Question

If sec A + tan A = p and sec A - tan A = 1/p then what are the values of cos A and sin A​

Answer 1

Answer:

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Step-by-step explanation:

sec@+ tan@ = p -(1) sec²@-tan²@=1 (sec@ + tan@)x(sec@ - tan@) = 1 px(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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