Question

prove this
sin (90°- A) cos (90° - A) = tan A 1+ tan² A​

Answer 1

QUESTION:

$\\ \\ \\ \\ sin \: (90 - A) \cos(90 - A) \\ \\= tanA + \frac{1}{ { \tan }^{2}A }$

ANSWER:

$\\ \\ lhs \: \: \: \\ \\ \\ \\ = \frac{tanA}{1 + {tan}^{2}A } \\ \\ \\ = \frac{tanA}{ {sec}^{2} A} \\ \\ \\ = \frac{ \frac{sinA}{cosA} }{ \frac{1}{ {cos}^{2} A} } \\ \\ \\ rhs \\ \\ \\ = \frac{sinA}{cosA} \times {cos}^{2} A \\ \\ \\ = cosAsinA \\ \\ \\ \\ lhs = rhs \: (proved)$

Answer 2

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