If tan x = b/a... Then find whole under root (a+b)/a-b) + whole under rt (a- b)/(a+b)...
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tanx = b/a
or, tanx+1 = b/a+1
or, tanx+1 = (b+a)/a
or, a(tanx) = a+b -------------(1)
or, tanx-1 = b/a-1
or, tanx-1 = (b-a)/a
or, a(tanx-1) = b-a
or, a(1-tanx) = a-b-------------(2)
now,under root[(a+b)/(a-b)] + under root[(a-b)/(a+b)]
put the value of a+b & a-b from (1) & (2)
or, under root[a(1+tanx)/a(1-tanx)] + under root[a(1-tanx)/a(1+tanx)]
or, under root[(1+tanx)/(1-tanx)] + under root[(1-tanx)/(1+tanx)]
when we rationalize the terms
or, under root[(1+tanx)^2/(1^2-tan^2x)] + under root[(1-tanx)^2/(1^2-tan^2x)]
or, (1+tanx)/root(1-tan^2x) + (1-tanx)/root(1-tan^2x)
or, (1+tanx+1-tanx)/root(1-tan^2x)
or, 2/root(1-tan^2x)
or, tanx+1 = b/a+1
or, tanx+1 = (b+a)/a
or, a(tanx) = a+b -------------(1)
or, tanx-1 = b/a-1
or, tanx-1 = (b-a)/a
or, a(tanx-1) = b-a
or, a(1-tanx) = a-b-------------(2)
now,under root[(a+b)/(a-b)] + under root[(a-b)/(a+b)]
put the value of a+b & a-b from (1) & (2)
or, under root[a(1+tanx)/a(1-tanx)] + under root[a(1-tanx)/a(1+tanx)]
or, under root[(1+tanx)/(1-tanx)] + under root[(1-tanx)/(1+tanx)]
when we rationalize the terms
or, under root[(1+tanx)^2/(1^2-tan^2x)] + under root[(1-tanx)^2/(1^2-tan^2x)]
or, (1+tanx)/root(1-tan^2x) + (1-tanx)/root(1-tan^2x)
or, (1+tanx+1-tanx)/root(1-tan^2x)
or, 2/root(1-tan^2x)
Answered by
8
Answer:
2cosx/√cos2x
Step-by-step explanation:
Since,
tanx=b/a....(.i)
Here,
√(a+b)/(a-b)+√(a-b)/(a+b)
=>[√(a+b)²+√(a-b)²]/√(a²-b²)
=>[(a+b)+(a-b)]/√(a²-b²)
>[2a] /a√ [1-b²/a²]
>2/√(1-tan²x)
>2cosx/√(cos²x-sin²x)
>2cosx/√cos2x
[Soon.]
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