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#