Question

sinA(tanA)/1-cosA=1+secA​

Answer 1

Answer:

Here is your answer

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Answer 2

${ \huge{ \mathbb{ \overbrace{ \underbrace{ \purple{ \boxed{ \fbox{ \bold { \large{ \mathtt{ANSWER}}}}}}}}}}}$

$\large \green{ \mathtt{to \: prove = > \frac{sinA.tanA}{1 - cosA} = 1 + sec A }}$

$\large{ \green{ \mathtt{ \frac{sinA. \frac{sinA}{cosA} }{1 - cosA} }}} \\ \\ = > \large{ \green{ \mathtt{ \frac{ {sin}^{2}A }{cos(1 - cosA)} }}} \\ \\ = > \large{ \green{ \mathtt{ \frac{1 - co {s}^{2} A}{1 - cosA} }}} \\ \\ = > \large{ \green{ \mathtt{ \frac{(1 - cosA)(1 + cosA)}{cos(1 - cosA)} }}} \\ \\ = > \large{ \green{ \mathtt{ \frac{1 + cosA}{cosA} }}} \\ \\ = > \large{ \green{ \mathtt{ \frac{1}{cosA} + \sf{ \cancel{ \dfrac{cosA}{cosA}}} }}} \\ \\ = > \large{ \blue{ \boxed{ \fbox{ \mathtt{secA + 1}}}}} \: \\ \\ \large \purple{ \underline{ \underline{\mathtt{hence \: proved}}} }$

$\mathtt{ \red{hops \: this \: may \: help \: you}} \\ \huge { \red{\ddot{ \smile}}}$