Question

a+b=90° then prove √tanAtanB+tanAcotB/sinAsecB_sin^2B/cos^2=tanA

Answer 1

Given, LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a

      = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a

      = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a

      = root 1+tan^2 a/1 - 1

     = root tan^2 a

    = tan a.

Hope this helps!

Answer 2

CONCEPT:

1+tan²A =sec²A

sin A/Cos A= tanA

sec(90-A)=cosecA

cot(90-A)=tanA

Given:

LHS = √( tan a tan b + tan a cot b/sin a sec b - sin²b/cos²b)

FIND:

LHS=RHS=tan a

Solution:

LHS = √( tan a tan b + tan a cot b/sin a sec b - sin²b/cos²b)

     = √(tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin²(90-a)/cos² a

     = √(tan a.cot a + tan a.tana/sin a.cosec a - cos² a/cos² a)

     = √((1+tan² a)/1 - 1)

    = √ tan² a

     = tan a.

Hence proved.

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