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Question

If SecA = x +1/4x, then prove that sec A+tan A=2x or 1/2x


Solution

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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