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TanX=b/a, then find the value of √a+b/a-b +√a-b/a+b

TanX=b/a, then find the value of √a+b/a-b +√a-b/a+b

Grade:11

2 Answers

Arun
25763 Points
4 years ago
Tytanx = 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)
Shailendra Kumar Sharma
188 Points
4 years ago
√a+b/a-b +√a-b/a+bSqrt{(a+b)/(a-b)} + Sqrt{(a-b)/(a+b)}=>(a+b+a-b)/Sqrt{(a+b)*(a-b)}=>2a/sqrt(a2-b2)Divide with a in numerator and denominator2/sqrt{1-(b/a)2}=2/sqrt(1-tan2x)

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