Question

A+B
If cosec A + sec A = cosec B + sec B , prove that tan(A+B)/2
= cot A cot B
​

Answer 1

Answer

Given : csc A + sec A = csc B + sec B ... (1)

To Prove : ( tan A )( tan B ) = cot [ (A+B)/2 ] ....... (2)

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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.