Prove that :

tan (55° + x) = cot (35° - x)

sec (70° - θ) = cosec (20° + θ)

$\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A)

If A and B are complementary angles, prove that :

(i) cot B + cos B = sec A cos B (1 + sin B)

(ii) cot A cot B - sin A cos B - cos A sin B = 0

(iii) cosec² A + cosec² B = cosec² A cosec² B

(iv) $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$