Prove the following trigonometric identities:

$(s e c A - t a n A)^{2} = \frac{1 - s i n A}{1 + s i n A}$

Open in App

Solution

$(s e c A - t a n A)^{2} = \frac{1 - s i n A}{1 + s i n A}$

$L H S = (s e c A - t a n A)^{2}$

$= (\frac{1}{c o s A} - \frac{s i n A}{c o s A})^{2}$

$= (\frac{1 - s i n A}{c o s A})^{2}$

$= \frac{(1 - s i n A)^{2}}{c o s^{2} A}$

$= \frac{(1 - s i n A)^{2}}{1 - s i n^{2} A}$

$= \frac{(1 - s i n A) (1 - s i n A)}{(1 - s i n A) (1 + s i n A)}$

$= \frac{1 - s i n A}{1 + s i n A}$

$= R H S$

Suggest Corrections

Similar questions

{(\sec A - \tan A)}^{2} = \frac{1 - \sin A}{1 + \sin A}

Prove the following trigonometric identities:

$\frac{1}{1 + s i n A} + \frac{1}{1 - s i n A} = 2 s e c^{2} A$

Prove the following trigonometric identities:

$\frac{c o s A}{1 - t a n A} + \frac{s i n A}{1 - c o t A} = s i n A + c o s A$

\frac{1 - s i n A}{1 + s i n A} = (s e c A - t a n A)^{2}

{(s e c A - t a n A)}^{2} = \frac{1 - s i n A}{1 + s i n A}

Join BYJU'S Learning Program