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if `tantheta+sectheta=l ` then prove that `tantheta=(l^2+1)/(2l)` |
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Answer» Given, `tantheta+sectheta=l` [multiply by `(sectheta-tantheta)` on numerator and denominator LHS] `((tantheta+sectheta)(sectheta-tantheta))/((sectheta-tantheta))=l rArr (sec^(2)theta-tan^(2)theta)/(sectheta-tantheta)=l` `1/(sectheta-tantheta)=l` `[therefore sec^(2)theta - tan^(2)theta=1]` `rArr sectheta-tantheta=1/l` ...............(ii) On adding Eqs. (i) and (ii), we get `2sectheta= l + 1/l` `sectheta=(l^(2)+1)/(2l)` Hence Proved. |
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