If a > 0 and sec x = a + 1/4a , prove that sec x + tanx = 2a or 1/2a
Answers
Given that:;
secA = x + 1/4x
we know that:-
1 + tan²A = sec²A
tan²A = sec²A �� 1
, tan²A = (x + 1/4x)² – 1
=> x² + 2 × x × 1/4x + 1/16x²-1
= > x² + 1/2 + 1/16x² – 1
=> x²+ 1/16x² – 1/2
=> (x – 1/4x)^2
, tan²A = x – 1/4x or tan²A = - (x – 1/4x)
put this 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
Answer:
Given that:;
secA = x + 1/4x
we know that:-
1 + tan²A = sec²A
tan²A = sec²A �� 1
, tan²A = (x + 1/4x)² – 1
=> x² + 2 × x × 1/4x + 1/16x²-16x² – 1
=> x²+ 1/16x² – 1/2
=> (x – 1/4x)^2
tan²A = x – 1/4x or tan²A = - (x – 1/4x)
put this 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
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