Question

What is (cosec $x-\sin x\left)\right(\sec x-\cos x\left)\right(\tan x+\cot x)$ equal to?

• A. $\sin x+\cos x$
• B. $\sin x-\cos x$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

$(\csc x-\sin x)(\sec x-\cos x)(\tan x+\cot x)$

$=(\frac{1}{\sin x}-\sin x)(\frac{1}{\cos x}-\cos x)(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})$

$=\left(\frac{1-{\sin }^{2}x}{\sin x}\right)\left(\frac{1-{\cos }^{2}x}{\cos x}\right)\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right)$

$=\left(\frac{{\cos }^{2}x}{\sin x}\right)\left(\frac{{\sin }^{2}x}{\cos x}\right)\left(\frac{1}{\sin x\cos x}\right)$

$=\frac{{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}=1$