Question

(1 + cotA + tanA)(sinA - cosA) = secA/cosec²A - cosecA/sec²A = sinA.tanA - cotA.cosA​ (prove it)

Answer 1

Answer:-

We have to prove:

(1 + cot A + tan A)*(sin A - Cos A) = (sec A /cosec² A) - (Cosec A /sec² A) = sin A tan A - cot A Cos A

→ 1 (sin A - Cos A) + cot A (sin A - Cos A) + tan A (sin A - Cos A)

→ sin A - Cos A + cot A sin A - cot A Cos A + tan A sin A - tan A Cos A

Putting the value of Cot A = Cos A / sin A and tan A = sin A / Cos A we get,

→ sin A - Cos A + Cos A/sin A * sin A - Cos A/sin A *Cos A + sin A * sin A/Cos A - sin A /Cos A * Cos A

→ sin A - Cos A + Cos A - (cos² A/sin A) + (sin² A/Cos A) - sin A

→ sin² A/Cos A - cos² A/sin A -- equation (1)

Putting the value of sin A = 1/Cosec A and Cos A = 1/sec A we get,

→ [ (1/cosec² A) / (1/sec A) ] - [ (1/sec² A) / (1/Cosec A) ]

→ [ (1/cosec² A) * (sec A) ] - [ (1/sec² A) * (Cosec A) ]

→ sec A / cosec² A - Cosec A/sec² A

From equation (1):

Taking sin A and Cos A common we get,

→ sin A * (sin A/Cos A) - Cos A * (Cos A/sin A)

→ sin A tan A - Cos A cot A

→ sin A tan A - cot A Cos A

Hence, Proved.

Answer 2

Answer:

U answer for rosie even They doing mine so I'm gonna ask them for rosie if you know any Mod

give their user name plz