निम्नलिखित समीकरण को हल कीजिए -
`sec theta - 1 = (sqrt2-1) tan theta`

Explanation

दिया है - `sec theta -1 = (sqrt2-1) tan theta`
`rArr" "(sec theta -1)^(2) = (sqrt2 -1)^(2) tan ^(2) theta`
`rArr " "(sec theta -1)^(2) - (3-2sqrt2)(sec^(2)theta-1) = 0`
`rArr (sec theta -1)^(2) - (3-2sqrt2)(sec theta -1)(sec theta+1) = 0`
`rArr (sec theta -1)[(sec theta -1)-(3-2sqrt2)(sec theta +1)] =0`
`rArr (sec theta -1)(sec theta -sqrt2)(2sqrt2-2) = 0`
`rArr (sec theta -1)(sec theta -sqrt2) = 0`
यदि `sec theta=1`
`rArr cos theta =1` तब `theta =0^(@)`
`=cos 2 n pi`
`:. theta = 2 n pi`
और यदि `sec theta -sqrt2 = 0 rArr sec theta = sqrt2`
`rArr cos theta = 1/sqrt2 = cos . pi/4`
`:. theta = 2 n pi pm pi/4`
इसलिए , दी गयी समीकरण के हल `theta = 2n pi` तथा `theta = 2 npi pm pi/4`होंगे ।