Can somebody please solve question 11 for me?

Mathematics

1 answer:

love history [14]3 years ago

6 0

1+tan^2(A) = sec^2(A) [Pythagorean Identities]

tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)

*see photo for complete solution*

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$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppp\displaystyle\ \Large \boldsymbol{} h(x)=\frac{f(x)}{g(x)} =? \\\\\\f(x)=x^2-8x+15\\\\\\x^2-8x+\underbrace{15+1-1}_{15}=0 \\\\\\\underbrace{x^2-8x+16}_{(x-4)^2}-1=0 \\\\\\(x-4)^2-1\Longrightarrow f(x)=(x-5)(x-3)$ $\displaystyle\ \Large \boldsymbol{} h(x)=\frac{(x-3)(x-5)}{(x-3)}=\boxed{x-5}$

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