This Trigonometry Question

How do I prove the following identity:

(COS beta / 1 - TAN beta) + (SIN beta / 1 - COT beta) = SIN beta + COS beta

If you could include the steps as to how to get the proof, I would greatly appreciate it, thanks!

Answer 1

Hi,

(COS β / 1 - TAN β) + (SIN β / 1 - COT β) = SIN β + COS β

Replace TAN β with sin β/cos β and COT β with cos β/sin β.

(COS β / 1 - sin β/cos β) + (SIN β / 1 - cos β/sin β) = SIN β + COS β

...cos β................sin β
------------------.+.------------------
1-sin β/cos β....1-cos β/sin β

Multiply the first fraction by cos β in the numerator and denominator. This is it:

cos β*cos β
-----------------------------
cos β[1 - sin β/cos β]

Simplifying it, it becomes:

cos² β
-----------------
cos β - sin β


Then multiply the second fraction by sin β in the numerator and denominator.

sin β * sin β
---------------------------
sin β(1 - cos β/sin β)

Simplifying it, it becomes:

sin² β
-----------------
sin β - cos β

Multiplying this by -1 makes it change to:
......sin² β
- -----------------
....cos β - sin β

With these revised fractions, the left side is:

cos² β...................sin² β
-----------------..-..----------------
cos β - sin β.....cos β - sin β

With a common denominator these fractions combine into:

cos² β..-..sin² β
---------------------
cos β - sin β

The top can be factored into:

(cos β.-.sin β)( cos β.+.sin β)
---------------------------------------
cos β - sin β

Cancel out the cos β.-.sin β factor and you end up with:

cos β.+.sin β

This is the same as the expression on the right side. therefore the identity has been proved.

I hope this helps!! :-)

Answer 2

If you don`t mind, I will use x instead of beta.
LHS
cos x / (1 - sin x / cos x) + sin x / (1 - cos x / sin x)
= cos² x / (cos x - sin x) + sin² x / (sin x - cos x)
= cos² x / (cos x - sin x) - sin² x / (cos x - sin x)
= 1 / cos x - sin x).[ (cos² x - sin² x) ]
= (cos x - sin x).(cos x + sin x) / (cos x - sin x)
= cos x + sin x
RHS
= cos x + sin x

Thus LHS = RHS

Answer 3

Mutiply the first term by cos(β)/cos(β) and the second term by sin(β)/sin(β). Then simplify, combine, and simplify more.

cos(β) / (1 - tan(β)) + sin(β) / (1-cot(β)) =
cos^2 (β) / (cos(β) - cos(β)tan(β)) + sin(β) / (1-cot(β)) =
cos^2 (β) / (cos(β) - sin(β)) + sin(β) / (1-cot(β)) =
cos^2 (β) / (cos(β) - sin(β)) + sin^2 (β) / (sin(β)-sin(β)cot(β)) =
cos^2 (β) / (cos(β) - sin(β)) + sin^2 (β) / (sin(β)-cos(β)) =
cos^2 (β) / (cos(β) - sin(β)) - sin^2 (β) / (cos(β)-sin(β)) =
[ cos^2 (β) - sin^2 (β) ] / (cos(β)-sin(β)) =
(cos(β)+sin(β))(cos(β)-sin(β)) / (cos(β)-sin(β)) =
cos(β)+sin(β)