1.

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

Answer»

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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