Help me I did not get this part so can you All help me

Mathematics

1 answer:

klemol [59]3 years ago

7 0

D. 10 because if you replace x with 10, subtract it with 1 it equals 9. Then 9 divided by 3 equals three so k=3 which is correct

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Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$