Question

if secA=(1+4x^2)/4x then cosecA=? and tanA=?​

Answer 1

secA=x+1/4x

∴, sec²A=(x+1/4x)²

=x²+2.x.1/4x+1/16x²

=x²+1/2+1/16x²

Now, sec²A-tan²A=1

or, tan²A=sec²A-1

or, tan²A=x²+1/2+1/16x²-1

or, tan²A=x²+1/16x²-1/2

or, tan²A=x²-2.x.1/4x+1/16x²

or, tan²A=(x-1/4x)²

or, tanA=+-(x-1/4x)

∴, either, secA+tanA

=x+1/4x+x-1/4x [when tanA=x+1/4x]

=2x  

or, secA+tanA

=x+1/4x-x+1/4x [when tanA=-(x+1/4x)]

=1/4x+1/4x

=2/4x

=1/2x (Proved)

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Answer 2

Answer:

Step-by-step explanation:we know. That sec²A-tan²A=1

Tan²A=sec²A-1

Tan²A=(1+4x²)-1=4x²

So tan²A=4x²

TanA=√4x²=2x

So cotA =1/tanA

So cotA=1/2x

We know cose²A-cot²A=1

Cosec²A=1+cot²A

=1+1/(2x)²

=(1+1/4x²)or(4x²+1/4x²)