Question

Simplify each expression. (by using the reciprocal identities, quotient identities, and pythagorean identities.)

Answer 1

37)

$\frac{\sec x}{\tan x+\cot x}$

First, we will simplify the denominator

tanx + cot x

tanx = sinx/cosx and cotx = cosx / sinx

$\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin ^2x+\cos ^2x}{\text{cosxsinx}}$

sin²x +cos²x = 1

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

substitute back to the original expression

$\frac{\sec x}{\cos x\sin x}$

but secx = 1/cosx

$\frac{1}{\cos x}dividedby\frac{1}{\cos x\sin x}$$=\frac{1}{\cos x}\times\cos x\sin x$

$=\sin x$

39)

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

1 - sin²x = cos²x

substitute the above into the original expression

$\frac{\cos ^2x}{\cos x}$

$=\cos \text{ x}$

40)

$\frac{\csc ^2x-1}{\csc ^2x}$

csc²x - 1 = cot²x

substitute the above into the original expression

$\frac{\cot^2x}{\csc^2x}$

But 1 + tan²x = cot²x

1 + tan²x

= 1 + sin²x/cos²x

= sin²x+cos²x /cos²x

=1 /cos²x

substitute into the expression

$\frac{\frac{1}{\cos^2x}}{\csc ^2x}$

1/cos²x ÷ csc²x

but csc²x = 1/sin²x

1/cos²x ÷ 1/sin²x

1/cos²x (sin²x)

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

$=\tan ^2x$

38)

$\frac{1+\sin x}{\cos x}+\frac{\cos x}{\sin x-1}$

Find the lcm

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

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

Re-arrange the numerator

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

But, sin²x+cos²x=1

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

= 0