Question

1 -tan4α/1+tan4α=cosα+sinα/cosα-sinα​

Answer 1

Answer:

Step-by-step explanation:

tana+cosa=2seca

Multiplying with cosa

Sina+(cosa)^2=2

Using (sina)^2+(cosa)^2=1

Sina+1-(sina)^2=2

Sina-1-sina^2=0

(Sina)^2-sina+1=0

It is a quadratic function with variable sina.

For any quadratic equation ax^2+bx+c=0, to have real roots,

b^2–4ac must be greater than 0.

So for the given equation

1–3 must be greater than 0 which is not true.

The solution will be sina=imaginary number

Which cannot be true. So no solution for a(or in this case alpha)