In a - ABC of A -tan -1 and B -tan -1 3 then C is equal toA- -3B- -4C- -6D- None of these

The correct option is B A+B+C=π⇒ C=π (A+B) Now A+B=tan^ 12+tan^ 13 π+tan^ 1(2+3/1 2× 3 ) [∵ tan^ 1x+tan^ 1y =π+tan^ 1(x+y/1 xy ) ] For x gt;0,y gt;0 and xy ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

In a Δ ABC of...

Question

In a $Δ A B C$ of A $- t a n^{- 1}$ and B $= t a n^{- 1} 3$ then C is equal to

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None of these

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Solution

The correct option is B

$A + B + C = π \Rightarrow C = π - (A + B)$
Now $A + B = t a n^{- 1} 2 + t a n^{- 1} 3$
$π + t a n^{- 1} (\frac{2 + 3}{1 - 2 \times 3})$
$[∵ t a n^{- 1} x + t a n^{- 1} y = π + t a n^{- 1} (\frac{x + y}{1 - x y})]$
For x>0,y>0 and xy>1
$= π + t a n^{- 1} (- 1) = π - t a n^{- 1} 1 = π \frac{π}{4}$
$= \frac{3 π}{4}$
Hence $C = π - \frac{3 π}{4} = \frac{π}{4}$

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In a ΔABC of A −tan−1 and B =tan−13 then C is equal to

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In a $Δ A B C$ of A $- t a n^{- 1}$ and B $= t a n^{- 1} 3$ then C is equal to