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निम्नलिखित समीकरण को हल कीजिए - `sec theta - 1 = (sqrt2-1) tan theta` |
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Answer» 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 । |
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