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If `2sin^2theta-cos^2theta=2,` find the value of `theta.` |
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Answer» Given, `2sin^(2)theta-cos^(2)theta=2` `rArr 2sin^(2)theta-(1-sin^(2)theta)=2` `[therefore sin^(2)theta+ cos^(2)theta=1]` `rArr 2sin^(2)theta + sin^(2)theta-1=2` `rArr 3sin^(2)theta=3` `rArr sin^(2)theta=` `rArr sintheta=1=sin90^(@)` `[therefore sin 90^(@)=1]` `therefore theta=90^(@)` |
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