if sinA+ sinB is l, cos A + cos B is m , and tan ( A/2)*tan (B/2) =n where n is not one prove that
cos (A+B) is (m square - l square )/(m square + l square)
also prove that (1+n)/(1-n) is ( l square + m square )/ 2 n
Answers
Answer: hope it work i have work hard on it
Step-by-step explanation:
Let’s agree again to the standard convention for labelling the parts of a right triangle. Let the right angle be labelled C and the hypotenuse c. Let A and B denote the other two angles, and a and b the sides opposite them, respectively.
Right triangle with parts standardly labelled
Solving right triangles
We can use the Pythagorean theorem and properties of sines, cosines, and tangents to solve the triangle, that is, to find unknown parts in terms of known parts.
Pythagorean theorem: a2 + b2 = c2.
Sines: sin A = a/c, sin B = b/c.
Cosines: cos A = b/c, cos B = a/c.
Tangents: tan A = a/b, tan B = b/a.
Let’s first look at some cases where we don’t know all the sides. Suppose we don’t know the hypotenuse but we do know the other two sides. The Pythagorean theorem will give us the hypotenuse. For instance, if a = 10 and b = 24, then c2 = a2 + b2 = 102 + 242 = 100 + 576 = 676. The square root of 676 is 26, so c = 26. (It’s nice to give examples where the square roots come out whole numbers; in life they usually don’t.)
Now suppose we know the hypotenuse and one side, but have to find the other. For example, if b = 119 and c = 169, then a2 = c2 – b2 = 1692 – 1192 = 28561 – 14161 = 14400, and the square root of 14400 is 120, so a = 120.
We might only know one side but we also know an angle. For example, if the side a = 15 and the angle A = 41°, we can use a sine and a tangent to find the hypotenuse and the other side. Since sin A = a/c, we know c = a/sin A = 15/sin 41. Using a calculator, this is 15/0.6561 = 22.864. Also, tan A = a/b, so b = a/tan A = 15/tan 41 = 15/0.8693 = 17.256. Whether you use a sine, cosine, or tangent depends on which side and angle you know.
Inverse trig functions: arcsine, arccosine, and arctangent
Now let’s look at the problem of finding angles if you know the sides. Again, you use the trig functions, but in reverse. Here’s an example. Suppose a = 12.3 and b = 50.1. Then tan A = a/b = 12.3/50.1 = 0.2455. Back when people used tables of trig functions, they would just look up in the tangent table to see what angle had a tangent of 0.2455. On a calculator, we use the inverse trig functions named arctangent, arcsine, and arccosine. Usually there’s a button on the calculator labelled “inv” or “arc” that you press before pressing the appropriate trig button. The arctangent of 0.2455 is 13.79, so the angle A is 13.79°. (If you like, you can convert the 0.79 degrees to minutes and seconds.)
That’s all there is to it.
The other three trigonometric functions: cotangent, secant, and cosecant
For most purposes the three trig functions sine, cosine, and tangent are enough. There are, however, cases when some others are needed. In calculus, secant is frequently used. You might ask, “why six trig functions?” It’s a kind of symmetry. There are six ways of making ratios of two sides of a right triangle, and that gives the six functions:
sin A = a/c (opp/hyp)
cos A = b/c (adj/hyp)
tan A = a/b (opp/adj)
cot A = b/a (adj/opp)
sec A = c/b (hyp/adj)
csc A = c/a (hyp/opp)
You can see by the listing that cotangent (abbreviated cot, or sometimes ctn) is the reciprocal of tangent, secant (abbreviated sec) is the reciprocal of cosine, and cosecant (abbreviated csc, or sometimes cosec) is the reciprocal of sine. They’re pretty much redundant, but it’s worthwhile to know what they are in case you come across them. Note that cotangents are tangents of complementary angles, which means that cot A = tan B, and cosecants are secants of complementary angles, and that means that csc A = sec B.