Question

1 upon sec theta -tan theta= sec theta +tan theta

Answer 1

$\huge \boxed{ \mathbb{ \ulcorner ANSWER: \urcorner}}$

$\huge \boxed{Prove \: that :-}$

$\large\bf = > \frac{1}{ \sec\theta - \tan \theta } = \sec \theta + \tan \theta.$

$\large \boxed {\mathbb{SOLVING \: LHS.}}$

$\huge \bf = \frac{1}{ \sec \theta - \tan \theta} .$

$\large \bf = \frac{1}{ \sec \theta - \tan \theta} \times \frac{ \sec \theta + \tan \theta}{\sec \theta + \tan \theta}.$

$\huge \bf = \frac{ \sec \theta + \tan \theta}{ { \sec}^{2} \theta - { \tan}^{2} \theta } .$

[ → sec²x - tan²x = 1.]

$\huge \boxed{ = \sec \theta + \tan \theta}$

✔✔Hence, it is proved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$

$\huge \bf{ \mathbb{ \#BEBRAINLY.}}$

Answer 2

$\huge{hey}$

$\huge \bf{ \mathbb{ANSWER}}$


$\bf{ \frac{1}{sec \theta - tan \theta} = sec \theta + tan \theta}$



LHS

$\bf{ \frac{1}{sec \theta - tan \theta} \times \frac{sec \theta + tan \theta}{sec \theta + tan \theta}}$

$\bf \huge{ \frac{sec \theta + cos \theta}{ {sec}^{2} \theta - {tan}^{2} \theta} }$


$\huge{ {sec}^{2} \theta + {tan}^{2} \theta}$
(since sec²-tan²=1)


$\huge{hope \: it \: helps}$

$\huge{ \mathfrak{thanks}}$