Question

tan theta + tan 90 - theta is equal to sec theta + sec 90 - theta​

Answer 1

we have to prove that secβ×sec(90-β)

tanβ+tan(90-β)

tanβ+ cotβ

$\frac{sin}{cos}$ +$\frac{cos}{sin}$

$\frac{sin^{2}+cos^{2} }{sin*cos}$

$\frac{1}{sin cos}$

cosecβ×secβ

secβ×sec(90-β)

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Answer 2

Answer:

i don't know the answer