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∘