solve this question ( maths )
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Solution -
Given –
secA = x + 1/4x
As 1 + tan^2A = sec^2A
tan^2A = sec^2A – 1
Therfore, tan^2A = (x + 1/4x)^2 – 1
= x^2 + 2*x*1/4x + 1/16x^2 – 1
= x^2 + 1/2 + 1/16x2 – 1
= x^2 + 1/16x^2 – 1/2
= (x – 1/4x)^2
Therefore, tan^2A = x – 1/4x or tan^2A = - (x – 1/4x)
Substitute the value of secA and tanA in the given equation secA + tanA
LHS = secA + tanA
= x + 1/4x + x – 1/4x
= 2x
= RHS
Or
LHS = secA + tanA
=x + 1/4x -x + 1/4x
= 2/4x
= 1/2x
= RHS
Hence proved.
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