Question

Prove that: `"sec"((3pi)/2-theta)"sec"(theta-(5pi)/2)+t a n((5pi)/2+theta)"tan"(theta-(3pi)/2)=-1.`

Answer 1

`L.H.S. = sec(((3pi)/2) - theta)sec(theta - ((5pi)/2))+ tan (((5pi)/2)+theta)tan(theta-((3pi)/2))`
`= sec((3pi)/2 - theta)sec(-((5pi)/2- theta))+ tan ((5pi)/2+theta)tan(-((3pi)/2-theta))`
As, `sec(-theta) = sec theta and tan (-theta) = -tan theta`
So, our expression becomes,
`= sec((3pi)/2 - theta)sec((5pi)/2- theta)- tan ((5pi)/2+theta)tan((3pi)/2-theta)`
`=-cosecthetasec(2pi+(pi/2-theta))-tan(2pi+(pi/2+theta))tantheta`
`= -cossecthetacosectheta+cotthetacottheta`
`=cot^2theta - cosec^2theta`
`=-(cosec^2theta - cot^2theta)`
`=-1 = R.H.S.`