Question

If cosec A + sec A = cosec B + sec B, Prove that $tanA.tanB=cot\frac{A+B}{2}$

Answer 1

Given : csc A + sec A = csc B + sec B ... (1)
To Prove : ( tan A )( tan B ) = cot [ (A+B)/2 ] ....... (2)
___________________________________
Proof : From (1),
... sec A - sec B = csc B - csc A
∴ (1/ cos A) - (1/ cos B) = (1/ sin B) - (1/ sin A)
∴ ( cos B - cos A ) / ( cos A. cos B ) = ( sin A - sin B ) / ( sin A. sin B )
∴ rearranging the terms on two sides,
... ( sin A / cos A )( sin B / cos B ) = ( sin A - sin B ) / ( cos B - cos B )
∴ ( tan A )( tan B ) = [ 2. cos (A+B)/2. sin (A-B)/2 ] / [ 2. sin ((A+B)/2. sin (A-B)/2 ]
∴ ( tan A )( tan B ) = cot [ (A+B) / 2 ] .......... Q.E.D.