If in a - ABC- - C - 900 then value of tan A-B is

The correct option is D |a^2 b^2|/2ab (A B) =1cos (A B) 1cos A cos B + sin A sin B #160; = 1ba . ac +ac, bc =c^22ab Now tan (A B) = √(sec ...

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Standard XII

Mathematics

Second Fundamental Theorem of Calculus

If in a Δ A...

Question

If in a $Δ$ ABC, $∠ C = 90^{0}$ then value of tan (A-B) is

\frac{a^{2} + b^{2}}{2 a b}

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\frac{| a^{2} - b^{2} |}{2 a b}

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\frac{c^{4}}{2 a b}

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\frac{c a}{2 b}

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Solution

The correct option is D $\frac{| a^{2} - b^{2} |}{2 a b}$
$sec (A - B) = \frac{1}{c o s (A - B)}$

$\frac{1}{c o s A c o s B + s i n A s i n B} = \frac{1}{\frac{b}{a} . \frac{a}{c} + \frac{a}{c}, \frac{b}{c}} = \frac{c^{2}}{2 a b}$

Now $tan (A - B) = \sqrt{s e c^{2} (A - B) - 1}$

$\sqrt{\frac{c^{4}}{4 a^{2} b^{2}} - 1} = \sqrt{\frac{c^{4} - 4 a^{2} b^{2}}{4 a^{2} b^{2}}} = \frac{| a^{2} - b^{2} |}{2 a b}$

(using $c^{2} = a^{2} + b^{2}$ )

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If in a Δ ABC, ∠C=900 then value of tan (A-B) is

The correct option is D |a2−b2|2absec(A−B)=1cos(A−B)1cosAcosB+sinAsinB=1ba.ac+ac,bc=c22abNow tan(A−B)=√sec2(A−B)−1√c44a2b2−1=√c4−4a2b24a2b2=|a2−b2|2ab (using c2=a2+b2)

If in a $Δ$ ABC, $∠ C = 90^{0}$ then value of tan (A-B) is