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25 votes
Sec²x + cosec²x ≡sec²xcosec²x​

proving qn pls i nd it by tdy

User Sebastian Rosch
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1 Answer

18 votes
18 votes

Explanation:

sec²x + cosec²x = sec²x.cosec²x

1/cos²x + 1/sin²x = 1/cos²x.1/sin²x

or, (sin²x+cos²x)/sin²x.cos²x = 1/(sin²x.cos²x)

or, 1/(sin²x.cos²x) = 1/(sin²x.cos²x)

Hence,

sec²x + cosec²x = sec²x.cosec²x proved!!

User Joshua Wolff
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