Conditional Trigonometric Identities on $A+B+C=π$.

Asked Aug 21 '16 at 17:07

Active Aug 21 '16 at 17:16

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If $A+B+C=\pi$ then prove that: $$\tan A\tan B+\tan B\tan C+\tan C \tan A=1+\sec A\sec B\sec C$$

My Approach:

Given, $$A+B+C=\pi$$ $$A+B=\pi-C$$ $$\sin(A+B)=\sin(\pi-C)=\sin C$$ $$\cos(A+B)=\cos(\pi-C)=-\cos C$$

Now,

$$\text{L.H.S.}=\tan A \tan B+\tan B\tan C+\tan C\tan A$$ $$=\frac {\sin A\sin B}{\cos A\cos B}+\frac {\sin B\sin C}{\cos B\cos C} + \frac {\sin C\sin A}{\cos C\cos A}$$

I got struck at here. Please help me to move on.

edited Aug 21 '16 at 17:16

Michael Hardy

asked Aug 21 '16 at 17:07

pi-π

7,416

You might try putting $\pi-A-B$ in place of $C$. $\qquad$ – Michael Hardy Aug 21 '16 at 17:18
Note proper MathJax usage, as in my edits. You had $A$+$B$+$C$=$\pi$ and I changed it to $A+B+C=\pi$, and I did changes more extensive than that in the later parts. $\qquad$ – Michael Hardy Aug 21 '16 at 17:19

Conditional Trigonometric Identities on $A+B+C=π$.

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