Question

9. Show that : 1 + tan2A tanA = tan2A cotA - 1 = sec2A​

Answer 1

Answer:

Step-by-step explanation:

First:

$1 + \tan 2A \tan A = 1 + \cfrac{2 \tan A}{1 - \tan^2 A} \times \tan A$

$=1 + \cfrac{2 \tan^2 A}{1 - \tan^2 A}$

$= \cfrac {1 - \tan^2 A + 2 \tan^2 A}{1 - \tan^2 A}$

$=\cfrac{1+\tan^2 A}{1 - \tan^2 A}$                       ...(1)

Second:

$\tan 2A \cot A - 1 = \cfrac{2\tan A}{1 - \tan^2 A} \times \cfrac{1}{\tan A} - 1$

$= \cfrac{2}{1 - \tan^2 A} - 1$

$= \cfrac{2 - 1 + \tan^2 A}{1 - \tan^2 A}$

$= \cfrac{1 + \tan^2 A}{1 - \tan^2 A}$                     ...(2)

Third

$\sec 2A = \cfrac{1}{\cos 2A}$

$=\cfrac{1}{\cos^2 A - \sin^2 A}$

$=\cfrac{\cos^2 A + \sin^2 A}{\cos^2 A - \sin^2 A}$

Divide numerator and denominator by $\cos^2 A$

$=\cfrac{1 + \cfrac{\sin^2 A}{\cos^2 A}}{1 - \cfrac{\sin^2 A}{\cos^2 A}}$

$=\cfrac{1 + \tan^2 A}{1 - \tan^2 A}$          ...(3)

From Eqns. (1), (2) and (3), we prove

$1 + \tan 2A \tan A = \tan 2A \cot A - 1 = \sec 2A$