Find a numerical value of one trigonometric function of x for cscx = sinx tanx + cosx

Mathematics

1 answer:

Alex787 [66]3 years ago

6 0

Answer:

Step-by-step explanation:

we have

$cscx = sinx tanx + cosx\\\\\frac{1}{sinx}=sinx.\frac{sinx}{cosx} +cosx\\\\\frac{1}{sinx}.sinx.cosx= sinx.\frac{sinx}{cosx}.sinx.cosx+cosx.sinx.cosx\\ cosx= (sinx)^{3} +sinx.(cosx)^{2} \\cosx= sinx[(sinx)^{2}+(cosx)^{2})\\ cosx=sinx.1\\cosx=sinx\\tanx=1\\x=\frac{\pi}{4} +k\pi,k-integer.$

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$\text{D. }b^2-4ac>0$

Step-by-step explanation:

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