Question

If A, B, C, are the interior angles of a treangle ABC,show that cosec square A(B+C/2)- tan square A(A/2)=1

Answer 1

Answer on Question #42692 – Math - Trigonometry

If A, B, C, are the interior angles of a triangle ABC, show that cosec square A(B+C/2)- tan square A(A/2)=1.

Solution.

We need to prove that $\csc^2\frac{B+C}{2} - \tan^2\frac{A}{2} = 1$.

First, rewrite $\csc \frac{B + C}{2}$ as $\frac{1}{\sin \frac{B + C}{2}}$.

Also notice that $\tan^2\frac{A}{2} = \frac{\sin^2\frac{A}{2}}{\cos^2\frac{A}{2}} = -1 + \frac{1}{\cos^2\frac{A}{2}}$ (here we use that $\sin^2\alpha + \cos^2\alpha = 1$).

After these steps we get $\csc^2\frac{B + C}{2} - \tan^2\frac{A}{2} = \frac{1}{\sin^2\frac{B + C}{2}} + 1 - \frac{1}{\cos^2\frac{A}{2}}$.

So, it suffices to prove that $\frac{1}{\sin^2\frac{B + C}{2}} = \frac{1}{\cos^2\frac{A}{2}}$.

Since $\sin \alpha = \sin (90{}^{\circ} - \alpha)$ and $A + B + C = 180{}^{\circ}$, we get

$$\begin{array}{l} \sin \frac {B + C}{2} = \sin \frac {1 8 0 {}^ {\circ} - A}{2} = \sin \left(9 0 {}^ {\circ} - \frac {A}{2}\right) = \cos \frac {A}{2} \Rightarrow \frac {1}{\sin^ {2} \frac {B + C}{2}} = \frac {1}{\cos^ {2} \frac {A}{2}} \\ \Rightarrow \frac {1}{\sin^ {2} \frac {B + C}{2}} + 1 - \frac {1}{\cos^ {2} \frac {A}{2}} = 1. \end{array}$$

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