यदि ` tan theta- tan (theta)/(2) = sec .(theta)/(2)` तो `theta` का व्यापक मान ज्ञात कीजिए।

Explanation

`tan theta – tan .(theta)/(2) =sec .(theta)/(2)`
`(sin theta )/(cos theta) –(sin.(theta)/(2))/(cos .(theta)/(2)) = (1)/(cos .(theta)/(2))`
`(sin theta cos .(theta)/(2) -cos theta sin .(theta)/(2))/(cos theta cos .(theta)/(2)) = (1)/(cos.(theta)/(2))`
` (sin (theta-.(theta)/(2)))/(cos theta )-1`
` sin.(theta)/(2)-cos theta`
`sin .(theta)/(2)-cos theta=0`
` sin .(theta)/(2) -1 +2 sin^(2) .(theta)/(2)=0`
`(sin. (theta)/(2)+1) (2sin .(theta)/(2)-1)=0`
यदि ` 2 sin .(theta)/(2)-1 =0`
तो `2sin .(theta)/(2)=1`
या `sin. (theta)/(2)=(1)/(2)`
या `sin.(theta)/(2) = sin. (pi)/(6)`
` (theta)/(2)= (pi)/(6)`
अतः `theta = (pi)/(3)`
अभीष्ट हल : `theta =npi + (-1)^(n) .(pi)/(3)`