Question

if cosecA+secA=cosecB+secB then prove that tan(A+B)/2=cotAcotB​

Answer 1

Answer:

Step-by-step explanation:

A+B+C =180

A+B = 180 -C

DIVIDE BY 2 ON BOTH SIDES

THEN

(A+B)/2 =(90 - C/2)

NOW APPLY TAN ON BOTH SIDES

THEN

TAN((A+B)/2) =TAN (90 - C/2)

SINCE TAN(90 - TITA) = COT(TITA)

THERE FORE

TAN((A+B)/2)) = COT (C/2)

Answer 2

Step-by-step explanation:

• Separating like terms we get Cosec A – cosec B = sec B – sec A
•       Substituting cosecA and sec A in terms of sin and cos we get  1/sin A – 1/sin B =  1/cos B – 1/ cos A      
• Sin B – Sin A / sin A sin B = cos A – cos B / cos B cos A
•   Cos A cos B / sin A sin B = cos A – cos B / sin B – sin A
• Cot A cot B = cos A – cos B / sin B – sin A
• Now cos A – cos B = 2sin A + B/2 sin A – B / 2
•       Sin A – sin B = 2 cos A + B/2 sin A – B/2
•    Substituting we get
•     Cot A cot B = 2sin A + B/2 sin B – A / 2 / 2 cos A + B/2 sin B – A/2
•           Cot A cot B = sin A + B/2 / cos A + B/2
•            Cot A cot B = tan A + B/2
•                 Hence proved

Reference link will be

https://brainly.in/question/4536027?