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Show that (sec(x))^2 /(sec(x)+1)(sec(x)-1) can be written as (cosec(x))^2.

( sec 2 (x))/((sec(x)+1)(sec(x)-1))Then, by the rule of 'difference of two squares', we know that this equals= (sec 2 (x))/(sec 2 (x)-1)= (sec 2 x/tan 2 x)since 1+tan 2 (x)=sec 2 (x), we get sec 2 (x)-1=tan ...
RS
Answered by Rishi S. • Maths tutor
12037 Views

Find the values of A between and including 0 and 360 degrees for tan(2A) = 3tan(A)

You cannot work with this equation in the current form so you must use identities to find an equivalent form that you can work with. It is known that tan(2A) = 2tan(A) / 1-tan 2 (A) so set this equal to 3tan...
DM
Answered by Daniel M. • Maths tutor
5692 Views

Express the equation cosecθ(3 cos 2θ+7)+11=0 in the form asin^2(θ) + bsin(θ) + c = 0, where a, b and c are constants.

We must first use the identity cosecθ = 1/sinθ. Now the equation becomes (1/sinθ)(3 cos 2θ+7)+11=0. Since we know that the question is asking for the answer in the form of asin 2 θ + bsinθ + c = 0, we realis...
GL
Answered by George L. • Maths tutor
6920 Views

Explain why for any constant a, if y = a^x then dy/dx = a^x(ln(a))

So let's start with taking the natural log on both sides of y=a x , giving us ln(y) = ln(a x ). Using the laws of logarithms we can write this as ln(y) = xln(a).Next, we differentiate both sides with respect...
JM
Answered by James M. • Maths tutor
13104 Views

How does one find the equation of a line passing through 2 points of a graph?

The general equation for a line on a graph is: y = mx + b ,where a is called the slope of the graph and b is the y-intercept(or the point where the line crosses the y axis). Let's assume the 2 points have th...
DD
Answered by Dimitar D. • Maths tutor
5455 Views