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Featured 2 weeks ago

Please give a look here...

#color(red)((1+sinA)/cosA+cosB/(1-sinB)#

#=color(green)({cos (A/2)+sin(A/2)}^2/{cos^2(A/2)-sin^2(A/2)}+{cos^2(B/2)-sin^2(B/2)}/{cos(B/2)-sin(B/2)}^2#

#=color(red)((cos(A/2)+sin(A/2))/{cos(A/2)-sin(A/2)}+(cos(B/2)+sin(B/2))/(cos(B/2)-sin(B/2))#

#=color(green)([{(cos(A/2)+sin(A/2))cdot(cos(B/2)-sin(B/2))}+{(cos(B/2)+sin(B/2))cdot(cos(A/2)-sin(A/2))}]/{(cos(A/2)-sin(A/2))cdot(cos(B/2)-sin(B/2))}#

#=color(red)({cos(A/2) cdot cos(B/2)-cos(A/2) cdot sin(B/2)+cos(B/2) cdot sin(A/2)-sin(A/2) cdot sin(B/2)+cos(B/2)cos(A/2)-cos(B/2) cdot sin(A/2)+sin(B/2) cdot cos(A/2)-sin(B/2) cdot sin(A/2)}/{cos(A/2)cos(B/2)-cos(A/2) cdot sin(B/2)-sin(A/2) cdot cos(B/2)+sin(A/2) cdot sin(B/2)}#

#=color(green)((2{cos(A/2)cdot cos(B/2)-sin(A/2) cdot sin(B/2)})/({cos(A/2)cdot cos(B/2)+sin(A/2) cdot sin(B/2)}-{sin(A/2)cos(B/2)+cos(A/2)sin(B/2)}#

#=color(red)((2 cos((A+B)/2))/{cos((A-B)/2)-sin((A+B)/2)}#

#=color(green)((2 cos((A+B)/2)cdot2sin((A-B)/2))/{2 cdot cos((A-B)/2) cdot sin((A-B)/2)-2 cdot sin((A+B)/2)cdot sin((A-B)/2)#

#=color(red)((2(sinA-sinB))/(sin(A-B)+cosA-cosB)# Hope it helps...

Thank you...

Featured 2 weeks ago

Use the identity

#sin(2x)=2sin(x)cos(x)#

Applying this identity to your example

Featured 6 days ago

Applying Law of Cosine,

Featured 4 days ago

Throug the equivalence

If we have 45Âº degrees convert to radians is easy

By the other hand, if we have

Featured yesterday

We first want to get this equation in terms of only one trigonometric function. Let's say we want to put everything in terms of cosine, as that will work best in this case.

Let's square both sides:

Recall the identity

Solving for

Thus,

And everything is in terms of cosine. Let's move everything to one side:

We can factor out an instance of

We now solve the following equations:

For

We already see no values of

Featured 6 hours ago

I got

Consider the diagram:

The angles

in numbers:

so that:

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