Question

$\sec(x) + \tan(x) = p \: show \: that \: \sec(x ) - \tan(x) = 1 \div p.hence \: find \: cos \: and \: sin$


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Answer 1

Step-by-step explanation:

Sec@ + tan@ = p ———(1)

sec²@ – tan²@ = 1

(sec@ + tan@)×(sec@ – tan@) = 1

p×(sec@ – tan@) = 1

sec@ – tan@ = 1/p ——(2)

(1) + (2)

2 sec@ = (p+1/p) = (p²+1)/p

sec@ = (p²+1)/2p

cos@ = 2p/(p²+1)

(1) — (2)

2tan@ = p–1/p = (p²–1)/p

tan@ = (p²–1)/2p

tan@ ÷ sec@ = [(p²–1)/2p]÷[(p²+1)/2p]

sin@ = (p²–1)/(p²+1)

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