Question

Prove that $tan\{\frac{\pi}{4} + \frac{1}{2}cos^{-1}(\frac{a}{b})\} + tan\{\frac{\pi}{4} - \frac{1}{2}cos^{-1}(\frac{a}{b})\} = \frac{2b}{a}$.

Answer 1

Solution :

Let cos^-1 ( a/b ) = A

=> cosA = a/b ----( 1 )

LHS = tan( π/4+A/2 )+tan( π/4 - A/2 )

= [(1+tanA/2)/(1-tanA/2)]

+[(1-tanA/2)/(1+tanA/2)]

= [(1+tanA/2)²+(1-tanA/2)²]/(1-tan²A/2)

= 2[1+tan²A/2]/(1-tan²A/2)

= 2[ sec²A/2 ]/[ 1 - (sin²A/2/cos²A/2)]

= (2sec²A/2cos²A/2)/(cos²A/2-sin²A/2)

= 2/cosA

= 2/(a/b) [ from ( 1 ) ]

= 2b/a

= RHS

•••••

Answer 2

HELLO DEAR,




let cos-¹ (a/b) = x
cosx = a/b


now,
tan(π/4 + x/2) + tan(π/4 - x/2)

=> (1 + tanx/2)/(1 - tanx/2) + (1 - tanx/2)/(1 + tanx/2

=> {(1 + tanx/2)² + (1 - tanx/2)²}/{(1 - tanx/2)(1 + tanx/2)}

=> {1 + tan²x/2 + 2tanx/2 + 1 + tan²x/2 - 2tanx/2}/{(1 - tan²x/2)}

=> {2 + 2tan²x/2)}/{1 - tan²x/2}

=> 2{1 + tan²x/2}/{1 - tan²x/2}

=> 2/cosx

=> 2/(a/b)

=> 2b/a



I HOPE IT'S HELP YOU DEAR,
THANKS