If tan A + sec A = 4, what is CosA?

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Answer 1 · 2018-03-23T05:55:29

Given $tan A + sec A = 4$

$(secA + tan A) = 4.....[1]$

So

$(secA + tan A)(secA-tanA) = 4(secA-tanA)$

$=>(sec^2A-tan^2A) = 4(secA-tanA)$

$=>1 = 4(secA-tanA)$

$=> secA-tanA=1/4.......[2]$

Adding [1] and [2] we get

$2secA=4+1/4$

$=>secA=17/8$

$=>cosA=8/17$

Answer 2 · 2018-03-23T05:55:43

$cosA=8/17$

Here,

$secA+tanA=4.....to(1)$

We know that,

$color(red)(sec^2A-tan^2A=1$

$(secA+tanA)(secA-tanA)=1$

$=>(4)(secA-tanA)=1.....to$ From(1)

$:.secA-tanA=1/4.....to(2)$

Adding (1) and (2),

$secA+canceltanA=4$

$(secA-canceltanA=1/4)/$

$2secA=4+1/4=17/4$

$=>secA=17/8$

$=>cosA=8/17$