Although your question doesn’t make complete sense, I can tell you how to solve it using calculus.
Let y=f(x), then dy/dx=sec²(2x - /2)×d(2x - /2)/dx.
What you need to do here is to replace the meaningless 2x - /2 with what it’s supposed to be.
Then you differentiate 2x - /2 with respect to x. That gives you d(2x - /2)/dx to multiply the sec²( ) by.
So now you have dy/dx as an expression in x. To get the gradient or slope you need, plug in the value x=(3) /4 (whatever that is supposed to mean), so that you get an actual value for dy/dx which is a number, not containing x. I suspect that your question probably contained π somewhere. If you don’t have the symbol on your input device you can always use (pi) to represent it. sec is the same is 1/cos, and I think that whoever set the question would probably use π/3, π/2, π/4, π, π/6, etc., somewhere in the question, because cosine has a commonly known value for these. For example: cos(π/6)=√3/2, so cos²(π/6)=¾. Therefore sec²(π/6)=4/3. This is just an example, and not necessarily applicable to this question.
OK. So now you have a value for dy/dx, which is the “m” in the line y-q=m(x-p) where p=“(3) /4” and q=the y value for this value of x, that is, y=f(“(3) /4”). (p,q) is the tangent point which the tangent line must pass through.
To get the tangent line into standard form:
y-q=m(x-p), y=mx-mp+q. The constant term in the line is q-mp.
To summarise:
Differentiate f(x) to give you the slope for a general value of x
Plug in the given x value into the above so you have a slope value
Plug in the given x value into the expression for f(x) to find the y-coord for the tangent point (p,q)
Use the slope-intercept form y-q=m(x-p) to get the tangent line y=mx-mp+q