Question

1/secA- tanA = secA+tanA​

Answer 1

$lhs = \frac{1}{ \sec(a) - \tan(a) } \\ \: \: \: \: \: \: \: = \frac{1}{ \sec(a) - \tan(a) } \times \frac{ \sec(a) + \tan(a) }{ \sec(a) + \tan(a) } \\ \: \: \: \: \: \: = \frac{ \sec(a) + \tan(a) }{ { \sec }^{2}a - { \tan }^{2}a } \\ \: \: \: \: \: \: = \frac{ \sec(a) + \tan(a) }{1} \\ \: \: \: \: \: \: = \sec(a) + \tan(a) \\ \: \: \: \: \: \: = rhs$

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Answer 2

Answer:

lhs=

sec(a)−tan(a)

1

=

sec(a)−tan(a)

1

×

sec(a)+tan(a)

sec(a)+tan(a)

=

sec

2

a−tan

2

a

sec(a)+tan(a)

=

1

sec(a)+tan(a)

=sec(a)+tan(a)

=rhs

Here is your answer

HOPE IT IS HELPFUL........:)☆