<s>[INST] If $A+B+C=\frac {\pi}{2}$ then prove that.... If $A+B+C=\frac {\pi}{2}$ thrn prove that:$$\tan (A+B) [\tan A-\tan B]=\sqrt {1+\cot^2 C} (\sec A\cos B-\cos A\sec B)$$
My attempt
Here, $$A+B+C=\frac {\pi}{2}$$
$$A+B=\frac {\pi}{2} - C$$
Now,
$$L.H.S.=\tan(A+B) [\tan A-\tan B]
= \tan\left(\frac{\pi}{2} - C\right) \left(\frac{\sin A}{\cos A} - \frac{\sin B }{\cos B} \right)$$
$$=\cot C \frac{\sin (A-B)}{\cos A \cos B}$$
now, plz help me to continue from here.
 [/INST] \begin{align*}
\cot C \frac{\sin (A-B)}{\cos A \cos B} &= \frac{\cos C}{\sin C}\frac{\sin (A-B)}{\cos A \cos B} \\
&= \frac{\sin(A+B)}{\cos(A+B)}\frac{\sin (A-B)}{\cos A \cos B} \\
&=\frac{\frac{\cos 2B - \cos 2A}{2}}{\sin C(\cos A \cos B)}\\
&= \frac{\cos^2 B - \cos^2 A}{\sin C \cos A \cos B}\\
&= \frac{1}{\sin C}(\sec A\cos B-\cos A\sec B)\\
&= \text{cosec }C(\sec A\cos B-\cos A\sec B)\\
&= \sqrt{1+\cot^2 C}(\sec A\cos B-\cos A\sec B)
\end{align*}
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