SEC THETA = X+1/4X .THEN PROVE THAT SEC THETA+ TAN THETA =2X OR 1/2X
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Answered by
946
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.
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.
Answered by
725
SecФ = x + 1/(4x)
sec²Ф = x² + 1/(16 x²) + 1/2
Tan²Ф = x² + 1/(16 x²) - 1/2 as sec² - tan² = 1
= (x - 1/(4x) )²
tanФ = + (x - 1/(4x)) or - (x - 1/(4x) )
sec Ф + tanФ = 2x or 1/(4x) + 1/(4x) = 1/(2x)
sec²Ф = x² + 1/(16 x²) + 1/2
Tan²Ф = x² + 1/(16 x²) - 1/2 as sec² - tan² = 1
= (x - 1/(4x) )²
tanФ = + (x - 1/(4x)) or - (x - 1/(4x) )
sec Ф + tanФ = 2x or 1/(4x) + 1/(4x) = 1/(2x)
Anonymous:
best answer
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