Question

$\mathbb{\huge{\bold{\red{content\: quality}}}}$

$\mathcal{\bold{\huge{\blue{if\: cotAcotB=2\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ show\: that\: cos(A+B)/cos(A-B)=1/3}}}}$

Answer 1

Hii !!!

Solution !!!

from LHS cotA * CotB = 2

=> cosA / sinA * cosB / sinB = 2 or, cosA * cosB / SinA * SinB = 2 or, cosA * cosB = 2sinA * sinB or, cosA * cosB = cos ( A - B ) -cos ( A +B)

multiplying by two in both side .

or, 2cosA*cos B = 2 {cos ( A - B ) - cos ( A + B ) }or

,cos ( A + B ) + cos ( A - B ) = 2 cos (A - B ) - 2cos ( A +B ) or, 3cos( A +B ) = 1 *cos ( A - B ) or, cos ( A + B ) / cos ( A - B ) = 1/3 proved

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Hope it helps you !!!

@Rajukumar111 ❤❤