If A - ac - B - c - a - C - a 3 - c and C- a 2 -a-1- then

The correct option is B C=A+B #160;( A+B ) #160; = A B 1 B + A #160; #160;=√( ac ) .√( c a #160;) 1 √( c a #160;) +√( ac ) ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If A =√ ac ...

Question

If $cot A = \sqrt{a c}, cot B = \sqrt{\frac{c}{a}}, cot C = \sqrt{\frac{a^{3}}{c}}$ and $C = a^{2} + a + 1$ , then

A = B + C

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B = C + A

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C = A + B

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none of these

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Solution

The correct option is B $C = A + B$
$cot (A + B) = \frac{cot A cot B - 1}{cot B + cot A}$

$= \frac{\sqrt{a c} . \sqrt{\frac{c}{a}} - 1}{\sqrt{\frac{c}{a}} + \sqrt{a c}} = \frac{(c - 1) \sqrt{a}}{\sqrt{c} (1 + a)}$

$= \frac{(a^{2} + a) \sqrt{a}}{\sqrt{c} (1 + a)} [∵ c = a^{2} + a + 1, ∴ c - 1 = a^{2} + a]$

$= \frac{a \sqrt{a}}{\sqrt{c}} = \sqrt{\frac{a^{3}}{c}} = cot C$

$∴ A + B = C$

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If cotA=√ac,cotB=√ca,cotC=√a3c and C=a2+a+1, then

The correct option is B C=A+Bcot(A+B)=cotAcotB−1cotB+cotA=√ac.√ca−1√ca+√ac=(c−1)√a√c(1+a)=(a2+a)√a√c(1+a)[∵c=a2+a+1,∴c−1=a2+a]=a√a√c=√a3c=cotC∴A+B=C

If $cot A = \sqrt{a c}, cot B = \sqrt{\frac{c}{a}}, cot C = \sqrt{\frac{a^{3}}{c}}$ and $C = a^{2} + a + 1$ , then