In a ΔABC, if C is a right angle, then tan−1(b+ca)+tan−1(c+ab)=?

tan−1(b+ca)+tan−1(c+ab)We know that tan−1x+tan−1y=tan−1(1−xyx+y)Replace, x by b+ca and y by c+ab=tan−1⎝⎛1−(b+c)(c+a)abb+ca+c+ab⎠⎞=tan−1⎝⎛(b+c)(c+a)(b+c)(c+a)−ab(b+c)(c+a)ac+a2+b2+bc⎠⎞=tan−1⎝⎛(b+c)(c+a)bc+c2+ca(b+c)(c+a)a2+b2+bc+ac⎠⎞=tan−1bc+c2+caa2+b2+bc+acGiven:∠C=90∘⇒a2+b2=c2=tan−1bc+c2+cac2+bc+ac=tan−11=4π

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In a , if C is a right angle, then ^{-1}({a}{b+c})+^{-1}({b}{c

Question

In a $Δ$ ABC, if C is a right angle, then $tan^{- 1} (\frac{a}{b + c}) + tan^{- 1} (\frac{b}{c + a}) = ?$

\frac{π}{3}

\frac{π}{4}

\frac{π}{6}

\frac{5 π}{2}

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In a

Δ

ABC, if C is a right angle, then

tan^{- 1} (\frac{a}{b + c}) + tan^{- 1} (\frac{b}{c + a}) = ?

7 mins ago

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7 mins ago

Text solutionVerified

tan^{- 1} (\frac{a}{b + c}) + tan^{- 1} (\frac{b}{c + a})

We know that

tan^{- 1} x + tan^{- 1} y = tan^{- 1} (\frac{x + y}{1 - x y})

Replace,

x

\frac{a}{b + c}

and

y

\frac{b}{c + a}

= tan^{- 1} ⎝ ⎛ \frac{\frac{a}{b + c} + \frac{b}{c + a}}{1 - \frac{ab}{( b + c ) ( c + a )}} ⎠ ⎞

= tan^{- 1} ⎝ ⎛ \frac{\frac{a c + a ^{2} + b ^{2} + b c}{( b + c ) ( c + a )}}{\frac{( b + c ) ( c + a ) - ab}{( b + c ) ( c + a )}} ⎠ ⎞

= tan^{- 1} ⎝ ⎛ \frac{\frac{a ^{2} + b ^{2} + b c + a c}{( b + c ) ( c + a )}}{\frac{b c + c ^{2} + c a}{( b + c ) ( c + a )}} ⎠ ⎞

= tan^{- 1} \frac{a ^{2} + b ^{2} + b c + a c}{b c + c ^{2} + c a}

Given

: ∠ C = 90^{\circ} \Rightarrow a^{2} + b^{2} = c^{2}

= tan^{- 1} \frac{c ^{2} + b c + a c}{b c + c ^{2} + c a}

= tan^{- 1} 1 = \frac{π}{4}

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Question Text	In a $Δ$ ABC, if C is a right angle, then $tan^{- 1} (\frac{a}{b + c}) + tan^{- 1} (\frac{b}{c + a}) = ?$
Answer Type	Text solution:1
Upvotes	150

In a ΔABC, if C is a right angle, then tan−1(b+ca​)+tan−1(c+ab​)=?

Text solutionVerified

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In a $Δ$ ABC, if C is a right angle, then $tan^{- 1} (\frac{a}{b + c}) + tan^{- 1} (\frac{b}{c + a}) = ?$