How do you express #4 cos theta - sec theta + 2 cot theta # in terms of #sin theta #?

1 Answer
Mar 18, 2016

#[4(1-sin^2theta)sintheta-sintheta+2(1-sin^2theta)]/(sinthetasqrt(1-sin^2theta))#

Explanation:

We shall use the following identities:

#sectheta= 1/costheta#

#cottheta= costheta/sintheta#

Also, #cos^2theta+sin^2theta=1#

#=>cos^2theta=1-sin^2theta color(red)rarr costheta=+-sqrt(1-sin^2theta)#

The #+# or #-# depends on the quadrant in which the angle #theta#
is found.

Assuming #theta# acute we can simply ignore this.

Hence,
#costheta= sqrt(1-sin^2theta)#

#=>1/costheta=1/(sqrt(1-sin^2theta))#

#costheta/sintheta=sqrt(1-sin^2theta)/sintheta#

#=>4costheta-sectheta+2cottheta color(red)rarr 4*(sqrt(1-sin^2theta))-(1/(sqrt(1-sin^2theta)))+2*(sqrt(1-sin^2theta)/sintheta)#

#color(red)rarr [4(1-sin^2theta)sintheta-sintheta+2(1-sin^2theta)]/(sinthetasqrt(1-sin^2theta))#