3 years ago

Prove that : sinA/1+cosA + 1+cosA/sinA=2cosecA

Mathematics

1 answer:

tekilochka [14]3 years ago

3 0

Step-by-step explanation:

$\frac{sinA}{1+cosA} +\frac{1+cos}{sinA} =2cosec\\\\\\\frac{sin^{2}A+(1+cos)^{2} }{sinA(1+cosA)} \\\\$

$\frac{sin^{2}A+1+cos^{2}+2cos}{sinA(1+cosA)} \\\\$ and we have $sin^{2}+cos^{2} =1$

so $\frac{1+1+2cosA }{sinA(1+cosA)} \\\\$

$\frac{2+2cosA }{sinA(1+cosA)} \\\\$

$\frac{2(1+cosA)}{sinA(1+cosA)} \\\\$

$\frac{2}{sinA}$ we have $\frac{1}{sin}=cosec\\$

Finally $2cosecA$

I really hope this helps coz it took much time

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What is the sum of the squares of the lengths of the $\textbf{medians}$ of a triangle whose side lengths are $10,$ $10,$ and $12

FrozenT [24]

A picture can help.

The median to the long side divides the isosceles triangle into two right triangles with hypotenuse 10 and short leg 6. Thus the long leg (median of interest) is found by the Pythagorean theorem to be

... √(10² -6²) = √64 = 8

Then the midpoint of the short side is found to be 6 + (6/2) = 9 units to the side and 8/2 = 4 units above the opposite vertex. Hence the square of the length of that median is 9² + 4² = 97.

The sum of squares of interest is

... 8² + 2×97 = 258