Question

if cos0+sin0=√2cos0,prove that cos0-sin0=√2sin0.​

Answer 1

Step-by-step explanation:

cos0 + sin0 = √2cos0

dividing both sides by cos0, we get

1 + (sin0/cos0) = √2 [tan0 = sin0/cos0]

So, tan0 = (√2 - 1)

cot0 = 1/(√2 -1)

Now, cos0 - sin0 = sin0 × (cos0/sin0 - 1)

= sin0 × [1/(√2-1) - 1]

= sin0 × [1 - (√2 - 1)] / (√2 - 1)

= sin0 × [2 - √2] / (√2 - 1)

taking √2 common from the numerator, we get

cos0 - sin0 = sin0 × [√2 × (√2 - 1)] / (√2 - 1)

cancelling out (√2 - 1) from numerator and denominator, we get

cos0 - sin0 = sin0 × √2 = √2sin0

hence proved.