Question

5. If Tan (3y+30)=1, then find the value of y?​

Answer 1

Given, tan3x.tan3y = 1

So, tan3x, tan3y, sin3x, sin3y, cos3x, cos3y none of these can be zero. Or else we can't make it 1 at right hand side.

So, tan3x = 1/tan3y

Or, sin3x/cos3x = cos3y/sin3y

Or, sin3x.sin3y = cos3x.cos3y

Or, cos3x.cos3y - sin3x.sin3y = 0

Or, cos(3x + 3y) = 0

Or, cos3(x + y) = cos(π/2)

Or, 3(x + y) = π/2

Or, (x + y) = π/6

Or, sin(x + y) = sin(π/6)

So, sin(x + y) = 1/2

.

Process 2:

ABC is a right angle triangle. <ABC = π/2

So, (AB)^2 + (BC)^2 = (AC)^2

Now, tan3x.tan3y = 1, Means,

If, <BCA = 3x, then <CAB = 3y

As, tan3x = AB/BC and tan3y = BC/AB

So, <BCA + <CAB = π/2

Or, 3x + 3y = π/2

Or, (x + y) = π/6

Or, sin(x + y) = sin(π/6)

So, sin(x + y) = 1/2