Question

1/secA-tanA=secA+tanA​

Answer 1

Answer:

Hey mate here is your answer.//

Step-by-step explanation:

1/secA+tanA

Multiply by secA - tanA on both denominator and nominator

(secA - tanA) ÷ (sec²A - tan²A)

We know, sec²A - tan²A = 1

Then,

(secA - tanA) ÷ 1

SecA - tanA.

Hope it helps you.//✔✔✔

Please do mark me as Brainliest.//

Answer 2

$\huge{\mathcal{ \overbrace{ \underbrace{ \pink{ \fbox{ \green{ \blue{S} \pink{o} \red{l} \green{u} \purple{t} \blue{i} \green{o} \red{n}}}}}}}}$

$\underline { \underline{\blue {\tt{L.H.S.}}}} \\ \: \: \: \: \bf{ \frac{1}{secA - tanA } } \\ \bf{ = \frac{ {sec}^{2} A - {tan}^{2}A }{secA - tanA}} \\ \\ [\red{ \tt{∵ {sec}^{2} A - {tan}^{2}A = 1 }}] \\ \\ = \bf{ \frac{(secA + tanA)(secA - tanA)}{(secA - tanA)} } \\ \\ [ \red {\tt{From \: \: \: \: {a}^{2} - {b}^{2} = (a + b)(a - b)}}] \\ \\ = \bf{secA + tanA}$

∵ L.H.S. = R.H.S.

Hence proved