Question

Show that cot² (B+C/2) + 1 = sec² (A/2)
Where A, B,C are angle of triangle​

Answer 1

Step-by-step explanation:

use basic identity and formulas

$\: \: \: \: \: \: \: \: \: \cot {}^{2} ( \frac{b + c}{2} ) + 1 = 1 + tan {}^{2} ( \frac{a}{2} ) \\ \\ \implies \: \cot {}^{2} ( \frac{b + c}{2} ) + 1 = 1 + \cot {}^{2} ( \frac{\pi}{2} - \frac{a}{2} )$

compare

$\: \: \: \: \: \: \: \: \frac{b + c}{2} = \frac{\pi}{2} - \frac{a}{2} \\ \\ \implies \: b + c = \pi - a \\ \implies \: a + b + c = \pi$

Hence proved

Answer 2

Answer:

Step-by-step explanation:

In ΔABC

∠A + ∠B + ∠C = 180°

∠B + ∠C = 180° - ∠A

(∠B + ∠ C) / 2 = 90° - ∠A/2

cot² (B +  C / 2) + 1 = cot² (90° - A/2) + 1

                               = tan²A/2 + 1

                               = sec² (A/2)