Question

Sec
$\sin(?)$
​

Answer 1

Sec∆ = 1/cos∆

Sin∆ = 1/cosec∆

Hope it helps you...

Answer 2

$\underline\textsf{Trignometric Ratios For Acute Angles :---}$

→ cosθ = sin(90-θ)

→ sinθ = cos(90-θ)

→ tanθ = cot(90-θ) = 1/cotθ

→ secθ = cosec(90-θ) = 1/cosθ

→ cosecθ = sec(90-θ) = 1/sinθ

→ cotθ = tan(90-θ) = 1/tanθ

$\mathcal{BE\:\:BRAINLY}$

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