Find cosA

Geometry Level 3

If sec A - tan A = p, then cos A is given by....\

1 2 ( p + 1 p ) \frac{1}{2}(p+\frac{1}{p}) None of these 2 p p 2 + 1 \frac{2p}{p^{2} + 1} 2 p p 2 − 1 \frac{2p}{p^{2} - 1}

Ojasee Duble by Ojasee Duble , 19, India

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1 solution

Hana Wehbi
Oct 22, 2017

s e c A − t a n A + s e c A + t a n A = 1 p + p ⟹ 2 s e c A = 1 + p 2 p ⟹ c o s A = 2 p 1 + p 2 sec A- tan A + sec A + tan A= \dfrac 1{p} +p\implies 2secA =\dfrac{1+p^2}{p}\implies cosA= \dfrac{2p}{1+p^2}

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How do you know secA + tanA = 1/p?

Ojasee Duble - 3 years, 7 months ago

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l have updated the solution.

Hana Wehbi - 3 years, 7 months ago

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Thank you very much:)

Ojasee Duble - 3 years, 7 months ago

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@Ojasee Duble – You're welcome.

Hana Wehbi - 3 years, 7 months ago

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