How do you verify the identify #sinthetatantheta+costheta=sectheta#?

3 Answers
Mar 9, 2018

See the proof below

Explanation:

We need

#tantheta=sintheta/costheta#

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

#sectheta=1/costheta#

Therefore,

#LHS=sinthetatantheta+costheta#

#=sintheta*sintheta/costheta+costheta#

#=(sin^2theta+cos^2theta)/(costheta)#

#=1/costheta#

#=sectheta#

#=RHS#

#QED#

Mar 9, 2018

Apply the identities #tan(theta)=(sin theta)/(cos theta)# and #sec theta=1/(cos theta)# along with the Pythagorean theorem.

Explanation:

Apply the identity #tan(theta)=(sin theta)/(cos theta)#:
#L.H.S.=sin theta*sin theta/(cos theta)+cos theta#
#=sin^2 theta/(cos theta)+cos theta#
#=(sin^2 theta+cos^2 theta)/(cos theta)#

Apply the Pythagorean theorem #sin^2 theta+cos^2 theta=1#
#=(sin^2 theta+cos^2 theta)/(cos theta)#
#=1/(cos theta)#

By the definition of secants #1/(cos theta)=sec theta#:
#=sec theta#
#=R.H.S#

Mar 9, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)tantheta=sintheta/costheta" and "sectheta=1/costheta"#

#•color(white)(x)sin^2theta+cos^2theta=1#

#"Consider the left side"#

#rArrsinthetatantheta+costheta#

#=sinthetaxxsintheta/costheta+costheta#

#=sin^2theta/costheta+cos^2theta/costhetalarr"common denominator "costheta#

#=(sin^2theta+cos^2theta)/costheta#

#=1/costheta=sectheta=" right side "rArr" verified"#