Question

if A,B and c are the interior angles of triangle ABC the sec(B+C-A/2)=​

Answer 1

Answer:

option 2

Step-by-step explanation:

I hope it is helpful for you.......

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\sec \bigg(\frac{ B+ C- A }{2} \bigg) = \frac{1}{sin \:A}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies A,\: B\: and \: C \: are \: interior \: angle \: of \: \triangle ABC \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies sec \bigg( \frac{B + C- A}{2} \bigg ) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \angle A + \angle B + \angle C = 180 \degree \\ \\ \tt \circ \:Subtracting \: both \: side \: \angle A \\ \\ \tt: \implies \angle A+ \angle B + \angle C - \angle A = 180 \degree - \angle A \\ \\ \tt: \implies \angle B + \angle C- \angle A= 180 \degree - \angle A- \angle A\\ \\ \tt: \implies \angle B + \angle C- \angle A = 180 \degree - 2 \angle A \\ \\ \tt \circ \: Dividing \: both \: side \: by \: 2 \\ \\ \tt: \implies \frac{\angle B + \angle C- \angle A}{2} = \frac{180 \degree - 2 \angle A}{2} \\ \\ \tt \circ \:Taking \: both \: side \: sec \\ \\ \tt: \implies sec \bigg(\frac{B + C - A }{2} \bigg) = sec \bigg( \frac{180}{2} - \frac{2A}{2} \bigg) \\ \\ \tt: \implies sec \bigg(\frac{ B + C - A}{2} \bigg) = sec(90 - A) \\ \\ \tt \circ \: sec(90 - \theta) = cosec \: \theta \\ \\ \tt: \implies sec \bigg(\frac{ B + C - A }{2} \bigg) = cosec \:A \\ \\ \green{\tt: \implies sec \bigg(\frac{ B+ C - A }{2} \bigg) = \frac{1}{sin \:A } }$