Trigonometric Equations

Given the problem:
tan‌A+tan‌B+cot‌A+cot‌B=tan‌A‌tan‌B−cot‌A‌cot‌B
We can express the trigonometric functions in terms of sine and cosine. That gives:
‌

sin‌A

cos‌A

+‌

sin‌B

cos‌B

+‌

cos‌A

sin‌A

+‌

cos‌B

sin‌B

=‌

sin‌A

cos‌A

⋅‌

sin‌B

cos‌B

−‌

cos‌A

sin‌A

⋅‌

cos‌B

sin‌B

Simplifying each term, we get:
‌

sin‌2A+cos2A

sin‌A‌cos‌A

+‌

sin‌2B+cos2B

sin‌B‌cos‌B

=‌

sin‌2Asin‌2B−cos2Acos2B

cos‌A‌cos‌B‌s‌i‌n‌Asin‌B

Since sin‌2θ+cos2θ=1, this reduces to:
‌‌

1

sin‌A‌cos‌A

+‌

1

sin‌B‌cos‌B

=‌

sin‌2Asin‌2B−(1−sin‌2A)(1−sin‌2B)

cos‌A‌cos‌B‌s‌i‌n‌Asin‌B

‌⇒‌

1

2

(sin‌2A+sin‌2B)=sin‌2A+sin‌2B−1
Using the sine addition and subtraction formulas, sin‌(2x)=2sin‌x‌cos‌x, transform the equation:
sin‌(A+B)‌cos(A−B)=sin‌2A−cos2B
Simplifying further:
sin‌(A+B)‌cos(A−B)=−cos(A+B)‌cos(A−B)
This implies:
tan(A+B)=−1⇒A+B=135∘
Given the constraint 0∘<A+B<270∘, the valid solution is:
A+B=135∘