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5 th question please can you answer

Answer»

Answer:it is SOLVED below

Step-by-step explanation:

GIVEN, SecA={x+(1/4X)}

To PROVE: SecA+TanA=2x or (1/2x)

SecA=(h/b)=(4x^2+1)/4x

We KNOW , Sec^2A-1=tan^2A

Or, {(4x^2+1)/4x}^2-1=tan^2A

Or, (16x^4+8x^2+1-16x^2)/16x^2=Tan^2A

Or, (16x^4-8x^2+1)/16x^2=Tan^2A

Or, TanA=√(16x^4-8x^2+1) /4x

Now consider

LHS= SecA+TanA

={(4x^2+1)/4x}+{√(16x^4-8x^2+1)}/4x

={4x^2+1+(4x^2-1)} / 4x

= 8x^2/4x=2x


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